BAI TAP vu TUAN (Chu bien) – TRAN VAN HAO OAO NGOC NAM – LE VAN TIEN -IVU VIET YEN y,p7x&quot’,. vr”. T’ ai” a NHA X

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1 BAI TAP vu TUAN (Chu bien) – TRAN VAN HAO OAO NGOC NAM – LE VAN TIEN -IVU VIET YEN y,»;p7x*”^’,..* ;»v<»*?firfj^. 1,j»VTIJ»>r*»«”». T’ ai” a NHA XUAT BAN GIAO DUC VIET NAM

3 VU TUAN (Chu bien) TRAN VAN HAO – BAG NGOC NAM LEVANTI^N-VUVI^TYEN BAITAP DAIS6 VAGIAI TICH (Tdi bdn ldn thd tu) 9 r NHA XUAT BAN GIAO DUC VIET NAM

4 Ban quy^n thu6c Nha xu^t ban Giao due Vi6t Nam l/cxb/ /gd Ma s6′: CB103T1

5 m.’ huang L HAM SO Ll/ONG GIAC PHUONG TRINH Ll/ONG GIAC 1. Ham so laong giac 1. Ham so sin A. KIEN THCTC CAN NHd Ham s6′ j = sinx co tap xae dinh la M va y = sin X la ham s6′ le. -1 < sinjc < 1, Vx G R. y = sinx la ham s6′ tu^n hoan v6i chu ki 2jt. Ham s6 y = sinx nhan cae gia tri dac bi6t: sinx = 0 khi x = kn, k e Z. n sm X = 1 khi x = + k2n, k G Z. sinx = -1 khi x = – + k2n, k e Z. D6 thi ham s6 y = sinx (H.l) : Hinh 1

6 2. Ham so cosin Ham s6′ y = cosx eo tap xae dinh la R va -1 < cosx < 1, Vx G y = cosx la ham so ehsn. y = cosx la ham so tu^n hoan vdi chu ki 2n. Ham s6′ y = cosx nhan cac gia tri dac bi6t: cosx = 0 khi X = + kn, k ez. cos X = 1 khi X = k2n, k e Z. cosx = -1 khi X = {2k + l)7i, k e It. D6 thi ham s6′ y = cosx (H.2) : Hinfi 2 3. Ham so tang Ham sd V = tanx = eo tap xae dinh la cosx D = R{^ + kn,ke y = tanx la ham s6 le. y = tanx la ham sd tu5n hoan vdi chu ki n. Ham sd y = tar. v nhan eae gia tri dae biet: tanx = 0 khi x =kn, k e Z.

7 n tanx = 1 khi X = + kn, k e.z. 4 tanx = -1 khi x = – + kn, k G D6 thi ham sd 3^ = tanx (H.3): -37t 2 Hinh 3 4. Ham so cotang COSX Ham s6 y = coix = c6 tap xae dinh la smx D = y = cotx la ham sd le. R{kTi,keZ]. y = coix la ham sd tusn hoan vdi chu ki %. Ham sd y = cot x nhan cac gia tri dac bi6t: 71 cot X = 0 khi X = + kn, k e Z. 71 cot X = 1 khi X = + ^71, k ez. 4 It, cotx = -1 khi X = + ^7r, )t G Z.

8 D6 thi ham sd j = cotx (H.4): -27t O ] – 2 Hinh 4 B. Vi DU Vidul Tim tap xae dinh cua eae ham sd a) y = sin3x ; c) y = cosvx ; b) y = cos ; X d) y = sin 1 + X 1-x” Gidi a) Dat t = 3x, ta duoc ham sd y = sin r co tap xae dinh la D = R. Mat khae, rgr<=>x = – GR nfen tap xae dinh eua ham s6 y = sin3x la R. 2 ‘ 2 b) Ta CO e R <=> X ;^ 0. Vay tap xae dinh eiia ham sd y = cos la X… ^ D = R{0}. e) Ta CO Vx G R o x > 0. Vay tap xae dinh cua ham s6 y = cosvx la D = [0 ; +00).

9 d) Ta CO 1 +.^ ir» l + ^..,^ 1^ G R <^ > 0 «-1 < X < 1. 1-X 1-x 1 + X vay tap xae dinh eua ham sd j = sin J-j la D = [-1 ; 1). Vidul. Tim tap xae dinh eua cae ham sd a) y = ; b) y = cot 2x -,, ^ 2cosx ‘ ^ y A)’ cotx,^ sinx+ 2 Gidi 3, K a) Ham sd y = x^c dinh khi va ehi khi cosx ^ 0 hay x?t + kn, k G ‘ ^ 2cosx 2 vay tap x^e dinh cua ham sd la D = R { + itti, A: G I 71 I 7C b) Ham sd y = cot 2x – xae dinh khi va chi khi 2x – ^t kn, k G Aj, 4 hay x * + k, k e Z. o 2 vay tap xae dinh cua ham sd y = cot 2x – la D = R { + ^,A:G cotx. ^., [sinx 9^0 lx^kn,kez e) Ham sd y = xae dmh <:> < <:> < cosx-1 lcosx?tl Ix^t A:27i,;tGZ.

10 Tap {^27:, k &Z] la tap con eua tap [kn, k ez} cot X chan). vay tap xae dinh cua ham sd cosx-1 D = R{kn,k Z]. la (umg vdd cac gia tri k sinx + 2 d) Bieu thiie ludn khdng am va no eo nghla khi cosx + 15«t 0, hay cosx + 1 ” cosx 9t -1. vay ta phai c6 x ^ (2k + l)n, it G Z, do do tap xae dinh cua ^ smx+ 2 ham so y = J ^’cosx + 1 la D = R{(2A: + l)7i, A;GZ}. Vi dn.? Tim gia tri ldn nhs^t va gia tri nho nha’t cua cac h im sd : a) y = 2 + 3eosx ; l + 4cos^x c)y= 3 ; b) y = 3-4 sin X cos x ; d) y = 2sin x – cos2x. Gidi a) Vl -1 < cosx < 1 ndn -3 < 3eosx < 3, do do -1 < 2 + 3cosx < 5. vay gia tri ldn nha’t eua ham sd’ la 5, dat duoc khi cosx = 1 o X = 2kn, kez. Gia tri nho nha’t cua ham sd la -1, dat duoc khi cos x = -1 d’ x = {2k + l)7t, kez. b) y = 3-4sin^ xcos^ x = 3 – (2sinxcosx)^ = 3 – sin^ 2x. Ta ed 0 < sin^ 2x < 1 nen -1 < -sin^ 2x < 0. vay 2<y<3.

11 Gia tri nho nha’t cua ham sd la 2, dat dugfc khi sin^ 2x = 1 <» sin2x = ±1 <z> 2x = +y + k2n, k & Z <:> x = ±j +kn, k e Z. Gia tri ldn nha’t cua y la 3, dat duac khi sin^ 2x = 0 n «sin2x = 0 «2x = A:7t, ^ G Z <» X = k, k G Z cos^x. 5 c) Vi 0 < cos^ X < 1 nen – < < 3″ 1 n Gia tri nho nha’t cua y la -, dat dugc khi cosx = 0 «> x = + A:7t, ^ G 5 2 Gia tri ldn nha’t eua y la -, dat dugc khi cos x = 1 <^ cosx = ±1 <:> X = kn, k e Z. d) y = 2sin^x-eos2x = l-2cos2x. Vi -1 < cos2x < 1 nen -2 < -2eos2x < 2, dodo-1 < l-2cos2x<3. Gia tri nho nha’t eua y la -1, dat duge khi cos2x = 1 <» 2x = 2kn, k e Z <:> x ^ kn, k : Z. Gia tri ldn nh^t cua y la 3, dat duge khi cos 2x = -1 Vidtid «2x = {2k + )n,k G Z «x = + ^TC, A: G Z. Xae dinh tinh chan, le cua cac ham sd a) y = xeos3x ; e) y = X sin2x ; 1 + cos X b) y = -j ; 13 – cosx X -smx “‘ ^ ” eos2x

12 Gidi a) Kl hieu /(x) = xcos3x. Ham sd ed tap xae dinh D = R. Ta cd vdi X G D thi -x G D va /(-x) = (-x)eos3(-x) = -xcos3x = -/(x). vay y = xcos3x la ham sd le. b) Bi^u thiie /(x) = xae dinh khi va chi khi 1-eosx cosx 5″t 1 <» X 5t 2kn, k ^ Z. vay tap xae dinh eiia ham sd y = ] ^ ^ ^^ la D = R {2A:7t, 1 -cosx Vdi X e D thi -x G D va /(-x) = /(x). Do dd ham sd da cho la ham sd chan. kez}. e) Tap xae, dinh D = R, do dd vdi x G D thi -x G D. Ta cd /(-x) = (-x) sin2(-x) = X sin2x = /(x). vay y = X sin2x la ham sd chan., X sin X d) Bieu thiie /(x) = ed nghia khi va chi khi cos2x ^ 0 cos2x <:i>2x^ + kn,kez<ii>xit + k, it G Z. vay tap xae dinh cua ham sd la D = R (^ + it, it G ZJ. 10 _ 3 Vdi X G D thi -X G D va /(-x) = ~^ ^l^^ cos2x x^ -sinx,.,. ^,, y = la ham so le. eos2x = -/(x), do dd ham sd

13 Vidti^ 1 X a) Chiing minh rang cos (x + 4^7t) = cos vdi mgi sd nguyen k. Tit dd X ve dd thi ham sd y = cos ; b) Dua vao dd thi ham sd y = cos, hay ve dd thi ham sd y = X V X cos 2 Gidi 1 (X X a) Ta ed cos (x + 4^7c) = eosi + 2kn = cos vdi mgi k e Z,do dd ham sd y = cos tu&i hoan vdi chu ki 47t. Vi vay ta ehi efe ve dd thi cua ham sd X y = cos tren mdt doan ed dd dai 47t, rdi tinh tidn song song vdi true Ox cae X doan cd dd dai 47i ta se dugc dd thi ham sd y = cos. X Hon niia, vi y = cos la ham sd chsn, nen ta chi esn ve dd thi ham sd dp tren doan [0 ; 27i] rdi la’y ddi xiing qua true tiing, se duge dd thi ham sd tren doan [-27t; 27r]. Dd thi ham sd duoc bidu dien tren hinh 5. Hinh 5 11

14 b) Ta cd X cos 2 X cos, ndu cos > X -cos, ne’u cos < X X Vi vay, tit dd thi ham sd y = cos ta giii nguyen nhflng phsn dd thi nam phia tren true hoanh va l^y dd’i xiing qua true hoanh nhihig phsn dd thi nam phia dudi true hoanh, ta dugc dd thi ham sd y = X cos- (H.6). Hinh Tim tap xae dinh eiia cac ham sd 2x a) y = cos-, X -1 c) y = eot2x ; 1.2. Tim tap xae dinh eua cae ham sd C. BAi TAP b) y = tan- ; d) y = sin x^-r a) y = vcosx + 1 ; b) y = 2 2 ‘ sm X – cos X 2 e) y = d) y = tanx + cotx. cosx – cos3x 1.3. Tim gia tri ldn nha’t va gia tri nho nh& eua eae ham sd a) y = 3-2 sinx ; b) y = cosx + eos[ x – ; 12

15 c) y = cos^x + 2cos2x ; d) y = v5-2cos^xsin^x Vdi nhiing gia tri nao eiia x, ta cd mdi dang thiic sau? a) = cotx ; b) r = cos x ; tanx 1 + tan^x c) – = 1 + cot X ; d) tanx + cotx =. ^. sin^x 1.5. Xae dinh tfnh chan le cua cae ham sd sm2x. eos2x a) y = ; b) y = x – sinx ; c) y = Vl -cosx ; d) y = 1 + eosxsin – 2x a) Chiing minh rang cos2(x + kn) = cos2x, ^ G Z. Tii dd ve dd thi ham sd y = eos2x. b) Tilt dd thi ham sd y = eos2x, hay ve dd thi ham sd y = eos2x Hay ve dd thi ciia cac ham sd a) y = 1 + sinx ; b) y = cosx – 1 ; e) y = sinlx-l ; d) y = cosi x +-J Hay ve dd thi eua eae ham sd a) y = tani x + I ; b)y = eotlx- 2. Phaong trinh lapng giac co ban ^ <- A. KiEN THl/C CAN NHO 1. Pliirong trinh sinx = a (1) a > 1 : phuong trinh (1) vd nghiem.

16 a < 1 : ggi or la mdt cung thoa man sin or = a. Khi dd phuong trinh (1) cd cae nghiem la X = or + k2n, it G Z va X = 7t – a + ^27t, ^ G Z. n n Ne’u or thoa man di6u Icien < or < va sina = a thi ta vie’t or = aresina. 2 2 Khi dd cac nghiem cua phuong trinh (1) la X = arcsina + ^27i, va X = 7: – arcsina + ^27i, ^ G Z k e.z. Phuong trinh sin x = sin P cd cae nghiem la x = J3 + k360, it G Z va X = fi + it360, it G Z. ^ Chu y. Trong mot cong thcfc nghi m, khdng dodc dung dong thdi hai ddn vj do va radian. 2. Pliirong trinh cosx = a (2) 14 a > 1 : phuong trinh (2) vd nghiem. a < 1 : ggi a la mdt cung thoa man cos a = a. Khi dd phuong trinh (2) ed cac nghiem la X = ±Qr + ^27t, ^ G Z. Ne’u or thoa man di6u kien 0 < or < TI va coso; = a thi ta vie’t or = arccosa. Khi dd nghiem cua phuong trinh (2) la X = larccosfl + ^27C, k e Z. Phuong tiinh cosx = cos/3 ed eae nghiem la x = ±j3 + it360, it G Z.

17 3. Phirong trinh tanx = a (3) V Dieu kien eua phuong trinh (3) : x ^ + kn, k e Z. n n Ndu orthoa man dilu kien – < or < va tanor = a thi ta vie’t a = arctana. 2 2 Liic dd nghiem eua phuong tiinh (3) la X = aretana + kn, k e Z. Phuong tiinh tan x = tan /? cd cac nghiem la x = fi + itl80, it G Z. n 4. Phirong trinh cotx = a (4) Dilu kien cua phuong tiinh (4) la x vt kn, k e Z. Ndu or thoa man dilu kien 0 < or < 7i va cot or = a thi ta vie’t a – arceota. Liic dd nghiem cua phuong trinh (4) la X = arceota + kn, k e Z. Phuong trinh cot x = cot fi cd cac nghiem la x = /3 + itl80, it G Z. B. VI DU Vidu 1 Giai cac phuong trinh a) smx = Y ‘ e) sin(x – 60 ) = ; b) sin X = ; d) sin2x =

18 Gidi a) Vl = sin[-yj nen n sinx = «sinx = sm -. vay phuong trinh cd cac nghiem la n X = – + ^271, ^ G Z va X = I + 2^7t = + it27i, it G Z. 1 b) Phuong trinh sinx = cd eae nghiem la X = arcsin + 2^7t, k G 4 va X = 7t – arcsin + k2n, k e Z. c) Ta ed = sin 30, nen 1 sin(x – 60 ) = -» sin(x – 60 ) = sin30. x-60 =30 +it360, itgz X – 60 = it360, it G Z vay phuong trinh ed eae nghiem la d) Ta ed X = 90 + it360, it G Z va X = it360, it G Z. sin2x = -1 (gia tri dae biet). Phuong trinh cd nghiem la 37t 2x = + it27r, ^ G Z 37t hay X = -T- + kn, k e Z.

19 . Vidu 2 Giai cae phuong tiinh a) cos 3x – 7t^ V2 b) eos(x – 2) = ; e) cos(2x + 50 ) = ^ ; d) (1 + 2eosx)(3 – cosx) = 0. Gidi. – V2 371, f- 71 a) Vl = COS nen cos 3x – (. n^ 371 <» cosi 3x – = cos 2 O 3x – – = ± ^ + it27r, it G Z 6 4 7T 3TI: <» 3x = – ± ^ + it27t, it G Z 6 4 II7C,-, _ 3x = – + it27t, it G Z 3x = – + ^27t, k G <=> 2 2 b) eos(x -2) = -<»x-2 = +areeos + ^27i, k e Z 2 <» X = 2 ± arceos + ^27t, k e Z. e) Vi = cos 60 nen II71, 27t, x^ – + k -,kei 3o 3 7TC, 2n, x = k-,ke cos(2x + 50 ) = ^ <» cos(2x + 50 ) = cos60» 2x + 50 = ±60 + it360, it G 2 «<» 2x = it360,itg 2x = it360, it G X = 5 + /:180, it G Z X = A:180, it e Z. 2. BTBS&GT11-A 17

20 d) Ta ed (1 + 2cosx)(3 – cosx) = 0 <» 1 + 2eosx = 0 <:> 3 – cosx = 0 cosx = – COSX = 3. Phuong trinh cosx = – cd cae nghiem la 27t X = ± – + it27i, it G Z ; eon phuong trinh cosx = 3 vd nghiem. vay cae nghiem cua phuong trinh da cho la 2n X = + + it27t, it G Z. Vi du 3 Giai cac phuong trinh 2n a) tan2x = tan ; c) cot 4x- l 6J = S; b) tan(3x -30 ) = – ^ ; d)(eotf -ixcotf + l). -0. Gidi 2n 2n a) tan2x = tan <^ 2x = + kn, k e Z <» X = + k, k e Z. 7 2 b) tan(3x – 30 ) = – ^ <» tan(3x – 30 ) = tan(-30 ) o 3x – 30 = /tl80, it e» 3x = itl80, it e Z <=> X = it60, it G Z BTBS&GT11-B

21 c) cot 4x – n = ^i^ <» cotj 4x – J = cot <:> 4x – = + K7t, k e 6 6 7C TX TT <» 4x = + ^71, A: G Z <^ X = + k, k e V X X d) Dilu kien : sin?!: 0 va sin ^t 0. Khi dd ta cd 3 2 cot -ljfcot – + li = 0 cot-l = 0 cot- + l = 0 2 n + A:7C, ^ G cot- = 1 X cot- = -1 2 X = + ^371, A: G 2 = 7 + ^71, ^ G 2 4 Cac gia tri nay thoa man dilu kien. vay phuong trinh da cho cd cac nghiem la n, -, X = – + K27I;, k G X = + A:37C, k e Z va X = – + k2n, k e Z. Vidu 4 Giai cae phuong trinh a) sin2xcotx = 0 ; b) tan(x )eos(2x ) = 0; e) (3tanx + /3)(2sinx -1) = 0. 19

22 a) Dilu kien ciia phuong trinh la sinx ^ 0. Ta bie’n ddi phuong trinh da cho /ix o cosx – (1) <» 2 sinx cosx.-: sinx = 0 <=> 2 cos X = 0 Gidi sin 2x cotx = 0 (1) <» cos X = 0 =i> X = + kn, k G Cae gia tri nay thoa man dilu kien eua phuong trinh. Vay nghiem eua phuong trinh la y n X = + kn, k &Z. b) Dilu kien cua phuong trinh la cos(x-30 )^0. Ta bie’n ddi phuong trinh da cho tan(x – 30 )cos(2x ) = 0 (2) sin(x-30 ).-,,-^ON rx (2) <» ^^ ^.cos(2x ) = 0 cos(x-30 ) sin(x-30 ) = 0 eos(2x-150 ) = 0 X = 30 + itl80, it G Z 2x = it360, it G 2x = 60 + it360, it G Z x-30 =itl80,itg Z 2x = ±90 + it360, it G X = 30 + itl80, it G Z X = itl80, it G Z X = 30 + itl80, it G Z. Khi thay vao dilu kien eos(x – 30 ) ^^ 0, ta ihiy gia tri x = itl80 khdng thoa man, cdn gia tri x = 30 +^180 thoa man. Vay nghidm eua phuong trinh da cho la X = 30 + itl80, it G Z. c) Dilu kien ciia phuong trinh (3 tan X + N/3 )(2 sin x – 1) = 0 (3) la cosx ^ 0. Tacd 20

23 (3) <» ^ tanx = sinx X – + kn, k 6 BZ X = 5 + ^^271, it G Z X = -^ + it27r, A: G Z. 6 Cae gia tri nay diu thoa man dilu kien eua phuong tnnh, trong dd tap cac gia tri + k2n, k & z la tap con cua tap cac gia tri j + /7t, / G Z (ling vdi cae gia tri / chan). vay nghiem eua phuong trinh (3) la va X = – + kn, k G, 6 X = + k2n, k G 6 Vi du ^ Vdi nhiing gia tri nao cua x thi gia tri cua cac ham sd tuong ling sau bang nhau? a) y = sin3x b) y = cos(2x + 1) va va y = sin[x + j ; y = cos(x – 2) ; c) y = tan3x va y = tani-2×1. Gidi Trudc he’t, md rdng cdng thiie nghiem ciia cac phuong trinh lugng giac co ban, ta ed cac cdng thiic sau. Vdi M(X) va v(x) la hai bilu thiic cua x thi sinm(x) = sinv(x) «u(x) – v(x) + k2n, k e Z M(X) =7C – v(x) + k2n, k G COSM(X) = eosv(x) <» M(X) = ±v(x) + k2n, k e Z. tanm(x) = tanv(x) => u(x) = v(x) + kn, k ez. cotm(x) = cotv(x) => M(X) = v(x) + A:7r, it G Z. 21

24 Ap dung cac cdng thiic md rdng nay cho cac bai toan trong Vi du 5, ta cd a) sin3x = sinj x + n ) «- 3x = X + + k2n, k G 3x = 7t- x ^271, k G 2x = – + it27t, it G Z 4x = + k2n, k el <» X = + kn, k ez 8 37t, 7t, x = – + k-,ke K 3TC TZ vay vdi X = + ^71 hoae x = – + it, k ez t^ y = sin3x va y = sin[ Jf + -j ) bang nhau. thi gia tri ciia hai ham sd b) eos(2x + 1) = eos(x – 2) <=> 2x + 1 = ±(x – 2) + k2n, k e X = -3 + k2n, k e 2x + 1 = X it27i, it G Z ^> <» 1,2u, 2x + 1 = -X it27t, it G X = + K-7-, k G i: vay vdi X = -3 + k2n hoae x = – + k-, A G Z thi gia tri eiia hai ham sd y = cos(2x + 1) va y = eos(x – 2) bang nhau. c) Dilu kien : cos 3x 9^ 0 va cos – 2x U 0. Khi dd f n ] n tan3x = tan – 2x <s> 3x = -2x + kn, k G n 5x = + kn, k e 71, 71, Cae gia tri nay thoa man dilu kien dat ra. vay vdi X = + k, k ez fn y = tani – 2x ] bang nhau. 7t n thi gia tri eiia hai ham sd y = tan3x va 22

25 2.1. Giai cac phuong trinh C. BAI TAP R J2 a) sin3x = – ^ ; b) sin(2x- 15 ) = ^ c) sinf ^ = ~ ; d) sin4x =., 2 ^ 2 ‘ ‘ Giai eae phuong trinh 4^ a) eos(x + 3) = – ; b) eos(3x – 45 ) =, c) cosf 2x + -^ J = -y ; d) (2 + eosx)(3cos2x – 1) = Giai cac phuong trinh a) tan(2x + 45 ) = -l ; b) cotf x + j = >/3 ; e) tanl^-^utanj ; 2 4J 8 ‘ d) cot ^ + 20 U “‘ “T3 “” ; 3 ‘ 2.4. Giai cac phuong trinh a) 7 = 0 ; b) cos2xeot x – = 0 ; eos3x-l V 4y e) tan(2x + 60 )cos(x + 75 ) = 0 ; d) (cotx + l)sin3x = Tim nhiing gia tri cua x dl gia tri cua cae ham sd tuong ling sau bang nhau a)y=cos 2x- va y = eosl -x b) y = sin 3x – j va y = sin x + c) y = tan 2x + va y = tan – x J ; f n d) y = cot3x va y = cot x Giai efic phuong trinh a) cos3x – sin2x = 0 ; b) tanxtan2x = -1 ; c) sin3x + sin5x = 0 ; d) cot2xcot3x = 1. 23

26 3. Mot so phuong trinh lapng giac thudng gap A. KIEN THUC CAN NHO 1. Phirong trinh bae nhat doi vdi mot ham so lirong giac Cac phuong trinh dang at + b = 0 (a ^ 0), vdi t la mdt trong cac ham sd lugng giac, la nhimg phuong trinh bae nhat ddi vdi mdt ham sd lugng giac. Sii dung eae phep bie’n ddi lugng giac, cd thi dua nhilu phuong trinh lugng giac vl phuong trinh bae nha’t ddi vdi mdt ham sd lugng giac. 2. Phirong trinh bae hai doi vdi mot ham so’ lirdng giac Cae phuong trinh dang at^ + bt + c – 0 (a ^ 0), vdi r la mdt trong cac ham sd lugng giac, la nhiing phuong trinh bae hai ddi vdi mdt ham sd lugng giac. Cd nhilu phuong trinh lugng giac cd thi dua vl phuong trinh bae hai ddi vdi mgt ham sd lugng giac bang eae phep biln ddi lugng giac. Mdt sd’ dang chinh se duge neu trong vi du. 3. Phirong trinh bae nhat doi vdi sinx va cosx Xet phuong trinh asinx + feeosx = c. (1) Bie’n ddi vl trai cua phuong trinh (1) vl dang a + b sin(x + or), A’ a. b trong do cos or =, sin or = Va^ + b’^ ‘ 4a 2+^2 ta dua phuong trinh (1) vl phuong trinh bae nha’t ddi vdi mdt ham sd lugng giac. 24

27 B. VI DU Vidul Giai eae phuong trinh a) sin2x – 2cosx – 0 ; e) tan2x – 2tanx = 0 ; b) 8cos2xsin2xcos4x = yu ; d) 2eos X + cos2x = 2. Gidi a) Ta cd sin2x – 2cosx = 0 <» 2sinxcosx – 2cosx = 0 <=> 2cosx(sinx – 1) = 0 <=> cosx = 0 sinx = 1 «n X = + kn, k e Z n X = + k2n, k G Z. Tap l^ + k2n, A; G Z la tap con cua tap y + itti, it G Z 71 vay nghiem cua phuong trinh da cho a x = + kn, k e Z. b) Ta cd 8cos2xsin2xcos4x = v2 <» 2sin8x = V2 <=> sin8x = – 4sin4xeos4x = v2 8x = J + it27t, it G Z 371 8x = + it27r, it G Z n, n, ^ = 32^^4’^^ 371, 71,.^=32+^4’^^ vay nghiem cua phuong trinh la 71, 71,, 371, 7t, ^”32^ 4^ va x = + A:-,/:G c) Dilu kien : cos2x ^ 0 va cosx ^ 0 Tacd 2tanx tan2x – 2tanx = 0 <=> 1 – tan^ X 2tanx = 0<=> 2tanx l-tan^x J = 0 «2tan^x = 0 <=> tanx = 0 => x = it7:, A G 25

28 Cae gia tri nay thoa man dilu kien eua phuong tiinh. Vay nghiem eua phuong tnnh la x = ^71, ^ G Z. d) Ta ed -y 1 2cos X + cos2x =2 «> 1 + 2cos2x = 2» cos2x = – «> 2x = ± + ^27t, A G Z <» X = + -kn,k & n vay nghiem cua phuong tnnh a x = ± ->r kn,k &Z. 6 i Vidu2 Giai eae phuong trinh a) cos3x – cos4x + cos5x = 0 ; 1 1 c) cos X – sin X = sin3x + cos4x ; b) sin7x – sin3x = cos5x ; ^. 2 3x d) eos2x – cosx = 2sm -. 2 Gidi a) eos3x – cos4x + cos5x = 0» eos3x + eos5x = eos4x <» 2cos4xeosx = eos4x <=> cos4x(2cosx – 1) = 0 cos4x = 0 1 <^ COSX = 4x = + ^Tt, it G Z X = ±^ + it27l, it G Z 71, 71, _ X = -^ + k, it G Z 8 4 X = ±^ + it27t, it G Z. vay phuong tnnh cd eae nghiem la x = – + k- va X = ±^ + it27t, it G Z. b) Ta cd sin7x – sin3x – eos5x = 0 <» 2eos5xsin2x – eos5x = 0 «> eos5x(2sin2x – 1) = 0 26

29 eos5x = 0 sin2x = 2 n 5x = + kn, k e Z 2x = 5 + ^^271, it G Z 6 2x = -^ + it27t, it G ; 6 7c n Vay phuong trinh ed cac nghiem la x = + it,itg 57T:,, X = + kn,k ez. c) Ta ed X = n + kn va cos X – sin^ X = sin3x + eos4x <=> eos2x – eos4x – sin3x = 0 <=> -2sin3xsin(-x)-sin3x = 0 <?:> sin3x(2sinx – 1) = 0 3x = kn, k e Z sin 3x = 0 <=> 1 «X = – + A:27t, it G Z sinx = , -, _ X = – + ^271, it G Z. 6 vay eae nghiem cua phuong trinh la X = k, k ez ; X = – + k2n a x =-^ + k2n, k G J O 6 d) Ta ed cos2x – cosx == 2sin 3-^ ^ ^. 3x. X x -z- «>-2sin -sm -2sm = 0.. 3xf. X. 3x^ ^ ^. 3x ^. X ^ -2sm I sin- + sin 1 = 0 <=> -2sm.2smxcos-= 0 “. 3x. sm- = 0 3x,, = K7r, ^ G Z sinx = 0 cos = 0 X = k-^,k <t> GZ X = ^71, A G Z X 7t,, -;:^^-^-kn,ke <=> X = ^71, ^ G Z X = 71 + A;27i, it G Z. 27

30 Tap {n + it27t, it G Z} la tap con cua tap {it7i, it G Z}. 2n vay cac nghiem ciia phuong trinh la x = ^ va x = it7i, it Vidu^ Giai cac phuong trinh a) 2cos2 2x + 3sin x = 2; b) cos2x + 2cosx = T 2-4 e) 2 – cos X = sm x ; j 4 4 ^ 2X 2sm ; 2 sm2x. d) sm X + cos X = 2 Gia7 a) Ta ed COS 2x 2eos” 2x + 3sin” x = 2 <=> 2cos” 2x + 3. = 2 <=> 4cos2 2x – 3eos2x – 1 = 0 <=> 2x = ^27t, k G / 1 2x = ± arceos I + it27i, it G Z V 4 X = A;7:, ^ G Z eos2x =1 1 eos2x = 4 28 ^ 1^ X = ± arceos + kn,k e Z. 2 V ^; vay cac nghiem cua phucmg trinh la 1 f iv X = ^7t, ^ G Z va X = ± arceos + it7t, it G Z. 2 t 4J b) Ta cd cos2x + 2eosx = 2sin2 – <=> 2eos2 x eosx = 1 – cosx <» 2cos2 X + 3cosx -2=0 1 cosx = 2 cosx = -2.

31 Phuong trinh cosx = -2 vd nghiem, eon phuong trinh cosx = cd nghiem X = +- + it27r, it G Z. 3 n Vay nghiem cua phuong trinh la x = ± + ^27r, ^ G Z. c) Ta cd 2 – cos x = sin x<:>2-(l- sin^ x) = sin’^ x «> sin X – sin x – 1 = 0. Dat t – sin’^ x, vdi dilu kien 0 <? < 1, ta dugc phuong trinh t^ -t- = 0..,.., -,. 1-^5 l + yfe Phuong tnnh nay eo hai nghiem t^ =, ^2 = Vi?! < 0, ^2 > 1 nen hai gia tri nay Ichdng thoa man dilu kidn. vay phuong trinh da cho vd nghiem. d) Ta cd sin X + cos X = sin2x» (sin^ x + cos^ x)^ – 2sin2 xeos x = sin2x 2-2 ^ sin 2x 1. ^. 2.-.,-. ^ «<» 1-2. = -sin2x <» sm 2x + sm2x -2 = sin2x = 1 sin2x = -2. Phuong tnnh sin2x = -2 vd nghiem, cdn phuong trinh sin2x = 1 cd nghiem 2x = + ^27c, k & Z., n vay nghiem eua phuong trinh la x = + kn, k e Z. Vidu 4. Giai cac phuong trinh.2. i) 3tanx + v3cotx v3 = 0 ; b) = tan^ x ; sin 2x-4eos x e) 2tanx + cotx = 2sin2x + sin2x 29

32 Gidi a) 3tanx + v3eotx v3 = 0 Dilu kien ciia phuong trinh (1) la cosx ^ 0 va sinx 9^ 0. (1) (1)»3tanx + ^^ 3->/3=0 tanx <^ 3tan2 X – (3 + V3)tanx + V3 = 0 tanx = 1 X = + A^7i:, ^ G Z tanx = >/3 4 X = 1- ^711, ^ G Z. 6 Cac gia tri nay thoa man dilu kien ciia phuong trinh (1). Vay cac nghiem eua phuong trinh (1) la b) X = l-^7t vax = I- kn, k e 4 6 sin^ 2x = tan X. sin 2x-4cos x 9 9 Dilu kien cua phuong trinh (2) la cosx?t 0 va sin 2x-4cos Tacd sin 2x-4eos x = 4sin xeos x-4eos x = 4eos x(sin x-1) = -4cos’^x. (2) x^o. Vi vay sin^ 2x – 4cos2 x?t 0 <=> cosx ^ 0. Do dd dilu kien ciia phuong trinh (2) la cosx ^ 0. Theo bie’n ddi tren, ta co (2) «9 9 sin 2x – 2 A 4 2-4cos X sin X <=> sin^ 2x – 2 = -4eos2 xsin^ x cos X <^ 2sin2 2x = 2 <» sin2x = ±1 <=> cos2x = 0 ^ 71, 71, 71, 2x = + kn => X = + k, k e Cae gia tri nay thoa man dilu kien cua phuong trinh (2). Vay nghiem ciia phuong trinh (2) la x = + it, it G Z. 30

33 1 c) 2tanx + eotx = 2sin2x + sin2x Dilu kien eua phuong trinh (3) la sinx?t 0 va cosx -t^ 0. Ta cd 2tanx + cotx = Dodo (3>: 2(sin2x + l) sin2x 2sinx cosx + : COSX sinx 2sin X + cos x _ sin x + 1 sinxcosx 1. T sin2x 2 2sin2 2x + l sin2x O 2sin2 2x – 2%^ x – 1 = 0 <^ 2(1 – cos^ 2x) – (1 – cos2x) -1 = 0 <=> -2eos2 2x + eos2x = 0 «eos2x(l – 2eos2x) = 0 (3) <» eos2x = 0 1 cos2x = 2 2x = ^ + it7c, ^ G Z 2 2x = ± + ^27C, * G 3 X = + k, k 4 2 n X = ± h kn, k ez ez. 6 Cac gia tri nay diu thoa man dilu kien eua phuong trinh (3). Vay cac TU 7C TZ nghilm ciia phuong trinh (3) la x = + k, k e Z a x = ± + kn, k e Z. Vidu 5 Giai cae a) 4 cos X + 3sinxeosx – sm X = 3 ; b) 2sin2 X – sinxcosx -cos^ X = 2 ; c) 4sin2 phuong trinh 2 9 X – 4sinxcosx + 3cos x = 1. o 31

34 Gidi a) Vdi cosx = 0 thi ve trai bang -1 cdn vl phai bang 3 nen cosx = 0 Ichdng thoa man phuong trinh. Vdi cosx ^ 0, chia hai vl eiia phuong trinh cho cos X ta duge 4 + 3tanx-tan2x = 3(l + tan2x)o 4tan2x – 3tanx – 1 = 0 <» tanx = 1 tanx = 1 <» X = + A:7:, ^ G Z 4 r X = arctan + kn, k e Z. 4 vay cae nghiem cua phuong trinh la / 1 ^ n X – + kn, k e Zva x = arctan + kn, k G Z. V ‘ty b) Vdi cosx = 0 ta tha’y ea hai vl eua phuong trinh bang 2. Vay cosx = 0 n thoa man phuong trinh, hay x = + ^7C, A: G Z la nghiem. 32 Vdi cos ^ 0, chia ca hai vl cua phuong trinh cho cos^ x ta dugc 2tan2 X – tanx – 1 = 2(1 + tan^ x) «tanx = -3 <=> X = aretan(-3) + ^71, ^ G Z. vay cac nghiem ciia phuong trinh la n X = + kn, k ez va x = arctan(-3) + it7i, it G Z. e) Vdi cosx = 0 thi vd trai bang 4, cdn vl phai bang 1, nen cosx = 0 Ichdng thoa man phuong trinh. Vdi cosx ^ 0, chia hai vl cua phuong trinh cho cos X ta dugc 4tan2 X – 4tanx + 3 = 1 + tan^ x <» 3tan2x – 4tanx +2 = 0. Phuong trinh nay vd nghiem. Vay phuong trinh da cho vd nghiem.

35 Vidu 6 Giai cac phuong trinh a) v3cosx + sinx = -2 ; e) 4sinx + 3cosx = 4(1 + tanx) – 1 ‘ cosx b) eos3x – sin 3x = 1 ; a) Ta cd Gidi S 1 v3cosx + sinx =-2 <:i> cosxh sinx =-1.71 n. <=> sm cosx + cos sinx <» sin x + – =-1 <=> X + = + k2n, k G 3 2 <=> X = + k2n, k G vay nghiem cua phuong trinh la x = + ^27t, ^ G Z. 6 b) Ta cd >/2, >/2.. cos 3x – sin3x = 1 «v2 cos3x sin3x 2 2 = 1 _ n.. 7t. V2 <«cos cos 3x – sm sin3x = r o cos 3x + V n 4y = COS <:> 3x + = ± + k2n x = ^27t, ^ G Z n <^ 3x = – + ^271, ^ G Z 2 X = ^, k G Z 3 71, 271, X = 1- k, k e BTDS&GT11-A 33

36 vay cae nghiem eua phuong trinh la, 2 n ^ , X = k, ^ G Z va X = + it, it G c) Dilu kien cua phuong trinh la cosx ^ 0. Tacd 1 4sinx + 3cosx = 4(1 + tanx) cosx <^ cosx(4sinx + 3cosx) = 4(sinx + cosx) – 1 <» cosx(4sinx + 3cosx) – cosx = 4sinx + 3eosx – 1 <» cosx(4sinx + 3cosx – 1) = 4sinx + 3eosx – 1 «> (cosx – l)(4sinx + 3eosx – 1) = 0 <=> cosx = 1 4sinx + 3cosx = 1 X = ^27t, ^ G Z smx + cosx = Kl hieu or la cung ma sin or =, cos or = ta duoc ^ 5 5 (2) <=> cos(x – or) = 1 (1) (2) <^ X – a = ±arecos + k2n «> x = or ± arceos + k2n. 5 5 vay cac nghiem ciia phuong trinh (1) la 1 3 X = k2n, k e Zva X – a ±arccos + k2n, k e Z, trong dd a = arceos. C. BAI TAP Gidi cdc phucmg trinh sau ( ) : 3.1. a) eos2x – sinx – 1 = 0 ; c) 4 sinx cosx cos 2x = -1 ; 3.2. a) sinx + 2sin3x = -sin5x ; e) sinx sin 2x sin 3x = sin4x ; 4 b) cosxeos2x = 1 + sinxsin2x ; d) tanx = 3cotx. b) cos5xcosx = eos4x ; d) sin X + cos x = cos^ 2x BT0S&GT11-B

37 3.3. a) 3eos2 x – 2sinx + 2 = 0; b) 5sin X + 3eosx + 3 = 0; e) sin X + cos x = 4eos 2x ; 3.4. a) 2tanx – 3cotx -2 = 0; c) cotx – eot2x = tanx + 1. j d) 1- sm X = cos X. 4 b) cos X = 3sin2x + 3 ; a) cos X + 2sinxcosx + 5sin x = 2 ; b) 3eos X – 2sin2x + sin^ x = 1 ; 1 1 e) 4cos X – 3sinxcosx + 3sin x = a) 2cosx – sinx = 2 ; b) sin5x + cos5x = -1 ; e) 8cos’^ X – 4eos2x + sin4x -4 = 0; d) sin^ x + eos^ x + sin4x = a) 1 + sinx – cosx – sin2x + 2eos2x = 0 ;,, b) sm X = sin x sin^x sinx c) cosxtan3x = sin5x ; d) 2tan2x + 3tanx + 2cot2x + 3eotx + 2 = 0. Bai tap on chuong I 1. Tim tap xae dinh cua cac ham sd 2-cosx a) y = 1 + tan X – n b)-y = tan X + cot X 1 – sin2x 2. Xae dinh tinh chan le cua cac ham sd a) y = sin x – tan x ; b)y = cos X + cot X sinx 3. Chia cac doan sau thanh hai doan, tren mdt doan ham sd y = sinx tang, cdn trdn doan kia ham sd dd giam : a) 7t ^ -;27t 2 b) [-n ; 0] ; c) [-271; -n]. 35

38 4. lim gia tri ldn nha’t va gia tri nho nha’t cua eae ham sd a) y = 3-4sinx ; b) y = 2 – Vcosx 5. Ve dd thi cua cac ham sd a) y = sin2x + 1 ; b) y = cos ^ n^ X V 6 Gidi cdc phucmg trinh sau (6-15) : sin X – cos x = cos4x. 7. eos3x – eos5x = sinx. 8. 3sin2x +4cosx-2 = sin^ X + sin 2x = sin 3x tanx + 3cotx = eos2 X – 3sin2x + sin x = sin X + sinxcosx – cos x = sinx – 4cosx = sin3x + sin5x – 2 sinx cos 2x = tan2 x – 3tanx + 2eot2 x + 3eotx -3 = 0. i- 1 LOI GIAI – HUONG DAN – DAP SO CHUONG I a)d = R{i: Y X Ti 3% b) cos ^0 ‘i^ ^ + kn >» x?t + k3n, ki VayD = R<{ + A;37T, it G 36

39 n c) sin2x ^0 <:^ 2x^ kn <^ X ^ k, k e 2 n VayD = RU-, it GZ d)d = R{-l ; 1} a) cosx + 1 > 0, Vx G R. Vay D = R. : V 2 2 7C b) sin X – cos x = -cos2x 9^ 0 <» 2x ^^ f- kn, k G 2 <» X ^ – + it-, it G Z. vay D = R I- + it-, it G 4 2 -^ [4 2 c) cosx – cos3x = -2sin2xsin(-x) = 4sin xcosx. Do dd cosx – eos3x?t 0 <^ sinx :?t 0 va cosx ^ 0 «> X?t it7t va X 5t – + it7t, it G Z. vayd = R\kj, ks. d) tanx va cotx cd nghia khi sinx 5^ 0 va cosx ;^ 0. vay tap xae dinh nhu trong cau c) a) 0 < sinx < 1 nen -2 < -2 sinx < 0. vay gia tri ldn nha’t ciia y = 3-2 sinx la 3, dat dugc khi sinx = 0 ; gia tri nhd nha’t cua y la 1, dat dugc khi sinx = ±1. b) cosx + cos f X – V A f 3j = 2 cos X – K A 6j / cos- = v3eos 6. V I In /TL vay gia tri nhd nha!t eua y la -v3, dat duge chang han, tai x = ; gia tri ldn nha’t cua y la v3, dat duoc chang han, tai x =. o c) Ta ed 2 ^ ^ l + cos2x ^ ^ l + 5cos2x cos X + 2cos2x = 1-2cos2x =. n 37

40 Vi -1 < cos2x < 1 nen gia tri ldn nh^t eua y la 3, dat dugc khi x = 0 n gia tri nhd nha’t ciia y la -2, dat duge khi x =. d) HD : 5-2eos2 xsin^ x = 5 – -sin^ 2x. 3>/2 Vi 0 < sin^ 2x < 1 nen < sin^ 2x < 0 => 2 2 ^ ‘<>/5. Suy ra gia tri ldn nha’t ciia y la Vs tai x = k-, gid tri nhd nh^t la tai n, n x= +k a) Dang thiic xay ra khi cac bieu thurc d hai vd cd nghla, tiic la sinx ^^ 0 va 7C COSX ^ 0. vay dang thiie x ty ra khi x ^-^ ^, ^ G Z. b) Dang thiic xay ra khi cosx ^ 0, tiic Vakhi x ^it – + kn, k e Z. c) Dang thiie xay ra khi sinx ^ 0, tvtc la x * kn, k e Z. n d) Dang thiic xay ra khi smx ^^ 0 va cosx ^^ 0, tiie la x ^^ A;, ^ G 1.5. a) y = ^^^ la ham sd le. b) y = X – sinx la ham sd le. e) y = Vl -cosx la ham sd chsn. d) y = 1 + eosxsin 37C -2x = 1 – cosx cos 2x la ham sd chan a) eos2(x + it7t) = cos(2x + k2n) = eos2x, k e Z. Vay ham sd y = cos2x la ham sd chan, tusn hoan, cd chu ki la n (H.7). 38 Hinh 7

41 ln 371 -‘571-7t 37t _il /LJLO ” 4 4 ‘ 371 /77C A- 2,/ 4 b) Dd thi ham sd y = cos2x (H.8) a) Dd thi ham sd y = 1 + sinx thu dugc tii dd thi ham sd y = sinx bang each tinh tidn song song vdi true tung len phia tren mdt don vi (H.9). ‘”‘ 37t 2 “”’ 1 N. y^ -71 71 ‘N^ 2 / 2 y ^.’^ y- = sinx “^ . y ”^ / 0-1 N^ y=+ sirur E. T^ 371 /27l X 2 ^ – / Hinh 9 b) Dd thi ham sd y = cosx – 1 thu dugc tii dd thi ham sd y = cosx bang each tinh tiln song song vdi true tung xudng phia dudi mdt don vi (ban dgc tu ve hinh). f n. c) Dd thi ham sd y = sin x thu dugc tii dd thi ham sd y = sin x bang 71 each tinh tien song song vdi true hoanh sang phai mdt doan bang (H.IO). ^^x ^^N. ^ : ^ y = sinjc 1-1 y 0 / ** /% / 3 n 2 y = %va{x \ T^ 47t 6 3 .:>.;-2>: 3’ 6 / /^7l X Hinh 10 39

42 d) Dd thi ham sd y = cos x H n thu dugc tit dd thi ham sd y = cosx n bang each tinh tiln song song vdi true hoanh sang trai mdt doan bang 6 (ban dgc tu ve hinh) a) Dd thi ham sd y = tan ^ X + 71^ thu duge tii dd thi ham sd y = tanx V 4y bang each tinh tidn song song vdi true hoanh sang trai mdt doan bang. n b) Dd Jhi ham sd y = cot X V 6y thu dugc tit dd thi ham sd y = cotx bang 71 each tinh tiln song song vdi true hoanh sang phai mdt doan bang n 2n 2.1. a) X = + k, k t va X = h k, k G 9 3 b) X = 30 + itl80, it = Z va X = 75 + itl80, it G Z. c) X = it720, it G Z va X = it720, itg Z. JN 1 – ^. T T, ^,, , 7 t, d) X = arcsin i- k, k e Z va x arcsin i- it, k G a) X = -3 ± arceos- + ^27t, it G Z. 3 b) X = 25 + itl20, X = 5 + itl20, it G e) X = + ^71, X = i- kn, k : d) X = ± arceos- + ^TC, ^ G 2 3

43 2.3. a) x = -45 +it90, itg, 371, -, c) X = h ^271, k G a) Dilu kiln : eos3x ^^ 1. Ta ed 71 b) X = h ^7t, ^ G Z. 6 d) X = it540, it G sin3x = 0 => 3x = kn. Do dilu kien, cac gia tri A: = 2m, m G Z bi loai, nen n 3x = {2m + l)7t, m G Z. Vay nghiem ciia phuong trinh la x = {2m + 1), m G / b) Dilu kien : sin X I 9^ 0. Bidn ddi phuong trinh cos2x. cot V X – n = 0 => cos2x.cos / V 71 = 0 cos2x = 0 X = h k, k G Z 4 2 n cos = X 4 X = h kn, k e Z. 4 n n Do dilu kien, eae gia tri x = + 2m, m G Z bi loai. Vay nghiem cua phuong trinh la X = I- (2m + l), m G Z va X = i- kn, k e Z. 4 ^ ^2 4 e) Dilu kien : cos(2x + 60 ) ^ 0. Ta cd tan (2x + 60 )cos(x + 75 ) = 0 => sin(2x + 60 ) eos(x + 75 ) = 0 sin(2x + 60 ) = 0 => eos(x + 75 ) = 0 x = -30 +it90, itg x = 15 + itl80,it_g 2x + 60 =itl80,itgz x + 75 =90 +itl80, itg 41

44 Do dilu kien d tren, cac gia tri x = 15 + itl80, it G Z bi loai. vay nghiem eiia phuong tnnh la x = it90, it G Z. d) Dilu kien : sin x ^t 0. Ta ed (cot x + 1) sin 3x = 0 -» cotx = -1 sin3x = 0 X = 7 + kn, k G 4 X = k, k ez. V 71 Do dieu kien sinx ^^ 0 nen nhftng gia tri x = k vdi k = 3m, mez loai. vay nghiem ciia phuong trinh la 7C, 7t, ^ 271,, _ X = – + kn ; x = + kn wa x = -^ + kn, k ez. hi 2.5. a) cos 2x – = cos -j – -“f n n 2x- = -x + ^271, ^ G Z 3x = + ^271, k G o 12 2x-^ = -j + x + k2n, k&z X = + k2n, kez TTT 271 7X vay cac gia tri csn tim la X = ^r- + A:- -, it G Z va x = + it27r, it G Z b) sin 3x- I = sin n ‘^6 42 TC TT 3x- = X + + ^27t, ^ G Z ,-, 3x – = 7t – X – + k2n, k G x = + A;27i, it G Z. 1371,-, _ 4x = it27t, it G Z 12 <» 571 X = + kn, k ez I3n, n,, ^ = -48-^^2’^^’

45 vay cac gia tri c&i tim la X = + ^7t, ^ G Z va x = + k, ke c) tan 2x + = tan ^n ^ cos <=> i 7t 2x + – 7t 0 va cos ^n 2x + = – X + ^TT, ^ G Z. ^ ^0 (1) (2) (2) «X = it^, it G Z. 71 Cae gia tri nay thoa man dilu kien (1). Vay ta ed x = k, k e Z. ^ d) cot 3x = cot V n^ ‘y ( sin 3x 5^0 va sin n^ ^0 <^ i V ^ J 3x = X + n + kn, k e (3) (4) (4) «x = + A:, A:GZ. Ndu it = 2m + 1, m G Z thi cac gia tri nay khdng thoa man dilu kien (3). Suy ra cac gia tri edn tim la x- n — mn, m e 2.6. a) eos3x – sin2x = 0 <» eos3x = sin2x <» cos3x = cos :r-2x n «3x = ± -2x 2 n 5x = – + it27i;, + ^271, k e kez n x = – + k2n,kez. 43

46 vay nghiem phuong trinh la x = – + k -,ke Zva x = -j + k2n,ke b) Dilu kien eua phuong trinh : cosx 5t 0 va cos2x ^0. tanx tan2x = -1 => sinxsin2x = -eosxeos2x => eos2xeosx + sin2xsinx = 0 => cosx = 0. Kit hgp vdi dilu kien, ta tha’y phuong tnnh vd nghiem. e) sin3x + sin5x = 0 «2sin4xcosx = 0 «> sin4x = 0 cosx = 0 <=> 4x = ^71, k e X = + kn, k e n n vay nghiem ciia phuong trinh la x = k,^gzva X = -^ + ^7I,^G ^ Zt d) Dilu kien : sin2x?t 0 va sin3x ^ 0. cot2xcot3x = l => eos2xcos3x = sin2xsin3x =» cos 2x cos 3x – sin 2x sin 3x = 0 TC => eos5x = 0 =>5x = + kn, k ez =^X = ^ + 4^GZ. Vdi A: = 2 + 5m, m G Z thi n n n 2n n X = + (2 + 5m) = -jtt + -p- + mn = + m7r, m G Z. Lue dd sin2x = sin(7x + 2m7i) = 0, khdng thoa man dilu kien Co the suy ra nghiem phuong trinh lax = + ^,^GZva^:?t2 + 5m, m e a) cos2x-sinx-l = 0 <» l-2sin^x-sinx-l = 0 «sinx(2sinx + l) = 0 44 <=> sinx = 0 1 <^ smx = – 2 X = ^71, A: G Z X = -^ + it27t, it G Z 6 77t X = – + it27c, it G Z. 6

47 b) cosxcos2x = l + sinxsin2x <=> cosxcos2x-sinxsin2x = <» eos3x = 1 <=> 3x = it27t «x = k, k e Z. e) 4 sin xcosx cos 2x = -1» 2sin2xeos2x = -1 <^ sin4x = -1 «> 4x = – + k2n,k G Z «x = ^ + k-t, k e Z. d) tanx = 3cotx. Dilu kien : cosx 9^ 0 va sinx # 0. Tacd tanx = <=> tan^x = 3 <=> tanx = ±yf3 ^^x = ± + kn, k e Z. tanx 3 Cac gia tri nay thoa man dilu kien eua phuong trinh nen la nghiem eua phuong trinh da cho a) sinx + 2sin3x =-sin5x -» sin5x + sinx + 2sin3x = 0 <:> 2sin3xcos2x + 2sin3x = 0 <=> 2sin3x(cos2x + l) = 0 <» 4sin3xeos x = 0 3x = kn, k e Z sin3x = 0 cosx = 0 <» X – + kn,k e b) cos5xcosx = eos4x (eos6x + eos4x) = eos4x <» 71 X = k, k ez X = + kn, k e <» cos6x = eos4x «- 6x = ±4x + k2n, k ez ‘2x – k2n, k e 2 mx = k2n,ke. X = kn, k ez x = k^,kez,n Tap {it7t, it G Z} chiia trong tap <{/-,/ G Z [> (ling vdi cac giatri / la bdi sd ciia 5) nen nghiem cua phuong trinh lax = A-^,A:GZ. 45

48 c) sinxsin2xsin3x = sin4x <:> sinxsin2xsin3x =-rsin2xeos2x ‘ 4 2 <5> sin2x(cos2x-2sinxsin3x) = 0» sin2x.eos4x = 0 <=> sin 2x = 0 2x = kn,kez cos4x = 0 <» Ax = kn,ke o X = k, k ez x = + k, kez. o 4 d) sin^ X + cos x = – cos 2x o (sin^ X + eos^ x)^ – 2sin^ xcos^ x = r-cos 2x <:> 1 – -sin 2x + -eos^ 2x = «1 + eos4x = 0 <» cos4x = Phuong trinh vd nghiem. ^ Chu y. C6 the nhan xet: Ve phai khong dudng vdi moi x trong khi vd trai duong v6i moi X nen phuong trinh da cho v6 nghiem a) 3eos^x-2sinx + 2 = 0 <» 3(l-sin^x)-2smx + 2 = 0 «3sin^x + 2sinx-5 = 0 «(sinx-l)(3sinx + 5) = 0 n <=> sinx = l<=>x = + ^271, k ez. b) 5sin^ X + 3cosx + 3 = 0 <» 5(1 – eos^ x) + 3cosx + 3 = 0 «> 5cos x-3cosx-8 = 0 <=> (cosx + l)(5cosx-8) = 0 <=> cosx = -1 <» X = (2A + l)7t, A G Z. 46

49 c) sin^ X + eos^ x = 4cos^ 2x <^ (sin x + cos x) -3sin^xeos^x(sin^x + eos^x) = 4eos^2x <=> 1 – -rsin^ 2x = 4cos^ 2x «1 – -(1 – cos^ 2x) = 4eos^ 2x 4 4^ -‘ 13 2o 1 o 13 l + eos4x o -rcos^ 2x = ^ <=> 1 + cos4x = <» cos4x = – <:> 4x = ± arceos + A;27i;, it G Z = 1 r <=> X = ± arceos, ‘ /-“,^ 4 l^ 13 + A;,^ G., l-eos2x f’l + cos2x d) r + sin x = cos X <» 7 + = <^ eos2x = 1 + 2cos2x + cos^ 2x <» cos 2x + 4eos2x = 0 cos2x = 0 n “» <^ 2x = + kn, k e cos 2x = – 4 (vd nghiem) 2 n, n, <» x = + ^, A: G Z a) 2 tanx – 3eot X – 2 = 0. Dilu kien : cosx 9^ 0 va sinx ^ 0. Tacd 2tanx- tanx -2 = ± V7 <=> 2tan x-2tanx-3 = 0 otanx = – X = arctan + kn, k G X = arctan + kn, k ez. C^c gia tri nay thoa man dilu kien nen la nghiem eiia phuong trinh. 47

50 b) cos X = 3sin2x + 3. Ta tha’y cosx = 0 khdng thoa man phuong tnnh. Vdi cosx ^ 0, chia hai vl 1 2 eua phuong trinh cho cos x ta dugc 1 = 6tanx + 3(l + tan^x)<» 3tan^x + 6tanx + 2 = 0 <:$ tanx -3±>^ <=> X = arctan X = arctan c) cotx – cot2x = tanx+ 1. Dilu kien : sinx ^t 0 va cosx * 0. Khi dd, cosx cos2x sinx (1) «> + 1 sinx sin2x cosx (1) e> 2cos X – cos2x = 2sin^ x + sin2x 9 9 <» 2(eos X – sin x) – eos2x = sin2x o cos2x = sin2x <:> tan2x = 1 2x = + kn, k G 4 X + / C, KG. Cae gia tri nay thoa man dilu kien nen la nghiem cua phuong trinh a) cos x + 2sinxeosx + 5sin^x = 2. Rd rang cosx = 0 khdng thoa man phuong tnnh. Vdi cosx ^ 0, chia hai vl cho cos X ta dugc 1 + 2tanx + 5tan^ x = 2(1 + tan^ x) <» 3tan^x + 2tanx-l = 0 >» 71,, _ tanx = -1 X = –kn,ke Z 4 1 <=> tanx = X = arctan + kn, kez. 48

51 b) 3eos^ X – 2sin*’x + sin^ x = 1. Vdi cosx = 0 ta thiy hai vl diu bang 1. Vay phuong trinh ed nghilm n x = + kn,kez. * Trudng hgp cosx 9^ 0, chia hai vl cho cos x ta dugc 3-4tanx + tan^x = l + tan’^x -o-4tanx = 2 <» tanx = O X = arctan + kn, k e Z. vay nghiem cua phuong trinh la n I x = + kn,ke Zva x = arctan + kn, k e Z. 9 9 c) 4cos x-3sinxeosx + 3sin x = l. Rd rang cosx * 0. Chia hai vd ciia phuong trinh cho cos x ta dugc 4-3tanx + 3tan^ X = 1 + tan^ X <:> 2tan^X-3tanx + 3 = 0. Phuong trinh cud’i vd nghiem (dd’i vdi tanx), do dd phuong trinh da cho vd nghiem a) 2cosx – sinx = 2 «- V^ J =reosx prsinx 2 1 Kl hieu or la sde ma cos or = -7=, sin or =?=, ta duoc phuong trinh Vs sis cos a cos X + sm or sm X = ^5 <» eos(x – or) = eosor <^ x – or = ±or + k2n, k e 2 «> X = 2or + k2n, k e x = it27i;, ^ G Z. 4. BTDS&GT11-A 49

52 b) sin5x + eos5x =-1 >» V2 -r-sm 5x + – cos 5x = -1 V J 71. c. 71 _ -^2 (~ Tt. o cos sin5x + sm cos5x = – o sm 5x + = sm ^ 4 v’4y <» 5x + ^ = -^ + it27t,itgz 4 4 5x +? = -^ + it27t,itgz 4 4 «> c) 8cos X – 4eos2x + sin4x -4 = 0 n,2n, x = – + k-^,ke 7t, 271, X = + ^-z-, ^GZ. ^ 71^ o 8 ri + cos2x 4eos2x + sin4x -4 = 0 y <:> 2(1 + 2eos2x + cos 2x) – 4cos2x + sin4x -4 = 0 <^ 2eos^ 2x + sin4x -2 = 0 <» 1 + cos4x + sin4x – 2 = <=> cos4x + sin4x = 1 <=> sin 4x + I 4^ = sm- n n, _, _ n 4x + = + k2n, kez X = k, k ez 4 4 <=> n 3n,-, _ 4x + =- + k2n, kez x-^ + k^,kez. 4 4 d) sin X + cos^ x + sin4x = 0 <^ (sin x + cos x) – 3sin xcos^ x(sin^ x + eos^ x) + sin4x = 0 <=> l-3sin xcos^x + sin4x = 0. T N2 <=> 1-3 sm2x^ + xsin4x = » 1- sin 2x + sin4x = BTBS&GT11-B

53 , 3 l-cos4x 1… <^ l-r- r + sin4x = “» cos4x + 4sin4x = 0 «3cos4x + 4sin4x = ,, <=> cos4x + sin4x = Ki hieu or Ik cung ma sin or =, cos or = T’ ta dugc sin a cos 4x +cos or sin 4x = -1 «sin(4x + or) = «> 4x + or = + ^271,keZ 3n a, n, <^x = – + k-,k e Z a) 1 +sinx-cosx-sin2x + 2cos2x = 0. Ta cd : 1 – sin2x = (sinx-cosx) ; (1) 2 cos 2x = 2(cos^ X – sin^ x) = -2(sin x – cos x)(sin x + cos x). vay (1) <:>,(sin X – cos x)(l + sin X – cos x – 2 sin x – 2 cos x) = 0 <» (sin X – cos x)(l – sin X – 3 cos x) = 0 <» smx = cosx 3cosx + sinx = 1 n X – + kn, k ez tanx = l X – a± eirccos j=r + it27i, k ez VlO 3. 1 trong do eosor = -7=^, sma = yflo’ yjlo’ « eosx+, smx = VlO >/io’ >/io 51

54 b) sm X : = sm X – smx sin^x Dilu kiln : sinx 5^ 0. Khi dd, ^ 1 (2) <=> (sinx-sin x) + vsin X sinx.,,. 1-sinx _ <=> sinx(l-smx) + :: = 0 sin X <» (1 – sinx)(sin x +1) = 0 c) cosx tan 3x = sin 5x. = 0 n => x = -^ + it7i,itg Z (thoa man dilu kien). sinx = -l 2 Dilu kien : eos3x ^t 0. Khi dd, (3) <^ cos X sin 3x = cos 3x sin 5x (2) (3) o (sin4x + sin2x) =.(sinsx + sin2x) sin8x = sin4x <=> ‘8x = 4x + it27i, it G Z 8x = 71-4x + A;27T, k e X = it, itgz 71^ I 7t, rs x^-^k-,kez. Kit hgp vdi dilu kien ta dugc nghiem ciia phuong trinh la 71 7t X = ^71, ^ G Z va X = -r::r + k, k ez d) 2tan’^ x + 3tanx + 2cot^ x + 3cotx + 2 = 0. Dilu kien : cosx *0 va sinx 9t 0. Khi dd, (4)» 2(tan^ x + cot^ x) + 3(tan x + cot x) + 2 = 0. (4) o 2 (tanx + cotx)^ (tanx + cotx) + 2 = 0. Dat f = tanx + cotx ta duge phuong trinh 2r + 3r – 2 = 0 t = -2,t^. 52

55 Vdi t = -2 ta cd tan X + cot x = -2 <:> tan^x + 2tanx + l = 0 => tanx = -1 =>x = – + kn,kez (thoa man dilu kien) Y6it = tacd tanx + cotx = o 2tan x-tanx + 2 = Phuong trinh nay vd nghiem. n vay nghiem cua phuong trinh (4) la x- + kn,ke Z. Bai tap on chirong i 1. a) Dilu kien : cos 71 r> V X – ht 0 va tan V ^ 7l^ ^-1 -^y 7t TC TC TC <::^ X- 9i + A:7:,AGZva X- it – + ^7r,A:G t n <:> x^-2- + kn,kezaxtt kn,kez. vay tap xae dinh eiia ham sd la D = R -^ + it7t,itgzlu -^ + it7i, itg b) Dilu kien : cosx ^ 0 ; sinx ^0 va sin2x * 1 71 n <:> X ^ k,k e Za x^ + kn,kez. Vay tap xae dinh cua ham sd la D = R U,A;Gzlu ^ + it7c,itg 2. a) y = sin x – tan x la ham sd le. cosx + cot X b) y = ^ la ham sd le. smx 3. a) Ham sd y = smx giam tren doan n _ 371 I’T va tang tren doan 371 ;27i 53

56 b) y = sinx giam tren n, tang tren -f;0 c) y = sinx tang tren -2n;- 3n, giam tren HD: a) -1 < 3-4sinx < 7. b) 1 < 2 – Vcosx < a) Dd thi ciia ham sd y = sin2x +1 thu dugc tii dd thi ham sd y = sin2x bang each tinh tiln song song vdi true tung len phia tren mdt don vi. b) Dd thi ham sd y = cos n V x – 6y thu duge tii dd thi ham sd y = cosx bang n each tinh tiln song song vdi true hoanh sang phai mdt doan bang. o 9 9′ 6. sin X – cos x = cos4x n <:>, n -cos2x, = cos4x o 2cos3xeosx = 0 x = -^k-,ke cos3x cosx = = 00 <:> n X = + kn, k ez. eos3x – eos5x = sinx <:> sinx(l -2sin4x) = 0 sinx = 0 sin4x = 2 X = ^71, k ez 71, n,, ^ = 2A^^r^^’ ^~2A^ 8. 3sin x + 4cosx-2 = 0 r 0 + #7 «-3eos X + 4cosx +1 = 0 <^ cosx = ^ o X = ± arceos 2-N/7 9 I rj + k2n,kez (gia tri r > 1 nen bi loai). 54

57 sin^ X + sin^ 2x = sin^ 3x o l-eos2x l-cos4x l-cos6x <:> l-cos4x + eos6x-cos2x = 0 <«2sin^2x-2sin4xsin2x = 0 <:> 2sin2x(sin2x-sin4x) = 0» 4sin2xcos3xsinx = 0. Ddpsd: x = it^,itgzva x = 5 + it-j,itgz tanx + 3eotx = 4. Dilu kien : cosx 9^ 0 va sinx 9^ 0. Ta cd 2tan^x-4tanx + 3 = 0. Phuong trinh vd nghidm ddi vdi tanx, do dd phuong trinh da cho vd nghiem eos^ X -3sin2x + sin^ x = 1. n cosx = 0 thoa man phuong tiinh => phuong trinh ed nghiem x = +, 2 1 Vdi cosx ^ 0, chia hai ve cho cos x, tim duoc tanx =. 6 kn,kez vay phuong trinh ed eae nghiem x = + kn,kez va x = arctan + kn,kez HD : 2sin x +sinxcosx-cos x = 3 => tan x-tanx + 4 = 0. Phuong trinh vd nghiem… o., sinx-4cosx = 1 <» sinx- cosx =, 1, , <=> sin(x – or) = – (voi eosor = -, sina = ) <=> X = or + arcsin + k2n, kez X = a + n-arcsin + A27t, k ez sin3x + sin5x-2sinxcos.2x = 0 <» 4sin3x + sin5x-sin3x + sinx = 0 c?- 3sin3x + sin5x + sinx = 0 <» 3sin3x + 2sin3xcos2x = 0 ‘ <» sin3x(3 + 2eos2x) = 0. 7i: Dapsd: x = k,k e 55

58 15. 2tan^x-3tanx + 2eot^x + 3cotx-3 = 0. (1) Dilu kien : cosx 5^ 0 va sinx ^t 0. (1) <» 2(tan^ X + cot^ x) – 3(tanx – cotx) – 3 = 0 «2(tanx-cotx)^ -3(tanx-cotx) + l = 0. Dat f = tanx – cotx ta duge phuong tnnh 2r^-3f + l = 0 => t=l, t = ^. Vdi f = 1 tacd tanx-cotx = 1 2 i + Vs -» tan x-tanx-l = 0 otanx = – X = arctan + kn, k e. X = arctan + kn,kez Voi t = ta CO tanx – cotx = l±>/l7 o 2tan x-tanx-2 = 0 <» tanx = : X = arctan i + Vn 4 + ^711, A G Z 1-Vl7 X = arctan, 4 + ^71, A G Z. Cac gia tri nay thoa man dilu kien nen chung la nghiem eua phuong trinh da cho. 56′

59 huang II. J TO HOP – XAC SUAT 1. Quy tac dem A. KIEN THUC CAN NHO 1. Quy tac cong Gia sit ddi tugng X cd m each chgn khae nhau, ddi tugng Y c6 n each ehgn khae nhau va Ichdng cd each chgn ddi tugng X nao triing vdi mdi each chgn dd’i tugng Y. Khi dd cd m + n each chgn mdt trong hai ddi tugng a’y. Gia sii A va 5 la cac tap hfiu han, khdng giao nhau. Khi dd n{a u fi) = n{a) + n{b). (1) ^ Chu y. Cong thiic (1) c6 the md rong theo hai hudng : a) Neu Avk Ba hai tap hcru han bat ki thi n(a u S) = n(a) + n(b) – n(a n S). (2) b) Ndu /^,…, A^ la cac tap hdru han tuy”}, doi mot khdng giao nhau thi 2. Quy tac nhan n(/i u >A2 u… u A^) = n(/li) + n(a2) n(a^). Gia sii A, fl la hai tap hiiu han. Ki hieu A x B la tap hgp tat ca cac cap cd thu: tu {a, b), trong dd a G A, 6 G fi. Ta cd quy tac n{a^b) = n{a).n{b). (3) Quy tac tren cd thi phat bilu nhu sau : Gia sii cd hai htinh ddng dugc thuc hien lien tilp. Hanh ddng thii nha’t ed m kdt qua. ijig vdi mdi kit qua cua hanh ddng thii nhit, hanh ddng thii hai cd n kit qua. Khi dd ed m x n kit qua ciia hai hanh ddng lien tilp dd. ^ Chu y. Quy tac nhan 6 tren c6 the md rdng cho nhieu h^nh ddng lien tiep. 57

60 B. VI DU V{ du 7 Trong mdt ldp ed 18 ban nam, 12 ban nii. Hdi cd bao nhieu each chgn a) Mdt ban phu trach quy ldp? b) Hai ban, trong dd cd mdt nam va mdt nii? 58 Giai a) Theo quy tac cdng, ta cd = 30 each chgn mdt ban phu trach quy ldp (hoae nam hoae nii). b) Mudn ed hai ban gdm mdt nam va mdt nfi, ta phai thuc hidn hai hanh ddng lua chgn: – Chgn mdt nam : Cd 18 each chgn ; – Khi da ed mdt nam rdi, cd 12 each ehgn mdt ban nii. vay theo quy tac nhan, taedl8.12 = 216 each chgn mdt nam va mdt nii. Vidu 2 ^ Tren gia sach cd 10 quyin sach tiing Viet khae nhau, 8 quyin tiing Anh khae nhau va 6 quyin tiing Phap khae nhau. Hdi ed bao nhieu each ehgn a) Mdt quyin sach? b) Ba quyin sach tiing khae nhau? c) Hai quyin sach tiing Ichae nhau? Gidi a) Theo quy tac cdng, cd = 24 each chgn mdt quyin sach. b) Theo quy tac nhan, ed = 480 each chgn ba quyin tiing khae nhau. c) Theo quy tie nhan, cd 10.8 = 80 each chgn mdt quyin tiing Viet va mdt quyin tiing Anh ; Cd 10.6 = 60 each chgn mdt quyin tiing Viet va mdt quyin tiing Phap ; Cd 8.6 = 48 each chgn mdt quyin tiing Anh va mdt quyin tiing Phap. Tii dd, theo quy tac cdng, ta cd sd each ehgn hai quyin sach tie’ng khae nhau la = 188 (each).

61 Vi du 3 Tii eae sd 1, 2, 3, 4, 5, 6, 7, 8, 9, cd bao nhieu each chgn mdt sd hoac la sd chan hoae la sd nguyen td? Gidi Ki hieu A la tap hgp cae sd chan (cd 4 sd) va 5 la tap hgp cae sd nguyen td (cd 4 sd) trong tap sd da cho. Khi dd, sd each chgn e^n tim la n{a u B). Nhung cd mdt sd nguyen td chan duy nhsft la 2, hie n(a n B) = 1. Vay theo (2), n{a u B) – n{a) + n{b) – n{ar^b) = A +A-I = 1. C. BAI TAP 1.1. Nam din cita hang van phdng ph&i dl mua qua tang ban. Trong cira hang cd ba mat hang : But, vd va thude, trong dd cd 5 loai but, 4 loai vd va 3 loai thude. Hdi ed bao nhieu each chgn mdt mdn qua gdm mdt but, mdt vd va mdt thude? 1.2. Trong mdt ddi van nghe cd 8 ban nam va 6 ban nii. Hdi ed bao nhieu each chgn mdt ddi song ca nam – ntt? 1.3. Cd bao nhieu sd tu nhien cd tinh chsft: a) La sd ehsn va ed hai chii sd’ (khdng nha’t thiit khae nhau) ; b) La sd le va ed hai chii sd (khdng nh^t thiit khad nhau) ; e) La sd le va ed hai chii sd khae nhau ; d) La sd chan va ed hai ehfl sd Ichac nhau Cd 10 cap vg ehdng di du tiec. Tinh sd each chgn mdt ngudi dan dng va mdt ngudi dan ba trong bfta tiec dl phat bilu y kiln, sao cho a) Hai ngudi dd la vg ehdng ; b) Hai ngudi dd khdng la vg ehdng Sd 360 ed bao nhieu ude nguyen duong? 1.6. Trong sd nguyen duong ddu tien, ed bao nhieu sd chiia mdt chii sd 3, mdt chii sd 4 va mdt chii sd 5? 1.7. Gifla hai thanh phd A va B ed 5 con dudng di. Hdi ed bao nhieu each di tii A din B rdi trd vl A ma khdng ed dudng nao duge di hai ldn? 59

62 1.8. Cd bao nhieu sd nguyen duong gdm khdng qua ba chii sd khae nhau? 1.9. Mdt ngudi vao eiia hang an. Ngudi dd mudn chgn thuc don gdm mdt mdn an trong 10 mdn, mdt loai hoa qua trang mieng trong 5 loai hoa qua v^ mdt loai nudc udng trong 4 loai nudc udng. Hdi ed bao nhieu each chgn thuc don cua biia an? Mdt ldp ed 40 hgc sinh, dang ki choi ft nhdt mdt trong hai mdn thi thao : bdng da va cdu Idng. Cd 30 em dang kf mdn bdng da, 25 em dang ki mdn cdu Idng. Hdi cd bao nhieu em ddng kf ea hai mdn thi thao? 2. Hoan vj, chinh hop, td hop A. KIEN THCTC CAN NHd Cho tap hgp A gdm n phdn tit {n > 1). 1. Kit qua eua su sdp xip n phdn tit ciia A theo mdt thii tu nao dd dugc ggi la mdt hodn vi cua tdp hgp A. Sd cac hoan vi ciia A dugc kf hieu la P, ta ed P = n.(n-1) =«!. Kit qua ciia viec Id’y k phdn tii eua A (1 < ^ < n) vd xip theo mdt thii tu nao dd dugc ggi la mdt chinh hgp ehdp k ciia n phdn tuf. Sd cae chinh hgp chap k eiia n ph^n tt dugc kf hidu la A^, ta cd (d day, quy ude 0! = 1). 3. Mdt tdp eon gdm k phdn tvt cua A (1 < it < n) dugc ggi la mdt ts hap ehdp k eiia n phdn th. Td hgp ehdp 0 ciia n phdn tii la tdp rdng. Sd cac td hgp ehdp k ciia n phdn tut dugc kf hieu la C*, ta cd 60

63 B. VI DU Vidu 1 Cd bao nhieu each xip bdn ban A, B, C, D vao bdn chile ghi kl thanh hang ngang? 9 Gidi Mdi each xdp cho ta mdt hoan vi ciia bd’n ban va ngugc lai. Vdy sd each xdp la B4 = 4! = 24 (each). Vidu 2 Cd bao nhieu sd nguydn duong gdm nam chii sd khae khdng va khae nhau (ddi mdt)? Gidi Mdi sd cdn tim cd dang 0^020^0^0^, trong dd a, ^ Oj vdi / ^j va ^ a, G [l,2,…,9},i=l,…,5. Nhu vdy, cd thi coi mdi sd dang tren la mdt chinh hgp ehdp 5 cua 9 (chii sd’). Do dd, sd cae sd’ cdn tim la A = II = = (sd^ Vi du 3 Cdn phan cdng ba ban ttt mdt td cd 10 ban dl lam true nhat. Hdi cd bao nhieu each phan cdng khae nhau? Gidi Kdt qua eua su phan cdng Id mdt nhdm gdm ba ban, tiic la mdt td hgp chap 3 cua 10 ban. Vay sd each phan cdng la Vi du 4 Trong mat phang cd 6 dudng thang song song vdi nhau va 8 dudng thang khae cung song song vdi nhau ddng thdi cat 6 dudng thang da cho. Hdi ed bao nhieu hinh binh hanh dugc tao nen bdi 14 dudng thing da cho? 61

64 Gidi Kl hieu A va B ldn Iugt la tdp hgp 6 dudng thang song song vdi nhau va 8 dudng thang song song eat 6 dudng thdng da cho. Mdi hinh binh hanh dugc tao bdi hai dudng thang ciia tap A va hai diidng thing ciia tap B. Vay sd hinh binh hanh cdn tim la C. Cg = = 420 (hinh). C. BAI TAP 2.1. Mdt cai khay trdn dung banh keo ngay Tdt cd 6 ngan hinh quat mau khae nhau. Hdi cd bao nhidu each bay 6 loai banh keo vao 6 ngan dd? 2.2. Cd bao nhieu each xdp 5 ban nam va 5 ban nii vao 10 ghd dugc ke thanh hang ngang, sao cho : a) Nam va nii ngdi xen ke nhau? b) Cdc ban nam ngdi liln nhau? 2.3. Cd bao nhieu each xdp chd ngdi cho 10 ban, trong dd cd An va Binh, vao 10 ghi ke thanh hang ngang, sao cho : a) Hai ban Anva Binh ngdi eaiih nhau? b) Hai ban An va Binh khdng ngdi canh nhau? 2.4. Thdy giao cd ba quyin sach Toan khae nhau cho ba ban mugn (mdi ban mdt quyin). Sang tudn sau thdy giao thu lai va tilp tuc cho ba ban mugn ba quyin dd. Hdi ed bao nhieu each cho mugn sach md khdng ban nao phai mugn quyin da dgc? 2.5. Bdn ngudi dan dng, hai ngudi dan ba va mdt diia tre duge xdp ngdi vao bay chile ghi dat quanh mdt ban trdn. Hdi ed bao nhieu each xdp sao cho : a) Diia tre ngdi giiia hai ngudi dan ba? b) Diia tre ngdi giiia hai ngudi dan dng? 2.6. Ba qua cdu duge dat vao ba cai hdp khae nhau (khdng nhdt thiit hdp nao ciing cd qua cdu). Hdi ed bao nhieu each dat, nlu a) Cac qua cdu gidng hdt nhau (khdng phdn bidt)? b) Cac qua cdu ddi mdt khae nhau? 2.7. Cd bao nhieu each chia 10 ngudi thanh a) Hai nhdm, mdt nhdm 7 ngudi, nhdm kia 3 ngudi? b) Ba nhdm tuong ling gdm 5, 3, 2 ngudi? 62

65 2.8. Mdt gia sach bdn tdng xip 40 quyin sach khae nhau, mdi tdng xip 10 quyin. Hdi cd bao nhieu each ehgn cac quyin sach sao cho tii mdi tdng cd a) Hai quyin sach? b) Tdm quyin sach? 2.9. Cd giao chia 4 qua tao, 3 qua cam va 2 qua chudi cho 9 chau (mdi chau mdt qua). Hdi cd bao nhidu each chia khdc nhau? Mdt doan dai bilu gdm bdn hgc sinh duge chgn tit mdt td gdm 5 nam va 4 nii. Hdi cd bao nhieu cdch chgn sao cho trong dd cd ft nhdt mdt nam va ft nhdt mdt nii? Cd bao nhieu tam “gidc ma eae dinh cua chiing thude tdp hgp gdm 10 diem ndm tren dudng trdn 7 Mdt da giac Idi 20 canh ed bao nhieu dudng cheo? Cd bao nhieu tdp con cua tdp hgp gdm 4 dilm phdn biet? Cd bao nhieu each Xip chd cho 4 ban nii va 6 ban nam ngdi vao 10 ghd ma khdng cd hai ban nii nao ngdi canh nhau, ndu a) Ghd sdp thanh hang ngang? b) Ghd sdp quanh mdt ban trdn? Chiing minh rang vdi 1 < ^ < n, f-ik+l _ /r^k, y^k,, f^k, f~<k ‘-n+l – ‘-/I + ^/i-l *-fc+l + ‘-A Sit dung ddng nhd’t thiic Ic’ = c+ 2Cl dl ehiing minh ring i^+2^+…+«^=i:ci+2i:c^”(»^^f»^^>. *=1 /t= Mdt ldp cd 50 hgc sinh. Cdn phdn cdng 4 ban quit sdn trudng va 5 ban xen eay. a) Tfnh sd each phdn cdng bing hai phuong phap dl nit ra ding thiic ^9 y~<a _ X-.4 ^5 *-50-^9 – ‘-’50-‘-46- b) Chiing minh cdng thiie Niu-ton CIC’^^CICZI {n>r>k>q). 63

66 2.18. Chiing minh ring nlu n la sd nguyen td thi vdi r = 1, 2,…, n – 1, ta ed C^ chia hit cho n Trong mdt da giac diu bay canh, ke cae dudng cheo. Hdi cd bao nhieu giao dilm cua cac dudng cheo, trit cdc dinh? Tim sd cac sd nguyen duong gdm ndm chii sd sao cho mdi ehfl sd ciia sd dd ldn hon chii sd d ben phai nd. / 3. Nhj thcrc Niu-ton A. KIEN THOC CAN NH6 1. Khi khai triln nhi thiie {a + b)”, ta nhan 3ugc cdng thiie {a + b)” = cy + Cy-^b C^-^ab”-^ + C>” (1) (cdng thiic Nhi thiie Niu-ton). 2. Trong vl phai cua cdng thiic (1) ta cd : a) Sd eae hang tit la n + 1 ; b) Sd hang (hang tii) thii it + 1 la C* a””’ b”, k = 0, l,…n (quy udc a = 1 vdia^to). c) Sd mu eua a giam ddn tvt n ddn 0, sd mii cua b tang ddn tii 0 din n, nhung tdng cdc sd mii eua avkb trong mdi hang tit ludn bing n. d) Cae hang tit each diu hang tit ddu va hang tit cud’i cd he sd bing nhau. Vidul B. VI DU Khai triln (x – a) thanh tdng cac don thiic. 64 Gidi Theo cdng thiic Nhi thiic Niu-ton ta ed (x – a)^ = [x + {-a)] = x^ + 5x’^(-a) + lox^(-a)^ + lox^(-fl)^ + 5x(-fl)’* + (-a)^ = x^ – 5x’^a + lox^a^ – lox^a^ + 5×0^^ – a^.

67 Vidu 2 Tim sd hang khdng ehiia x trong khai triln Gidi Sd hang tdng qudt trong khai triln la ^ C^{2xf-‘. – X J = cl2^–lf x^-‘^ Ta phai tim k sao cho 6-3A: = 0, nhdn duge it = 2. Vdy sd hang cdn tim la C 2^”^(-l)^ = 240. C. BAI TAP r 2V 3.1. Tim sd hang thii nam trong khai triln I x +, ma trong khai triln dd sd mii eiia x giam ddn Vie’t khai triln eua (1 + x)^ a) Dung ba sd hang ddu dl tfnh gdn dung 1,01^. b) Diing may tfnh dl kilm tra kit qua tren Bie’t he sd eua x trong khai triln cua (1 + 3x)” la 90. Hay tim n Trong khai triln cua (1 + ax)” ta cd sd hang ddu la 1, sd hang thii hai la 24x, sd hang thii ba la 252x. Hay tim a van Trong khai triln ciia (x + af{x – bf, he sd cua x^ la -9 va khdng cd sd hang chiia x^. Tim avkb. 5. BTBS&GT11-A 65

68 4. Phep thijt va bi^n co A. KIEN THOC CAN NH6 Tdp hgp mgi kit qua cd thi xay ra cua mdt phep thit dugc ggi la khdng gian mdu cua phep thit va duge kf hiiu la Q. Ta chi xet cdc phep thit vdi khdng gian mdu Q la tdp huu han. Mdi tdp con A eiia Q duge ggi la bien co. Tdp 0 dugc goila bien c6 khdng the, tdp Q duge ggi la bie’n cd’chdc chan. Nlu khi phep thit dugc tiln hanh ma kit qua cua nd la mdt phdn tit cua A thi ta ndi ring A xay ra, hay phep thii la thudn Igi cho A. Biln cd A = Q A dugc ggi la bien co ddi ciia A. A va B dd’i nhau <^ A = B. A xay ra khi va ehi khi A khdng xay ra. Biln cd A u B xay ra Ichi va chi khi A hoae B xay ra. Biln c6 ArB xay ra khi va chi khi A va B ciing xay ra. Nlu A n B = 0 thi A va B dugc ggi la hai bien co xung khdc. B. VI DU Vi du 1 Gieo mdt con sue sdc cdn dd’i, ddng chdt va quan sdt sd chdm xud’t hien. a) Md ta Ichdng gian mdu. b) Xae dinh eae biln ed sau : A : “Xud’t hien mat chan chd’m” ;, B : “Xud’t hien mat le chd’m” ; C : “Xudt hien mat cd sd chdm khdng nhd hon 3”. e) Trong cdc biln ed trin, hay tim cdc biln ed xung khic BTBS&GT11-B

69 Gidi a) Kf hieu kit qua “Con siie sic xudt hien mat k chd’m” la k {k = I, 2,…, 6). Khi dd khdng gian mdu la Q = {1, 2, 3, 4, 5, 6}. b)tacd A = {2,4,6}; B={1,3,5}; C={3,4,5,6}. c) Cae biln ed A va B la xung Ichie, vi ArB =0 Vidu 2 Tit mdt hdp chiia 3 bi trang, 2 bi dd, Idy ngdu nhien ddng thdi 2 bi. a) xay dung khdng gian mdu. b) Xae dinh cdc biln cd : A : “Hai bi ciing mau tring” ; B : “Hai bi ciing mdu dd” ; C : “Hai bi ciing mau” ;. D : “Hai bi khae mau”. c) Trong edc biln cd tren, hay tim eae biln cd xung Ichde, cae biln cd dd’i nhau. Gidi a) Cdc bi trdng duge danh sd 1,2, 3. Cdc bi dd duge danh sd 4, 5. Khi dd Ichdng gian mdu gdm cae td hgp ehdp 2 ciia 5 (sd). Tiic la Q= {{1,2}, {1,3}, {1,4}, {1,5}, {2,3}, {2,4}, {2,5}, {3,4}, {3,5}, {4,5}}. b) Ta cd A = {{1,2},{1,3},{2,3}}, B= {{4,5}},C = AuB,D= C. c)taed AnB= 0, AnD =0, BnD =0, CnD =0. Do dd A va B xung khic ; D xung khae vdi cac biln cd A,B,C. Vi D = C nen C va D la hai biln ed dd’i nhau. 67

70 C. BAI TAP 4.1. Gieo mdt ddng tiln ba ldn va quan sat su xud’t hien mat sdp (5), mat ngiia (AO. a) Xdy dung khdng gian mdu. b) Xae dinh cdc biln cd : A : “Ldn gieo ddu xudt hien mat sdp” ; B : “Ba ldn xud’t hien cac mat nhu nhau” ; C : “Diing hai ldn xud’t hien mat sdp” ; D : “it nhat mdt ldn xud’t hien mat sdp” Gieo mdt ddng tiln, sau dd gieo mdt con sue sde. Quan sat su xudt hien mat sdp (S), mat ngiia (N) ciia ddng tiln va sd chdm xudt hien tren eon sue sic. a) Xdy dung Ichdng gian mdu. b) Xae dinh cac biln cd sau : A : “Ddng tiln xudt hien mat sdp vd con siie sic xud’t hien mat chan chdm”; B : “Ddng tiln xud’t hien mat ngita va eon siie sic xud’t hien mat le chdm”; C : “Mat 6 chd’m xudt hien” Mdt con siie sic dugc gieo ba ldn. Quan sat sd chdm xud’t hien. 68 a) Xdy dung khdng gian mdu.

71 b) Xdc dinh cdc biln cd sau : A : “Tdng sd chd’m trong ba ldn gieo la 6” ; B : “Sd chd’m trong ldn gieo thu” nhd’t bdng tdng cac sd chdm eiia ldn gieo thii hai va thii ba”. 5. Xae sudt cua bien co A. KIEN THUC CAN NHO 1. Nlu A la bie’n cd lidn quan din phep thii chi cd mdt sd hiiu han cac kit qua ddng Icha nang xudt hien thi ti sd P{A) = n{a) dugc ggi la xdc sudt eua n(q) bie’n c6 A. Xae suit ed cac tfnh ehdt sau : a) B(A) > 0, VA ; b) P{n) = 1 ; c) Nlu A va B la hai biln cd xung khic ciing lien quan din phep thit thi P{A u B) = P{A) + P{B) (cdng thiic cdng xdc sudt). Md rgng : Vdi hai biln cd A va B bdt ki ciing lien quan din phep thuf thi P{A VJB) = P{A) + P{B) – P{A n B). Hai bidn cd A vd B duge ggi la doc lap, nlu su xay ra cua mdt trong hai bie’n ed khdng anh hudng din xdc sud’t xay ra cua biln cd kia. Ngudi ta chiing minh dugc ring, A va B ddc ldp khi va chi khi P{A nb) = P{A)P{B). Ngoai ra, A va B ddc ldp «A va B ddc ldp <=> A va B ddc ldp <^ A va B ddc ldp. 69

72 B. VI DU Vi du 1 Ldy ngdu nhien mdt the tii mdt hdp ehiia 20 the dugc danh sd tii 1 ddn 20. Tim xdc sud’t dl the dugc ldy ghi sd a) Chan; b) Chia hit cho 3 ; c) Le va chia hit cho 3. Gidi Khdng gian mdu Q = {1,2,…, 20}. Kf hieu A, B, C la cac biln cd tuong ling vdi cdu a), b), c). Ta cd : a)a={2,4, 6,…,20}, «(A) = 10, n(q) = 20 => B(A) = ^ = 0,5. b) B = {3, 6, 9, 12, 15, 18}, F(B) = ^ = 0,3. c)c={3,9, 15},F(C)=^=0,15., Vi du 2 : Mdt ldp cd 60 sinh vien trong dd 40 sinh vien hgc tiing Anh, 30 sinh vien hgc tiing Phap va 20 sinh vien hgc ca tiing Anh va tiing Phap. Chgn ngdu nhien mdt sinh vien. Tfnh xae sudt cua cdc biln cd sau a) A : “Sinh vien duge ehgn hgc tiing Anh” ; b) B : “Sinh vien duge chgn chi hgc tidng Phap” ; c) C : “Sinh vien dugc chgn hgc ca tidng Anh ldn tiing Phap” ; d) D : “Sinh vien duge ehgn Ichdng hgc tiing Anh va tiing Phdp “. Gidi Rd rang P{A) = = =-, P{B) = ^ = – wap{a n B) = =-. ^ Tit dd P{A u B) = P{A) + P{B) – B(AnB) = = , 70

73 va P{D) = P(A n B) = P(A u B) = 1 -/'(A ub) = = Dd la xdc sudt ehgn dugc sinh vien khdng hgc ca tidng Anh ldn tiing Phdp. Vidu 3 Gieo mdt con siie sic edn dd’i va ddng ehd’t hai ldn. Tfnh xae sud’t sao cho tdng sd chdm trong hai ldn gieo la sd chin. Gidi Kf hidu A : “Ldn ddu xud’t hien mat chin chdm” ; B : “Ldn thii hai xudt hien mat chin chdm “; C : “Tdng sd chdm trong hai ldn gieo la chin”. Ta ed C = AB u A B. Dl thd’y AB va A.B xung khic nen P{C) = P{AB) + p(jjy Vi A va B ddc ldp nen A va B ciing ddc ldp, do dd /'(C) = /'(A)B(B) + />(A)B(B) = i.l + l.l = i C. BAI TAP 5.1. Mdt td ed 7 nam va 3 nii. Chgn ngdu nhien hai ngudi. Tim xdc sud’t sao cho trong hai ngudi dd : a) Ca hai diu la nii ; b) Khdng ed nii nao ; c) It nhd’t mdt ngudi la nii; d) Cd diing mdt ngudi la nii Mdt hdp ehiia 10 qua cdu dd duge danh sd tii 1 ddn 10, 20 qua cdu xanh dugc danh sd tit 1 dl’n 20. Ld’y ngdu nhien mdt qua. Tim xdc sud’t sao cho qua duge chgn : a) Ghi sd chan ; b) Mau dd ; 71

74 c) Mau dd va ghi sd chin ; d) Mau xanh hoac ghi sd le Cd 5 ban nam va 5 ban nir xip ngdi ngdu nhien quanh ban trdn. Tfnh xae sud’t sao cho nam, nii ngdi xen ke nhau Kit qua (b, c) eua viec gieo con siie sic cdn ddi va ddng chdt hai ldn, trong dd b la sd chd’m xud’t hien trong ldn gieo ddu, e la sd chdm xud’t hidn d ldn gieo thii hai, dugc thay vao phuong trinh bdc hai Tinh xdc sud’t dl a) Phuong trinh vd nghiem ; 2 X + bx + c = 0. b) Phuong trinh cd nghiem kep ; c) Phuong trinh cd nghiem Mdt hdp chiia 10 qua cdu duge danh sd tit 1 ddn 10, ddng thdi eae qua tit 1 din 6 dugc son mau dd. Ld’y ngdu nhien mdt qua. Kf hieu A la biln cd : “Qua ld’y ra mau dd”, B la biln ed : “Qua ldy ra ghi sd chan”. Hdi A va B cd ddc ldp khdng? 5.6. Mdt eon sue sic cdn dd’i va ddng chdt dugc gieo hai ldn. Tfnh xdc sudt sao cho a) Tdng sd chd’m ciia hai ldn gieo la 6. b) It nhd’t mdt ldn gieo xud’t hien mat mdt chdm Trong ki kilm tra ehd’t lucmg d hai khdi ldp, mdi Ichdi cd 25% hgc sinh trugt Toan, 15% trugt Lf va 10% trugt ca Toan ldn Lf. Tii mdi khdi chgn ngdu nhien mdt hgc sinh. Tfnh xae sudt sao cho a) Hai hgc sinh dd trugt Toan ; b) Hai hgc sinh dd diu bi trugt mdt mdn nao dd ; c) Hai hgc sinh dd khdng bi trugt mdn nao ; d) Cd ft nhdt mdt trong hai hgc sinh bi trugt ft nhd’t mdt mdn Cho A va B la hai bien cd ddc ldp vdi F(A) = 0,6 ;/'(B) = 0,3. Tfnh a)f(aub); b) F(AuB). 72

75 5.9. Tit mdt ed bai tu lo kho gdm 52 con, ldy ngdu nhidn ldn Iugt cd hoan lai tiing con cho din khi ldn ddu tien ld’y duge con at thi diing. Tfnh xae sud’t sao cho a) Qua trinh ld’y dufng lai d ldn thii hai; b) Qua trinh ld’y ditng lai sau khdng qua hai ldn. Bai tap on chudng II 1. XIp ngdu nhien ba ngudi dan dng, hai ngudi dan ba va mdt diia be vdo ngdi tren 6 cai ghi xip thanh hang ngang. Tfnh xae sud’t sao cho a) Diia be ngdi giiia hai ngudi dan ba ; b) Diia be ngdi giiia hai ngudi dan dng. 2. CQng hdi nhu bai 1 nhung 6 ghd dugc xdp quanh ban trdn. 3. Cd bao nhieu each xdp 7 ngudi vao hai day ghi sao cho day ghd ddu cd 4 ngudi va day sau cd 3 ngudi. 4. Chiing minh ring : a)cr/=^c “‘, (l<m<n); b) C^+ = C+ -l + C+n-l ‘ (1 ^ ‘”‘W)- 5. Tfnh xae sud’t sao cho trong 13 con bai tu lo kho dugc chia ngdu nhien cho ban Binh cd 4 eon pich, 3 eon rd, 3 con ca va 3 con nhep. P( A VJ R) 6. Gia sii A va B la hai biln cd va -^rrr: ^^TT^Z = ^- Chiing minh ring P{A) + P{B), P{AnB). K^ 1 ^ ^r ^>P(A) + B(B)=^-“‘ b)-<a<l 7. Hai hdp chiia eae qua cdu. Hdp thii nhd’t chiia 3 qua dd va 2 qua xanh, hdp thii hai chiia 4 qua dd va 6 qua xanh. Ld’y ngdu nhien tii mdi hdp mdt qua. Tfnh xde sud’t sao cho a) Ca hai qua diu dd ; b) Hai qua ciing mau ; c) Hai qua khae mau. 73

76 Ldl GIAI – HUdNG DAN – DAP SO CHUONG II Theo quy tic nhdn, cd 5 x 4 x 3 = 60 each chgn Ap dung quy tie nhdn, ed 8 X 6 = 48 each chgn a). Cd 5 each chgn chii sd hang don vi la sd chdn. Cd 9 each chgn chii sd’ hang chuc. Theo quy tic nhan, cd 5 x 9 = 45 sd chin gdm hai chii sd. b) Cd 5 each ehgn ehfl sd hang don vi la sd le. Cd 9 each chgn ehfl sd hdng chuc. Vdy cd 5 X 9 = 45 sd le gdm hai ehfl sd (cd thi gidng nhau). e) Cd 5 each chgn ehfl sd hang don vi la sd le ; Cd 8 each ehgn ehfl sd hang chuc ma khae ehfl sd hang don vi. Vdy cd 5 X 8 = 40 sd le gdm hai ehfl sd Ichae nhau. d) Sd cdc sd chin ed hai ehfl sd, tan ciing bing 0 la 9. Dl tao nen sd ehsn khdng chan chuc, ta ehgn ehfl sd hang don vi khdc 0. Cd 4 each chgn. Tilp theo ehgn ehfl sd hang chuc. Cd 8 each chgn. Vdy theo quy tie cdng va quy tie nhdn, ta cd 9 + 8×4 = 41 sd chin gdm hai ehfl sd khdc nhau a) Cd 10 each chgn ngudi dan dng. Khi da ehgn ngudi dan dng rdi, chi’co 1 each chgn ngudi dan ba la vg eua ngudi dan dng dd. Vdy cd 10 each. b) Cd 10 each chgn ngudi dan dng. Khi da ehgn ngudi dan dng rdi, co 9 each chgn ngudi dan ba khdng la vg cua ngudi dan dng dd. Vdy co 10 X 9 = 90 each chgn Phdn tich sd 360 thanh tfch cae thfla sd nguydn td 360 = 2^. 3^. 5. SdrfId ude ciia 360 phai cd dang d = 2′”. 3″. 5” vdi 0 < m < 3, 0 < «< 2, 0 <p < 1. Vdy theo quy tie nhdn, ta cd (3 + 1) (2 + 1) (1 + 1) = 24 udc nguyen duong ciia

77 1.6. Nlu vie’t thi ta hiiu dd la sd ed ba chu sd 345. Vdi quy udc nhu vdy ta If ludn nhu sau : Tfl day hinh thiic ***** ta ldn Iugt thay dd’u * bdi cac ehfl sd. Chfl sd 3 cd 5 each dat, khi da dat sd 3, cd 4 each ddt sd 4, cd 3 each dat sd 5. Khi da dat xong cac sd 3, 4, 5 rdi cdn hai chd nfla. Ta cd 7 each dat mdt trong 7 sd cdn lai vao chd ddu * ddu tien tfnh tfl ben trai va 7 each ddt chfl sd vao ddu * cdn lai. Vdy theo quy tic nhdn, cd = 2940 sd nguyen duong khdng vugt qua ma chfla mdt chfl sd 3, mdt ehfl sd 4 va mdt ehfl sd Cd 5 each di tfl A din B. Din B rdi, cd 4 each trd vl A ma khdng di qua con dudng da di tfl A din B. Vdy cd 5. 4 = 20 each di tfl A den B rdi trd vl A ma khdng dudng nao di hai ldn Cd 9 sd nguyen duong gdm mdt chfl sd ; Cd 9.9 sd nguyen duong gdm hai chfl sd khae nhau ; Cd sd nguyen duong gdm ba chfl sd’ khae nhau. Vdy sd cac sd edn tim la = Theo quy tic nhdn cd = 200 each ehgn Kf hieu A vd B ldn Iugt la tdp cac hgc sinh dang ki mdn bdng da va cdu Idng. Ta cd A u B = 40. Theo quy tic cdng md rdng ta ed n{a nb) = n{a) + n{b) – n{a u B) = = 15. Vdy cd 15 em ddng Icf choi hai mdn thi thao Cd 6! = 720 each bay banh keo Dl xae dinh, cdc ghd duge danh sd thfl tu tfl 1 ddn 10 tfnh tfl trai sang phai. a) Ndu cdc ban nam ngdi d cac ghi ghi sd le thi cac ban nfl ngdi d cac ghi edn lai. Cd 5! cdch xdp ban nam, 5! each xdp ban nfl. Tdt ca cd (5!) each xip. Nlu cac ban nam ngdi d cae ghi ghi sd chin, cdc ban nu ngdi d cae ghd cdn 2 ^ 2 ‘ lai thi ed (5!) cdc lai thi ed (5!) each xdp nam va nfl. Vdy cd tdt ca 2. (5!) each xip nam nfl ngdi xen ke nhau. 75

78 b) Cac ban nam dugc bd tri ngdi d cae ghd tfl k din k + 4, k = 1, 2, 3, 4, 5, Trong mdi trudng hgp cd (5!) each xip nam va nfl. Vdy cd 6.(5!) each xdp ma cae ban nam ngdi canh nhau a) Cd 2. 9 = 18 each xep chd cho An va Binh ngdi canh nhau, 8 ban kia dugc xdp vao 8 chd edn lai. Vdy cd 8! each xdp 8 ban edn lai va do dd cd 18.8! each xdp sao cho An, Binh ngdi canh nhau. b) Cd 10! each xdp chd ngdi cho 10 ban. Tfl dd cd 10! ! = 72.8! each xdp chd cho 10 ban ma An va Binh khdng ngdi canh nhau Dl xdc dinh, ba ban duge danh sd 1, 2, 3. Kf hieu A,- la tdp hgp cdc each cho mugn ma ban thfl / dugc thdy giao cho mugn lai cudn da dgc ldn trudc (/ = 1, 2, 3). Kf hieu Xld tdp hgp cac each cho mugn lai. Theo bai ra edn tfnh «[X(AiUA2uA3)]. Ta ed n(ai u A2 u A3) = n(ai) + n(a2) + n(a3) – «(Ai n A2) – – n(ai n A3) – n{a2 n A3) + n(ai n A2 n A3) = 2! + 2! + 2!-l =4, n{x) = 3=6. Tfl dd n[x (Aj u A2 u A3)] = 6-4 = a) XIp hai ngudi dan bd ngdi canh nhau. Cd 2 each. Sau dd xdp dfla tre ngdi vao gifla. Cd 1 each. XIp 4 ngudi dan dng vao 4 ghd cdn lai. Cd 4! each. Theo quy tie nhdn, ed 2. 4! = 48 each. b) Ddu tien ehgn hai ngudi dan dng. Cd C each. Xdp hai ngudi dd ngdi canh nhau. Cd 2 each. Sau dd xip dfla tre vao giua. Cd 1 each. Xdp 4 ngudi cdn lai vao 4 ghd cdn lai. Cd 4! each. Vdy theo quy tic nhdn, cd C4.2.4! = 288 each a) Trong trudng hgp nay, sd each dat bing sd cdc nghidm (xj, X2, X3) nguyen, khdng dm ciia phucmg tiinh Xj + X2 + X3 = 3. Tfl dd, dap sd cdn tim la c = b) Qua thfl nhd’t cd 3 each dat; Qua thfl hai cd 3 each ddt; Qua thfl ba ed 3 each ddt. Vdy sd each ddt la 3^ = 27.

79 2.7. a) Chgn 7 ngudi tfl 10 ngudi dl lap mdt nhdm, ba ngudi cdn lai vao nhdm khae. Vdy sd each chia la CJQ. b) Tuong tu, kit qua la C^Q.CI 2.8. a) Cd CiQ cdch chgn hai quyin tfl tdng tm k, k = 1, 2, 3, 4. Vdy cd tdt ca (Cjo) each chgn. b) Tuong tu, ed (Cfo)”^ = {C^Q)’^ each chgn Ddu tien coi eae qua la khdc nhau. Do vdy ed 9! each chia. Nhung cac qua cung loai (tao, cam, ehudi) la gid’ng nhau, nen nlu cdc chau ed cung loai qua ddi cho nhau thi vdn ehi la mdt each chia. Vi vdy, sd each chia la 9! = !3!2! Cd thi giai theo each khdc nhu sau : Chgn 4 trong 9 chau dl phat tao. Cd Cg each. Chgn 3 trong 5 chau edn lai dl phat cam. Cd C each. Chudi se phat cho hai chau cdn lai. Vdy cd Cg.Cl =1260 each Kf hieu Xld tdp hgp cac doan dai bilu. A, B ldn Iugt Id tdp eae doan dai bilu gdm toan nam va toan nfl. Theo bai ra, cdn tim «[X (A u B)] = n{x) – n{a u B) = n{x) – n{a) – n{b) Ta ed n{x) = Cg, n{a) = Cg, n{b) = C4. Vdy n[x (A u B)] = C^ – C^ – C^ = Cfl ba dilm dung dugc mdt tam giac. Vi vdy cd thi dung dugc cfg = 120 tam gidc Sd doan nd’i hai dinh cua da giac da cho la C20, sd canh cua da giac la 20. vay sd dudng cheo la C o – 20 = Sd tdp con eua tdp hgp gdm bd’n dilm la Cl + C + Cl+Cl + Ct=6. 11

80 2.14. a) Xdp 6 nam vdo 6 ghd canh nhau. Cd 6! each. Gifla cac ban nam cd 5 khoang trdng cung hai ddu day, nen cd 7 chd cd thi dat ghd cho nfl. Bay gid ehgn 4 trong 7 vi trf dl dat ghi. Cd C^ each. XIp nfl vdo 4 ghd dd. Cd 4! each. Vdy cd 6!. Cj. A = 120.7! each xip ma khdng cd hai ban nfl nao ngdi canh nhau.. b) XIp 6 ghi quanh ban trdn rdi xip nam vao ngdi. Cd 5! each. Gifla hai nam ed khoang trdng. Xdp 4 nfl vao 4 trong 6 khoang trdng dd. Cd Ag each. Theo quy tic nhdn, cd 5! A^ = each Ta cd /- k+l _ pk r’k+l _ f^k f^k+, y^k+1 Tfldd y~ik+l _ /^k, /~<k+ ^k+2 – ^k+l + ^k+l C’ tl=c’ +Cl, Ci,+cl:l = ^«+ ^/i-l Q+l + ^k n n n Ta cd A = 2^ it^ = 2^ C[ + 2^ C^. Kit hgp vdi bdi tdp 2.15, ta dugc k= k= k=2 A-r^.^nr’i – “(” + 1), (” -!)»(» + 1) _»(» + 1)(2«+ 1) ^- ^n+^ -^^n+x – 2 ~ 3 ~ 6 ~’ a) Cdch thii nhdt. Chgn 9 ban trong 50 ban dl lam true nhdt. Cd cfg each. – Khi da ehgn dugc 9 ban rdi, ehgn 4 trong 9 ban dd dl quit sdn. Cd Cg each. Tfl dd, theo quy tic nhdn, cd C^Q. Cg each phdn cdng. 78 Cdch thii hai. Chgn 4 trong 50 ban dl quit sdn, sau dd ehgn 5 trong 46 ban cdn lai dl xen cdy. Vdy cd C^Q. C^g each phdn cdng. Tfl dd ta cd dang thflc cdn chiing minh. b) Ldp ludn tuong tu.

81 2.18. Cd thi chiing minh dl dang ding thflc sau rc ‘=«C ^:; (r=l,2,…,n-l). Vi «la sd nguydn td va r < n, nen n la ude cua C^ Mdi giao dilm cua hai dudng cheo flng vdi mdt va chi mdt tdp hgp gdm 4 dilm tfl tdp hgp 7 dinh ciia da gidc. Vdy cd C^ = 35 giao dilm Cd cfo each chgn 5 chfl sd khdc nhau di ldp sd cdn thiet. Nhung khi da cd 5 chfl sd khae nhau rdi, chi cd mdt each xdp 5 chfl sd dd dl tao nen sd cdn thiit. Vdy cd Cfo = 252 sd Sd hang thfl k+ 1 trong khai triln la Vdy ^5 = CiVi -lf-j = 210.x^ X ^ = 3360x^ Dapsd: ^5 = 3360x^. :, 3.2. (1 + xf = 1 + 6x + I5x^ + 20x^ + 15x’^ + 6x^ + x^. ; a) 1,01^ = (1 + 0,01)^ «1 + 6 X 0, X (0,01)^ = 1,0615. b) Diing may tinh ta nhdn dugc 1,01^ «1, Sd hang thfl it + 1 eua khai triln la h+ = C^Oxf. Vdy sdhang chfla x^ la tj, = Cj9.x^. Theo bai ra ta cd 9.C 2 = – 90 “‘^ <:> C; ^^ = – 10 lu <:> w,. «- = ^. 5 79

82 3.4. Ta cd (1 + axf = 1 + Clax + C^a^x^ + Theo bai ra Cla = 24 CU^ = 252 na = 24 n{n -l)a 252 {n – l)a = 21 ^ U = 8. na = 24 la = Sdhang chfla x^ la {c^.c}{-b)^ + Cla.cl{-b) + Cla^C^x”- Sd hang chfla x^ la [cl.cl{-b) + Ca.Cl)x^. Theo bai ra ta cd [l56^ – %ab + 3a^ = -9 I a = 0 a = 2b b^ = l a = 2 [b = l a = -2 b = -l a) Khdng gian mdu cd dang Q = {SSS, SSN, SNS, NSS, SNN, NSN, NNS, NNN}. b) A = { 555,5iV5,55A^, SNN } ; B= {SSS, NNN) ; C= {SSN, SNS, NSS} ; D= { NNN } = Q{ iviviv } a) Q = {51, 52, 53, 54,55,56, Af 1, N2, N3, NA, N5,N6]. b) A ={52,54,56}; B={iVl,iV3,iV5} ; C={56, iv6}. 80

83 43. a) Q = {(/, j, k)l<i,j,k<6}, gdm edc chinh hgp ehdp 3 eiia 6 (sd ch&i). b) A = {(1, 1, 4), (1, 4, 1), (4, 1, 1), (1, 2, 3), (2, 1, 3), (1, 3, 2), (2, 3, 1), (3, 1,2), (3, 2,1), (2, 2, 2)}; B = {(2, 1, 1), (3, 1, 2), (3, 2, 1), (4, 1, 3), (4, 3, 1), (4, 2, 2), (5, 1, 4), (5,4,1), (5,2,3), (5,3,2), (6,1,5), (6,5,1), (6,2,4), (6,4,2), (6,3,3)} Sd each chgn la C^Q. Kf hidu A^ la bidn cd : ” Trong hai ngudi da chgn, ed dung it nfl”, it = 0, 1,2. a)a tfri,p(a^.tac6m.)= ^ = J 4 4 ; MO c) /'(Ao) = l-p(ao) = l-j^ = ^. ‘-‘ Rd rang trong hdp cd 30 qua vdi 15 qua ghi sd chinr 10 qua mau dd, 5 qua mau dd ghi sd chdn, 25 qua mau xanh hodc ghi sd le. Vdy theo dinh nghia a) P{A) = = -; b) P{B) = = – ; c) P{C) = = – ; d) P{D) = = – ; ^ ^ trong dd A, B, C, D la edc bidn cd tuong flng vdi cac cdu a), b), e), d)..^ 5.3. Sd each xip quanh ban trdn la n{q.) = 9!. Kl hidu A la biln cd : “Nam nfl ngdi xen ke nhau”. Tacd n(a) = 4!5! va P(A) = «0,008. 9!.6. BTBS&GT11-A 81

84 5.4. Khdng gian mdu 1= {{b,c): 1 < &, c < 6}. Kf hieu A, B, C Id cac biln cd cdn tim xae sudt vtng vdi cac cdu a), b), e). Ta ed A = b – Ac. a) A = {{b, c) e n b^ – Ac <0} = {(1, 1), (1, 2),…, (1, 6), (2, 2),… (2, 6), (3, 3), (3, 4), (3. 5), (3, 6). (4, 5), (4, 6)}. n(a) = = 17, P{A) = ^. 36 h) B = {{b, c) e n b’^ – Ac = 0} = {(2,1),,(4,4)}. Tfl dd P{B) = =. 36, 18 c)c=a.v,yp(c)=l-il = Kf hieu A la biln cd : “Qua ldy ra mau dd”; B la biln ed “Qua ldy ra ghi sd chan”. a) Khdng gian mdu Q = {1, 2,…, 10} ; A={1,2,3, 4,5,6}. Tfldd Tidp theo, B(A)= = B = {2, 4, 6, 8, 10} va A n B = {2, 4, 6}. Do dd Tathd’y B(AB) = ^ = = P(A)P(B), vdy A vd B ddc ldp Rd rang fi ={(/,; ): 1 </,; < 6}. Kf hieu Al : “Ldn ddu xud’t hien mdt 1 chain”; Bl : “Ldn thfl hai xud’t hien mat 1 chdm” ; C : “Tdng sd chdm Id 6”; D : “Mat 1 chdm xudt hien ft nhdt mdt ldn”; 82 6.BTBS&GT11-B

85 a) Ta cd C = {(1, 5), (5, 1), (2, 4), (4, 2), (3, 3) },P{C) = ^. it) b) Ta cd Aj, Bi ddc ldp vd D = Aj u Bi nen /'(D) = />(Ai) + /'(Bi)-F(AiBi) – 1 i_l i-ii ~ 6^6 6 6 ~ Kf hieu Al, A2, A3 ldn Iugt la cdc bidn cd : “Hgc sinh dugc chgn tfl khdi / trugt Toan, Lf, Hod” ; Bj, Bj, B3 ldn Iugt la cac biln ed : “Hgc sinh duge chgn tfl khdi // trugt Todn, Lf, Hod”. Rd rang vdi mgi (/, j), cae biln cd A,, vd Bj ddc ldp. a) Cdn tfnh B(AIBI). Ta ed P(AiBi) = P{A)P{B,) = – = ^- b) Xae sudt cdn tfnh la P[{A^ U AJ U A3) n (Bi u B2 u B3)) = /'(AiUA2UA3).B(BiUB2uB3)= i.i = i. c) Ddt A = Al u A2 u A3, B = Bl u B2 u B3. Cdn tfnh P(A n B ). Do A vd B ddc ldp, ta cd B(A nb) = P^A^^B) = [l – P{A)f = (j 4″ d) Cdn tfnh P{A u B). Ta cd P{A u B) = P{A) + P{B) – P{AB) – i i _ i – 1 ~2’^24~4′ 5.8. a) P{A <JB) = P{A) + P{B) – P{AB) = P{A) + P{B) – P{A)P{B) = 0,6 + 0,3-0,18 = 0,72. b) P(A u B) = 1 – B(AB) = 1-0,18 = 0, Kf hidu A^ : “Ldn thfl k ldy dugc con dt”, ^ > 1. Rd rdng Ai, A2 ddc ldp. a) Ta cdn tfnh P(Ai n Aj). Ta ed P(Ai n A2) = B(AI)P(A2) = TI “^

86 b) Theo bdi ra cdn tfnh P(Ai)+P(AinA2) = : ^,^ = +. «0, Bai tap dn chitdng II 1. Khdng gian mdu gdm cac hoan vi cua 6 ngudi. Vdy M(Q) = 6!. Kf hidu A la biln cd : “Dfla be dugc xl’p gifla hai ngudi dan bd”; B la biln cd : “Dfla be dugc xdp gifla hai ngudi dan dng”. a) Di tao nen mdt each xdp ma dfla be dugc xip gifla hai ngudi dan ba, ta tidn hanh nhu sau : – XIp dfla bl ngdi vdo ghi thfl hai din ghd thfl ndm. Cd 4 each. – Lftig vdi mdi each xdp dfla be, ed 2 each xd’p hai ngudi dan ba. – Khi da xip hai ngudi ddn bd vd dfla be, xip ba ngudi ddn dng vdo cdc chd cdn lai. Cd 3! each. Theo quy tie nhdn, ta cd n{a) = 4.2.3! = Tfl dd P{A) = =. 6! 15 b) Dl tao nen mdt each xdp ma dfla be ngdi gifla hai ngudi dan dng.ta tiln hanh nhu sau : – Xdp dfla be vao cae ghd thfl hai ddn thfl nam. Cd 4 each. – Chgn hai trong sd ba ngudi ddn dng. Cd Cf = 3 each. – Xdp hai ngudi dan dng ngdi hai ben dfla be. Cd 2 each. – XIp ba ngudi cdn lai vao ba chd cdn lai. Cd 3! cdch. Theo quy tic nhdn, ta cd n(b) = 4. C.2.3! = Vdy P{B) = i-11 = -. 6! 5 84

87 2. Sd each xdp 6 ngudi quanh bdn trdn Id 5!. Vdy khdng gian mdu cd 5! = 120 phdn tfl. a) Tfnh n{a) : – Cd Teach xdp dfla bl; – Cd 2 cdch xdp hai ngudi ddn bd ngdi hai bdn dfla bl; – Cd 3! cdch xdp ba ngudi ddn dng. V4y n(a) = 2. 3! = 12. Tfldd P(A)= = b) Tuong tu n(b)=l. Cj.2. 3! = Chgn 4 ngudi dl xdp vao 4 ghd d day ddu : cd A^ each. Cdn lai 3 ngudi xdp vao 3 ghd d day sau : cd 3! each. Vdy cd tdt ca A^.3! = 5040 cdch xip. 4. HD. Dung cdng thflc tfnh sd td hgp. 5. Sd each rut ra 13 con bdi la C52. Nhu vdy n(q) = ch- Kl hidu A : “Trong 13 con bdi cd 4 con pfch, 3 con rd, 3 con co va 3 con nhep”. Tacd Vdy nta’i – C^ p3 j^3 _ IJ!. 4!(3!)^ P{A) = ^^l «0, !(3!)lcJ 6. a) Vi P{A nb) = P{A) + P{B) – P{A u B) nen P{A n B) P{A) + P{B) – P{A u B) = l-fl. P{A) + P{B) P{A) + P{B) 85

88 b) Vi P{A u B) = P{A) + P{B) – P{A nb)< P{A) + P{B) _P(AuB) nen a = -r- zrr-^ ^ 1. P{A) + F(B) Mat khdc, 2F(A u B) = ^(A u B) + P(A u B) > P(A) + P{B). P{A^B) ^l Vdv a = ^^ ^ ^. ^ P(A) + B(B) 2 Kit hgp vdi (1), ta cd – < a < Kf hieu A : “Qua ld’y tfl hdp thfl nhd’t mau dd” ; B : “Qua ldy tfl hdp thfl hai mau dd”. Ta thdy A vd B ddc ldp. a) Cdn tfnh P{A n B). Ta cd P{A nb) = P{A) P{B) = -. = 0,24. b) Cdn tfnh xdc sudt eua C = {A n B) J {A (~ B). Do tfnh xung khic va ddc ldp ciia cac bidn cd, ta ed P{C) = P{A) P{B) + P ( A) P (B) c) Cdn tfnh P{C). Ta cd P{C) = 1 – P{C) = 1-0,48 = 0,52. 86

89 e huang IIL DAY SO ^’ CAP SO CONG VA CAP SO NHAN 1. Phaong phap quy nqp toan hoc A. KIEN THQC CAN NHd 1. Dl chiing minh mdt menh dl Id dung vdi mgi n G N* bing phuong phap quy nap todn hgc, ta tiln hdnh hai bude : Bude 1 : Kilm tra ring mdnh dl dung vdi n = 1. Bude 2 : Gia thiit menh dl dung vdi mdt sd tu nhidn bdt ki n = it (^ > 1) (ta ggi Id gia thiit quy nap) vd chiing minh ring nd cung dung vdi n = ^ Trong trudng hgp phai chiing minh mdt menh dl Id dung vdi mgi sd tu nhien «> p (p la sd tu nhidn) thi : 6 bude 1, ta kilm tra menh dl dflng vdi n = p. 6 bude 2, ta gia thiit menh dl dung vdi mdt sd tu nhien bdt ki n = ^ (^ > p) vd chflng minh ring nd cung dflng vdi n = k Phip thfl vdi mdt sd hiiu han sd tu nhidn, tuy khdng phai la ehiing minh, nhung cho phep ta du dodn dugc kdt qua. Kdt qua ndy chi Id gia thiit, vd dl chflng minh ta cd thi dflng phuong phdp quy nap todn hgc. B. VI DU Vidul Chflng minh ring «(3n – 1) = nn + 1) vdi n G N*. (1) Gidi Bude i. Vdi «= I, 1, vd trdi bing ] 1.2 = 2, vd phai bing 1^(1 + 1) = 2. He thflc (1) dflng. 87

90 Buac 2 : Ddt vd trai bing 5. Gia sfl he thflc (1) dflng vdi «= ^ > 1, tflc la : 5^ = ^(3it – 1) = it^(jt + 1) (gia thidt quy nap). Ta phai chiing minh ring (1) cung dung vdi «= it + 1, tfle Id : S ^i = {k+l)k + 2). Thdt vay, tfl gia thidt quy nap ta ed Sk+i =Sk + {k+l) [3{k + 1) – 1] = it^(it + 1) + (it + 1) (3it + 2) = (it + 1) (it^ + 3it + 2) = (it + 1)^ (it + 2). vay he thflc (1) dflng vdi mgi n G N*. Vidu 2 Chflng minh ring n – n chia hdt cho 7 vdi moi n e 1 Ddt A = «- n. Gidi Khi n = 1 thi Al = 0, chia hdt cho 7. Gia sfl da cd A^. = {k – k) : 1 (gia thidt quy nap). Ta phai chflng minh A^+i : 7, tfle la (it + I)” – (it + 1) : 7. Thdt vdy, dp dung cdng thflc Nhi thflc Niu-ton ta ed >i/t+i = (* + 1)^ – (^ + 1) = ^^ + 7itV 21it^ + 35it’* + 35it^ + 21it^ + 7it k – 1 = it^ – it + 7 (;ts 3it^ + 5it’^ + 5it^ + 3k^ + it). (. Theo gia thidt quy nap thi A^ = ^ – k chia hdt cho 7, do dd I Vdy n – n chia hdt cho 7 vdi mgi «G N*. 88

91 Vfdu^ Chflng minh ring J2’+ V >/2 = 2cos ^. V ^. 2 n ddu cdn ” (3) Dat vd trai cua he thflc (3) bing C. Gidi Khi «= 1, vltrdi bkngyfl, vl phai bing 2eos = >/2 ; hi thflc (3) dflng. Gia sfl hi thflc (3) dung vdi n = it > 1, tflc la Ta phai chflng minh Ct = 2cos r-r- <^*+i=2cos^. That vay, tfl gia thidt quy nap ta cd Q+i=>/2 + C, =^2 + 2eos^ k+l =, 4COS’ 7 = 2cos p^ (vi cos,. > 0). y 2*+2 2*+2 2 vay he thflc (3) da dugc chflng minh. Vi du 4 Chflng minh ring vdi mgi sd tu nhidn n > 3 ta cd. 3″>n^ + 4n + 5. (4) Gidi Vdi n = 3, vl trai bing 27, cdn vd phai bang 26. Bdt dang thflc (4) dung. 89

92 Gia sfl bdt dang thflc (4) dung vdi n = it > 3, tfle Id 3* > it^ + 4it + 5. (4′) Ta phai chiing minh nd cung dung vdi n = A: + 1, tflc la 3”^^>(it+l)^ + 4(it+l) + 5. Thdt vdy, nhdn hai vl cua bdt dang thflc (4′) vdi 3 ta cd 3^*^ > 3^ + 12it + 15 = (it + 1)^ + 4(it + 1) it^ + 6it + 5. Vi 2it^ + 6it + 5 > 0 nin 3*^’>(it+l)^ + 4(it+l) + 5. Bdt dang thflc (4) da dugc chflng minh. ViduS Chflng minh bdt dang thflc a”+fe” Ja + b^ (5) trong dd a, b Id cdc sd duong vd n e N*. Gidi Trudc hit nhdn xet ring nlu a = & thi bdt dang thflc (5) xay ra ddu bing (=) vdi mgi n e N*. Gia sxta^b. Ndu n = 1 thi bdt ding thflc (5) dflng vd ddu bang (=) xay ra. Ta se chflng minh vdi «> 2 thi bdt ding thflc (5) dflng, bing phuong phdp quy nap. Thdt vdy : Vdi «= 2 thi (5) cd dang a^ +b^^ fa + b^ hay (a – b) > 0. Rd rang bdt ding thflc nay dung va ddu bing khdng xay ra. Gia sfl bdt ding thflc (5) dflng vdi a T^ b va n = k > 2, tflc Id a* + 6* fa + b * 90

93 Nhdn hai vd cua bdt dang thflc ndy vdi a + 6 > 0, ta cd k, ik / I uk+x.(«a + b, +ft),. > {a + b) a’^^^+a’^b + ab’^+b'”‘-^ {a + bf^^ hay 2 ^ Jk (5) Vi /^^ +fo*””^- {a’b + a6*) = /(a – b) – ba – b) = {a – b) {a” -b”)>0 nen /””^ +fr*””^ > a’^b + a6*. (5″) Tfl (5′) vd (5″) suy ra {a”^ + b”^^) + {a”^’+b”^’) a’^^’+b”^’ *+i^.*+i /^^^*+i a + b hay > 2 I, 2 y nghla Id bdt ding thflc (5) dung vdi n =fe + 1. Vdy, bdt dang thflc da dflge chiing minh. Vidu 6 ^ Vdi gid tri ndo cua sd nguydn duong n, ta cd ^{a + bf^^ 2″””^ >n^ + 3nl (6) Gidi Thuc chdt ddy la bai todn giai bdt phuong trinh tren tdp hgp N*. Tuy nhien, khd cd thi giai nd bing each thdng thudng. Ddt vd trdi bing A vd vl phai bing B. Bing phip thfl vdi n = 1, 2, 3, de dang thdy ring Al < Bl ; A2 < B2 ; A3 < B3. Ddn ddy, ndu vdi kit ludn bd’t phuong trinh (6) vd nghiem thi se Id sai 1dm, vi chi cdn thfl vdi n = 4 ta cd A4 = 32 > 28 = B4. Thfl tilp vdi n = 5, 6 ta cung cd A5 > B5, Ag > Bf,. Din ddy ta ed thi du dodn : Vdi mgi sd tu nhidn rt > 4 ta cd 2 > n + 3n. Ta se chiing minh dilu dfl dodn dd bing phflong phdp quy nap. 2* 91

94 Thdt vdy, gia sfl bd’t ddng thflc (6) dflng vdi n = it > 4, tflc la 2*^* > it^ + 3it. Nhdn hai vd cua (6′) vdi 2 ta dugc 2*””^ > 2it^ + 6it = (it + 1)^ + 3(it + 1) + it^ + it – 4. Vi it > 4 nen it^ + it – 4 > 0, do dd 2^+2 ^ 2(*+i>+i > (it + 1)2 + 3(it + 1); tflc la (6) dung vdi n =fe + 1. Vdy, vdi «> 4 thi 2″^^ > n^ + 3«. (6′) Vidu? Cho tdng 5 = ^ {2r r-l)(2 n + ) a) Tfnh 5i, 52, ^3, 54.. b) Hay du doan quy nap. cdng thflc tfnh 5 va chiing minh bing phuong phap Gidi a) Ta ed 5i = = =1 + J_.A _ 15 _ 3 ‘^ ” 7 “^ 7.9 ~. 63 ” 9 b) Tfl kit qua d cdu a) ta du doan Ta se chflng minh cdng thflc (7) bing phuong phap quy nap. 92

95 Vdi n=l, theo a) thi (7) Id dung. Gia sfl cdng thflc (7) dung vdi «=fe > 1, tflc la ^ ^’=”2fe + r Ta phai chflng minh nd cflng dflng vdi«=fe + 1. Thdt vdy, tfl gia thidt quy nap ta ed ^k+i = ‘^ifc + T^,,, ix iir^i/i., IX. ll – ^k ‘^ [2(fe.+ l)-l][2(fe + l) + l] * (2fe + l)(2fe + 3) fe. 1 _ fe(2fe + 3) + l 2fe + 1 (2fe + l)(2fe + 3) (2fe + l)(2fe + 3) 2fe2 +.3fe + 1 (fe + l)(2fe + 1) fe + 1 (2fe + l)(2fe + 3) (2fe + l)(2fe + 3) 2(fe + l) + l tflc la (7) cung dung vdi n =fe + 1. Vdy cdng thflc (7) da dugc chiing minh. Vidu 8 Chflng minh ring nlu tam gidc ABC vudng tai A, cd sd do cdc canh Id a, 6, c thi vdi mgi sd tu nhien «> 2, ta ed hsit ding thflc b” + c”<a”. (8) Gidi Vdi n = 2 thi theo dinh If Py-ta-go ta cd b^ + c^ = a’. Vdy bd’t ding thflc (8) dung. Gia sfl bd’t dang thflc (8) dung vdi «=fe > 2, tflc la h^^c'<a^. (8″) Khi dd h^^^ + c*””^ = b^.b + c^.c <b^a + c*a = {h^ + c*)a. Sfl dung gia thidt quy nap (8′), ta cd ^fc+i ^ ^*+i ^ ^fc+i^ ^^^ ^^ ^g^ ^jjj^g vdi n =fe + 1. Vdy, bdt ding thflc (8) da dflge chiing minh. Ddu “=” xay ra khi va chi khi n = 2. V 93

96 Nhdn xet 1. Cdc vf du neu tren thi hien rd hai bude cua phuong phdp quy nap : Bude 1 : Kilm tra menh dl (dilu cdn chflng minh) dung vdi n = 1 (hodc vdi n = p, p la sd tu nhien). Bude 2 : Gia sfl menh dl dung vdi n =fe > 1. Sau dd, phai chflng minh ring nd cung dung vdi n =fe + 1. Luu y ring phai thuc hiln ddy du ca hai bude, xong bude 1 mdi lam bude 2. Dae bilt, d bude 2 phai ddt ra duge bai todn, trong dd : Gia thiet (quy nap) la mdnh dl dflng vdi n =feva kdt ludn Id mdnh dl dung vdi n = fe+ 1. Hodn thanh xong hai bude phai neu kit ludn cudi cung. 2. Phuong phap quy nap cd thi dung dl giai cdc loai bdi todn sau : Loai 1 : Chiing minh mdt kit ludn cho san (xem cac Vf du 1,2, 3,4, 5,7, 8). Loai 2 : Tim dilu kien dl mdt kit ludn la dung, bing each sfl dung phip quy nap khdng hoan todn dl du dodn kit qua, sau dd mdi chiing minh bing phuong phdp quy nap. (Xem Vf du 6). C. BAI TAP 1.1. Chiing minh cae ding thflc sau (vdi n e N*) a) (3n – 1) = -^^ – ; b) ” =-(3″”^ – 3) Chflng minh cdc ding thflc sau (vdi n e N*) a)l2 + 3^ (2n-l)2^^^^^^; 94 b) 1^ + 2^ + 3^ n’ = “‘^”; ^^’. 4

97 1.3. Chflng minh ring vdi mgi n G N*, ta cd a) ll”””‘ + 12^” ^ chia hit cho 133 ; 3 2 b) 2n – 3«+ «chia hit cho Chflng minh cdc bdt dang thflc sau (n G N*) a) 2″””^ > 2/1 + 5 ; b)sin2”a + cos^”a< 1 ; 1.5. Vdi gia tri nao cua sd tu nhidn n ta cd a) 2″ > 2/1 + 1 ; h) 2″ > /i^ + 4/1 + 5 ; c)3″>2″ + 7/i? 1.6. Cho tdng a) Tfnh 5i, 52,53,54 ; c ” (4/z – 3)(4/i +1) b) Du dodn cdng thflc tfnh 5 va chflng minh bing phuong phap quy nap Cho n sd thuc a^, a2,…, a thoa man dilu kiln – 1 < a, < 0 vdi / = 1, /I. Chflng minh ring vdi mgi n e N*, ta cd (1 + fli) (1 + 02) – (1 +ci ) > 1 + ai a Chflng minh ring vdi cac sd thflc a^, ^2, 03,…, a {n e N*), ta cd ai + ^ a < ai + 1^ K. 95

98 2. Day so A. KIEN THQC CAN NHd 1. Djnh nghta a) Mdi ham sd u xdc dinh trin tdp sd tu nhidn N* dugc ggi la day sd vd han (ggi tit la day sd’) M : N*. n I > R > u{n) Ddt u{n) =M vd ggi nd Id sdhang tdng qudt cua day sd (M ). b) Mdi hdm sd u xdc dinh tren tdp M = {1,2, 3,…, m}, vdi m G N*, dugc ggi la ddy sdhctu han. 2. Cich cho mot day so 96 a) Ddy sd cho bang cong thdc cua sdhang tdng qudt Khi ddm =f{n), trong dd/ la mdt ham sd xdc dinh tren N*.» Ddy Id cdch khd thdng dung (gid’ng nhu ham sd) va ndu bid’t gia tri cua n (hay cung chfnh la sd thfl tu cua sd hang) thi ta cd thi tfnh ngay dugc u. b) Day sd cho bdng phuang phdp md td Ngudi ta cho mdt mdnh dl md ta each xdc dinh cdc sd hang lidn tidp cua day sd. Tuy nhien, khdng thi tim ngay dugc M vdi /I tuy y. c) Ddy sdcho bdng cong thjjcc truy hdi (hay quy nqp) Cho sd hang thfl nhdt u^ (hodc mdt vai sd hang ddu). Vdi n>2, cho mdt cdng thflc tfnh u ndu bid’t M _I (hodc mdt vdi sd hang dflng ngay trudc nd). Cac cdng thflc cd thi la : [u-i = a [u, = a,w, = b ^ hodc ^ ^ l”«= /K-i) vdi /z > 2 [M = /(M _I, M _2) vdi /2 > 3.

99 3. Day so tang, day so giam Day sd (M ) dugc ggi la tdng ndu M +I > M vdi mgi n e N* ; Day sd (M ) dugc ggi la gidmneu M +I 0 vdi mgi /i G N* thi day sd tang ; – Nlu // < 0 vdi mgi /J G N* thi day sd giam. Phuang phdp 2 NduM > 0 vdi mgi n G N* thi ldp ti sd -^^, rdi so sanh vdi 1. “«Nlu -^il- > 1 vdi mgi n e N* thi day sd tang ; Ndu -^^ < 1 vdi mgi /z G N* thi day sd giam. “«4. Day so bi chan Day sd (M ) dugc ggi la bi chan tren nlu tdn tai sd M sao cho u <M,/ne N*. Day sd (M ) dugc ggi la bi chan dudi ndu tdn tai sd m sao cho M >m,fne N*. Day sd dugc ggi la bi chan, ndu nd vfla bi chdn tren vfla bi chan dudi, tflc la tdn tai hai sd m, M sao cho m<u <M,/ne N*. ^Luu y : Cac da’u “=” neu tren khdng nhat thiet phai xay ra. 7 BT0S&GT11-A 97

100 B. VI DU Vi du 1. Cdc day sd (M ) dugc cho bdi edc cdng thflc : 2″ -1 * a)m =- {nen) 2″+l MI = 1 c) [M +1 = V”n + 1 vdi /2 > 1 ; h)u =^{ne 3″ d) Ml = 1 M xi = *n+ 1 + M N*); vdi/i > 1. Hay vie’t sau sd hang ddu cua mdi day sd’. Khao sdt tfnh tang, giam cua chflng. Tim sd hang tdng qudt cua cdc day c) va d). Gidi c’ xu A^ a) Sau sd hang ddu : >…. Du dodn day sd tdng Ta se chiing minh du doan dd. That vay, xet hieu 2″+! – 1 2″ – 1 ^ = “«+l ~”n = 2«+i +1 2″ + 1 ^ {2″^^ – 1)(2″ + 1) – (2″+^ + 1)(2″ – 1) (2″+^ + 1)(2″ + 1) 22/1+1, 2″+l _ O” _ 1 _ 92/1+1 jn+l _ 9″, 1 (2″+^ + 1)(2″ + 1) 2.2″+^ – 2.2″ 2″+^ (2″+^ + 1)(2″ + 1) (2″+^’+ 1)(2″ + 1) Suy ra M +I > u. Vay day (M ) tang, b) Sau sd hang ddu : I yji S 4A S >/6 ~i» >o BTBS&GT11-B

101 Ta se chiing minh day sd giam. v. u-. TI y/n + l 4n yln + 1-3yln Xet hieu H = M,, -u = ; =, n+l */! +i -n -jn+l Do 3″”^^ > 0, 3y/n = yf9n = yfntsn > y/n + l, nen // < 0. c) Sau sd hang ddu : 1, V2, V3, 4A, V5, V6. Ta se chiing minh day sd tang. Vdi n= 1,T6 rang MI = 1 < v2 = M2- Gia sflkhing dinh dflng vdi n = k> I, tflc la “)fc+l > “A:- Theo cdng thflc cua day sd vd gia thidt quy nap, ta cd «/t+2 = V”f > V”fc + 1 = “fc+1′ tflc la khing dinh dung 6in = k+ 1. Vdy day sd’ tdng. Ban dgc cd thi dl dang chiing minh M = y/n hang quy nap. d) Sdu sd hang ddu : I,,,,, Ta se chiing minh day sd giam bing quy nap. Vdi /? = 1, rd rang MI = 1 > = M2. Gia sfl da cd M^^+I <UJ^ (fe ^ 1), ta phai chiing minh “jt+2 < “*+! Thdt vdy, theo cdng thflc cua day sd va gia thiit quy nap, ta cd “fc Mi. < k+2-1 TTT. + M4+1 ~ 1 i ^ T ~ ” k+l 1 + “it+1 “it (vi 0 < M^+i < M^ nen > ). Vdy day (M ) giam. “/t+i “/t ‘ 1 Ban dgc cd thi chflng minh M = bing quy nap. 99

102 Vi du 2. 2n + (-1)” Cho day sd (M ) vdi u = ^^ ‘ -, n en*. An +.(-1)””^ a) Tinh sdu sd hang ddu cua day sd. Neu nhdn xet vl tfnh don dieu cua day sd. b) Tfnh M2 va M2 +I. Chiing minh 0 < M < vdi mgi n>l. 100 Gidi a) Sdu sd hang ddu cua day sd : ->1> >.. Day sd khdng tang va Ichdng giam. 3.2/2+ (-1)^” 6/1 + 1 b) U2n= 4.2/2 + (-l)^”””^ 8/t – 1 3{2n + 1) + (-1)^^^’ ^ 6″ + ^ “2n+l ~ 4(2n + 1) + (-l)2″+2 “8/2 + 5 Dl thdym > 0. Ta xet hai trudng hgp : 3/1 + 1 n chin An-I’ 3n-l 3/2 + 1 /2 le : M = < ; ” 4/ /2-1 i7«rv ^ 3/2 +1,.. Vdy : 0 <M < voi moi n. ^ ” 4 / 2-1 Vidu 3 Bie’t ndm sd hang ddu cua mdt day sd’ la 3,4,6,9,13,… a) Hay chi ra mdt quy ludt rdi vidt tidp 5 sd hang eua day sd da cho. b) Hay xet Ichoang each gifla hai sd hang lien tilp tfl trdi sang phai. Neu nhdn xet vd vilt tilp nam sd hang theo each dd. c) Ldp cdng thflc truy hdi cua day sd dugc cho theo quy luat neu d cdu b). d) Tim cdng thflc bilu diln M.

103 Gidi a) Cd nhilu quy ludt dl ed mdt day sd ma 5 sd hang ddu nhu da cho. Don gian nhd’t la day sd da cho tudn hodn vdi chu ki bing 5, ta cd day 3,4,6,9,13,3,4,6,9,13,… Tuy nhien, ndu nhan xet tdng = 35dl neu ra quy luat : “Day sd gdm lien tilp edc nhdm 5 sd hang cd tdng bing 35” thi theo dd ta se cd nhilu kit qua khdc nhau, do phuong trinh 5 dn sd Mg + uj +Mg +M9 + Mio = 35 la vd dinh. “Quy ludt” vfla neu da vi pham dinh nghla day sd. b) Ta cd 4-3 = = = = 4 Nhan xet: Khoang each gifla hai sd hang lien tidp tfl trdi sang phai la 4 sd hang ddu eua day sd tu nhien Ichdc 0. Tfl ddy cd thi neu kit ludn : Nam sd hang tren la cdc sd hang cua mdt day sd, trong dd cac khoang each gifla hai sd hang lien tidp tfl trai sang phai ldp thanh day sd 1,2, 3, 4,…,/2vdi/2 G N*. Vdy, nam sd’ hang tilp theo Id 18,24,31,39,48. c) Dl thdy theo quy ludt neu tren thi “n+i ~ “n = ” vdi «G N* vd cdng thflc truy hdi Id d) Dl timm ta vilt «i =3 [M +I = M + /2 vdi /2 > 1. Ml = 3 M2 = “l + 1 M3 = M2 + 2 M4 = M3 + 3 “n-l = “«-2 + «-2 z M = M _i +/

104 Cdng tflng ve n ding thflc tren va rut ggn, ta ed M = (/2-2) + (n – 1), – o {n-l)n 2 n -n + 6 Vdy M = ‘ z ‘ VOI n e N*. Vi du 4 Cho day sd «i=l 1 M ^i = M + 2/2 + 1 vdi /2 > 1. a) Vilt nam sd hang ddu cua day sd ; b) Du doan cdng thflc M va chflng minh bing phuong phap quy nap. Gidi a) Nam sd hang ddu la 1, 4, 9, 16, 25. b) Du doan cdng thflc u = n^ {*) vdi n e N*. Ta se chflng minh cdng thflc vfla neu bing quy nap. Hien nhien vdi n= I, cdng thflc dung. 2 Gia sfl da c6uj^ = k vdi fe > 1. Theo cdng thflc cua day sd va gia thidt quy nap ta cd “A:+1 ~ “A + 2fe + 1 =fe^ + 2fe + 1 = (fe + l)^ tflc la cdng thflc (*) dung vdi /2 =fe Vdy M = /2 vdi mgi n e N*. Vl ffii ^ Vdi gia tri nao day sd giam? cua a thi day sd (” ), vdi M = na–2 la day sd tdng? n + l ‘ 102

105 Gidi {n + l)a + 2 na + 2 Xet hieu H = M +I -U = n+l+l n+l a-2 {n + 2)(/2 + 1) Vi (/2 + 2) (/2 + 1) > 0 nen : Ndu a>2thih Ndu a<2thih Vi du 6 > 0, suy ra day sd (M ) la day sd tang. <0, suy ra day sd (M ) la day sd giam. Cho day sd (M ) vdi M = (1 – af + (1 + af, trong dd 0 < a < 1 va /2 G N*. a) Vilt cdng thflc truy hdi eua day sd; b) Khao sdt tfnh tdng, giam cua day sd. Gidi a) Vdi /2 = 1, ta ed Ml = 1 – a <3 = 2. Vdi /2 > 1 thi M +i = (1 – fl)”^’ + (1 + a)”*’ do dd M +i – M = (1 – af*^ + (1 + af*^ – (1 – af – (1 + af hay Vdy cdng thflc truy hdi la = (1 – fl)” (1 – a – 1) + (1 + af (1 + a – 1) = a[{+af-{l-af] (*) M +i = M + a[(l+a)”-(l-a)”]. rmi=2 [M +I =M + 4(1 + a)” – (1 – a)”] vdi/2 > 1. b) Vi 0 < a < 1 ndn 1 + a > 1 – a > 0, suy ra (1 + a)” > (1 – af hay (1 + a)” – (1 – a)” > 0. Tfl kdt qua (*), ta cd “n+l – Uri = a {{I + af – {I – af] > 0, tflc la day sd da cho la day sd tang. 103

106 Nhdn xet Cach giai cua cdu a) cho ta mdt phuong phdp dl tim cdng thflc truy hdi khi bie’t sd hang tdng quat M, dd la – Tim Ml. – Tinh M +i rdi tim hieu M +I – M (cung cd thi tim tdng M ^.I + M ). Vidu? Cho phuong trinh x^ – X – 1 = 0. Ggi a, p la hai nghiem cua no {a> /3). Chiing minh ring day sd (M ) xae dinh bdi M = 1, n -/3″),din>l, la day Phi-bd-na-xi. Gidi Ta CO a = va p =. Tu cdng thuc u suy ra Mi=-^(«i-;0i) = l ; M2 = ^{a^ – J3^) = ^{a- J3) {a + /3) = -^.l.yfs = I. Dl tfnh M, ta chfl y ring a = a+ I va /3 =/?+!, do dd = -^[a”-2(«+ 1) – y5″-^(y5 + 1)] = -^(a”-^ – P”-^ + a”-2 – fi”-‘^) 104 “Jjl ~^ ‘ 11^ ~ J = “n-l+”«-2- Vdy ta cd cdng thflc truy hdi cua day Phi-bd-na-xi [MI = 1 ;M2 = 1 [M = M _] + M _2 vdi /2 > 3.

107 Vi du 8 Cho day sd xdc dinh bdi cdng thflc a) Tfnh M2, M3, M4 ; Ml = ,.. ^, “«+i =”2″” +2″”^ ^^ ” b) Chiing minh ring M +3 = M vdi mgi n en*. Gidi a)m2 = 2,M3 = 0,M4 = 1. Nlu tinh tilp ta lai cdm5 = 2,Mg = 0,M7 = 1. Nhu vdy day sd tren gdm cac nhdm 3 sd hang (1, 2, 0) dugc nd’i tilp nhau mdt each vd han. b) Ta ehiing minh bing quy nap. Vdi n= I, theo cdu a) thi cdng thflc dflng vim4 = 1 = MI. Gia sfl cdng thflc dung vdi mdt gid tri bd’t Id /2 =fe > 1, tflc la M^+3 = M^. Ta phai chiing minh nd cung dflng vdi /2 =fe + 1, tflc la “/t+4 = “jt+l- Thdt vdy, theo cdng thflc cua day sd thi _ 3 2-5, h+4 – “(/t+3)+l – ~^”A+3 + ^”/t+3 2′ + ^ Sfl dung gia thiit quy nap M^^.3 = Uj^, ta cd H = ~ 2 “* + 2 “* ^ “‘^+1” Vdy cdng thflc da dugc chflng minh. ^ Chu y. Day sd da cho dugc goi la day sd tuan hoan vdi chu ki la 3. Tdng quat, ta c6 djnh nghta sau : Day sd {u ) dugc goi la tuan hoan vdi chu ki p (p e N*), neu u +p = u vdi moi n en*. 105

108 C. BAI TAP 2.1. Viet ndm sd hang ddu va khao sat tfnh tang, giam cua cdc day sd (M ), bilt a) u = 10^”‘”; b)m = 3″ – 7 ; c) “n 2/2 + 1 n^ d) M =. 2″ 2.2. Cho day sd (M ) vdi M = /2 -An+ 3. a) Viet cdng thflc truy hdi cua day sd ; b) Chiing minh day sd bi chan dudi; c) Tinh tdng n sd hang ddu cua day da cho Cho day sd (M ) vdi M = 1 + (/2-1) 2”. a) Viet ndm sd’ hang ddu cua day sd ; b) Tim cdng thflc truy hdi; c) Chiing minh day sd tdng vd bi chdn dudi Day sd (M ) dugc xae dinh bing cdng thflc [M ^I = M + /2 vdi /2 > 1. a) Tim cdng thflc cua sd hang tdng qudt; b) Tinh sd hang thfl 100 cua day sd Cho day sd (M ) xae dinh bdi “i = 5 K+i =u +3n-2. a) Tim cdng thflc eua sd hang tdng qudt; b) Chflng minh day sd tang. 106

109 2.6. Tim cdng thflc sd hang tdng qudt cua cac day sd sau Ml =2 a) 1 b) “n+l = 2 – (/2>1); Ml = 2 [“n+l = “n – 1 ; C) “1 = 2 _ L”n+1 = 3M Day sd {x ) dugc bilu diln tren true sd bdi tdp hgp cdc dilm, kf hieu la A : A = [AQ, Al, A2,…, A,…}. Ggi B la mdt dilm nim ngoai true sd. Ngudi ta dung cac tam giac dinh B va hai dinh cdn lai thude tdp hgp A. Dat M la sd cdc tam gidc duge tao thdnh tit B van + I diim rdi ldp day sd (M ). ^0′ -^1′ -^2′ ‘ ^n a) Tfnh Ml, M2, M3, M4 ; b) Chiing minh ring, “n=cn+i vam +i =u +n + l Cho day sd (M ) thoa man dilu kien : Vdi mgi n e N* thi 0 < M < 1 vd M +i < 1 1 4M Chflng minh day sd da cho la day giam. 3. Cdp so cong ^ <- A. KIEN THUC CAN NHO 1. Djnh nghta (M ) la cd’p sd cdng <^ M +I = M + d, din en*,d la hing sd. (1) Hi qud : Cdng sai d = u^^^ – u. 107

110 2. So hang tong quat M = Ml + (/2 – l)d {n > 2). (2) d=^^f^- (2′) 3. Tinh chat ^^^M,_i+M,^l ^^k>2 (3) hay M;t-i+”it+i =2″^. (3′) 4. long n so hang dau s, = i!(!ip>, e N. (4) h,c,^,4^”.^ (“-‘Ml. ‘,4., ^ Luu y : Khi giai cac beii toan ve cdp sd cong, ta thudng gdp 5 dai ludng. 06 la u^, d, u^, n, Sp. Can phsi bie’t it nhat 3 trong 5 dai luong dd thi se tinh dugc cac dai lugng cdn lai. Vidul. Cho day sd (M ) vdi M = 9-5/2. a) Viet 5 sd hang ddu cua day ; B. VI DU b) Chflng minh day sd (M ) la cdp sd cdng. Chi rouivad; c) Tfnh tdng cua 100 sd hang ddu. 108

111 a) 4,-1,-6,-11,-16. Gidi b) Xet hieu M ^.I – M = 9-5 (/2 + 1) /2 = -5, do dd M +i =M – 5, suy ra day sd (M ) la cdp sd cdng vdi MI = 4 ; J = -5. n2u, +{n-l)d] c) Ap dung cdng thflc S = * ^ -^ (A’). ‘ c 100[2.4 + (100-l)(-5)] ta CO 5ioo = 5 = ^ Chu y : Neu sfl dung cdng thflc (4) ta phai tinh t;. oo- Vi du 2 a) Vie’t sau sd xen gifla hai sd 3 vd 24 dl dugc mdt cdp sd cdng ed tam sd hang. Tfnh tdng cdc sd hang cua cdp sd nay. b) Vie’t ndm sd hang xen gifla hai sd 25 va 1 dl dflge mdt cdp sd cdng ed bay sd hang. Sd hang thfl 50 eua cdp sd nay la bao nhieu? a) Ta cd MI = 3,Mg = 24. Gidi Tfl cdng thflc M = Ml + (/2 – l)d, suy rad = “n-“i n-l 24-3 Tim duoc d = = Vdy 6 sd hang cdn vilt them lad, 9, 12, 15, 18, 21. Tfnh tdng 58 = ^^It^ = 108. b) Ta cd Ml = 25, Uj = l,d = ^-^ = – 4. Vdy 5 sd cdn vilt them Id 21, 17, 13, 9, 5. Tfnh M5o = (-4) =

112 Vidu 3. Cho hai cdp sd cdng {x ): 4, 7, 10, 13, 16,… (3′ ):1,6, 11, 16, 21, Hdi tronj y 100 sdhang ddu tien cua mdi cdp sd ed bao nhieu sd’ hang chung? Gidi Ta ed : A: = 4 + (/2-1)3 = 3/2 + 1 vdi 1 < /z < 100, j^ = 1 + (fe – 1)5 = 5fe – 4 vdi 1 <fe < 100. Dl mdt sd la sd hang chung, ta phai ed 3/2 + 1 = 5fe – 4 <t^ 3/2 = 5(fe – 1) suy ra n ‘5, tflc h’^’5t va k = 3t + I {t e Z). Vi 1 </2 < 100 nen 1 < r < 20. Lftig vdi 20 gid tri cua t, ta tim dugc 20 sd hang chung. Ching han, vdi r = 1 thi /2 = 5,fe = 4, Ichi dd x^=y^= 16. Vidu 4 Tim x trong cae cdp sd cdng 1, 6, 11,… va 1, 4, 7,… bidt: a) A: = 970; b) (jc + 1) + (x + 4) (jc + 28) = 155. Gidi a) Ta cd cdp sd cdng vdi MI = 1, d = 5, 5 = 970 va M = x. Ap dung eong thflc 5^^”[2″i+^»-lM] t,,. ^^^^ [2 + (n-l)5] ^^y sn^-3/ = Giai ra tim dflge n = 20, suy ra JC = M20 = = 96.

113 b) Ta cd cdp sd cdng vdi u^ = x + I, d = 3, u = x + 2S va S = 155. Ap dung cdng thflc M = MI + (/2 – l)c?, ta cd x + 2S=x+l+{n-l)3, suyra /2 = 10. rr^. u. o /2(M,+M ),,^^ 10(2x + 29) Tu cdng thflc S = ^ ^ ” ta cd 155 = -^, tfl dd tim duge x = 1. Vi du 5 Chflng minh ring ba sd duong a, b, c theo thfl tu ldp thanh mdt cdp sd cdng khi va chi khi cae sd r= j=^ j= j=^ j= ^r theo thfl tu ldp y/b +ylc y/c +yla y/a +ylb thanh mdt cdp sd cdng. Gidi Ta se chflng minh bing phep bidn ddi tuong duong. Ba sd p T=r, = =, = = lap thdnh cdp sd’ cdng Ichi vd chi khi y/b +ylc yjc +yla y/a +yjb ‘ yfc + yfa yfb + yfc y/a + yfb yfc + yfa «> yfb-yfa {yfc + yfa){yfb + yfc) y/c -yfb {yfa + yfb){yfc + yfa) <* {yfb- yfa){yfb + yfa) = {yfc – yfb){yfc + yfb) <=> b – a = c – b <:> a, b, c l&p thanh cd’p sd cdng. Vi du 6 Chu vi cua mdt da gidc la 158em, sd do cac canh cua nd ldp thdnh mdt cdp sd cdng vdi cdng sai d = 3cm. Bilt canh ldn nhdt la 44cm, tfnh sd canh cua da gidc dd. «111

114 Gidi Ggi canh nhd nhd’t la MI vd sd’ canh cua da gidc la n. Tacd 44 = Mi+ (/2-l).3hayMi = 47-3/2. Tdng cac canh (tflc chu vi da gidc) la 158, ta cd,5,^n{aa + Al-3n) ^^^ 3/2^-91/ = 0. Giai phuong trinh vdi n e N* ta dugc n = A. Vidu? Cd thi cd mdt tam gidc ma sd do cdc canh vd chu vi cua nd ldp thanh mdt cdp sd cdng duge khdng? Gidi Gia sfl tdn tai mdt tam giac nhu vdy. Ggi sd do cdc canh cua tam gidc la MI, M2,M3 vd chu vi cua nd la M4, ta co Ui = X – d, U2 = X, U2 = X + d, u^ = 3x. Theo tfnh ehd’t cua cdp sd cdng thi MI +M4 =M2 + M3, nhung Ml +M4 = Ax – d, U2 + U2 = 2x + d ndn Ax- d = 2x + d, suy Tax = Tfl dd Ml = 0 (vd If). Vdy khdng thi cd tam gidc thoa man yeu cdu bdi todn. C. BAI TAP 3.1. Trong cac day sd (M ) sau ddy, day sd ndo la cdp sd cdng? a)m = 3/2-1 ; b) M = 2″ + 1 ; 9 9 imi = 3 c) u = {n + I)’ -n’; d) ^ l”n+l = 1 – “n Tfnh sd hang ddu MI vd cdng sai d cua cdp sd cdng (M ), bidt: [M, + 2M, = 0 [UA = 10 a) b) ^ K = 14 ; ‘ [M7 = 19 ; 112

115 . “l +”5-“3 =10. ^. j«7-“3=8 [MI +Mg = 7 ;. IM2.M7 = Cdp sd cdng (M ) cd 56 = 18 va 5io = 110. a) Ldp cdng thflc sd hang tdng qudt M ; b) Tfnh Tfnh sd cae sd hang cua cdp sd cdng (a ), ndu foj + ^ «2n = 126 l«2 + «2n = Hm cdpvsd cdng (M ), bidt [MI +M2 + M3 = 27 TMI “n. = ^ [M^ + M + M = 275 ; [M^ + M u^ = b^ 3.6. Chflng minh ring ndu 5, 52, 53 tuong ling la tdng cua n, 2/2, 3/2 sd hang ddu tien cua mdt cd’p sd cdng thi ^2n – 3(52-5 ) thi ;dng “n (M ), chflng 1 m2 n^ 2m-I 2/ Hm x tfl phuong tiinh a) x = 245, bilt 2, 7, 12,…, x la cdp sd cdng. b) (2x + 1) + (2x + 6) + (2JC + 11) {2x + 96) = 1010, bidt 1, 6, 11,… la cdp sd cdng. B. BTBS&GTII -A 113

116 4. Cdp so nhdn A. KIEN THGC CAN NH6 1. Djnh nghta (M ) la cdp sd nhdn <» M +I = M <7, vdi n G N*. He qud : Cdng bdi q = -^^ “,, 2. So hang tong quat 3. Tmh chat 2 _ “A: – “n = “l^” ^ “*-l”/fc+l hay uk = V”fe-l”it+i (^ – 2)- -*? 4. Tong n sd hang dau tien ^Luu y : Khi giai cac bai toan ve cap sd nhan, ta thudng gsp 5 dai luong. Dd la u ^, q, n, u, Sp.. Can phai biet it nhat 3 trong 5 dai lugng tren thi cd the tfnh duoc cac dai lugng cdn lai. B. VIDU Vidu 1. Cho day sd (M ) vdi M = 2 “”*”. a) Chflng minh day sd (M ) la cd’p sd nhdn. Ndu nhdn xet vl tinh tang, giam cua day sd ; b) Ldp cdng thflc truy hdi cua day sd ; c) Hdi sd 2048 la sd hang thfl mdy cua day sd nay? BTBS&GT11-B

117 Gidi M., 22(“+i)+’ a) Ldp tl sd – ^ = = 4, suy ra M ^I = 4M ; U^ ‘yin+l n hodc bidn ddi Yid= “n+l = 22(“+i)”i = 22″^^+’ = 4.2^”^^ = 4.M. -^±1- = 4 > 1 nen day sd (M ) tdng va la cdp sd nhdn. “n b) Cho/2 = 1, ta cd Ml = 8..Cdng thflc truy hdi la JMi = 8. l”n+l = 4M vdi /2 > 1. c) Ta cd M = 2048 = 2^^ = 2^”^ suy ra 2/2 + 1 = 11, tfl dd /2 = 5. Vdy 2048 la sd hang thfl nam. Vidu 2 a) Viet nam sd xen gifla cae sd 1 va 729 dl dugc mdt cdp sd nhdn cd bay sd hang. Tfnh tdng cac sd hang eua cdp sd nay. b) Vidt sdu sd xen gifla cdc sd -2 vd 256 dl dugc mdt cdp sd nhdn cd tam sd hang. Ndu vilt tilp thi sd hang thfl 15 la bao nhieu? a) Ta ed MI = 1, u-j = 129. Gidi Vi Uj = u^.q nen q^ =^ = 129 = 3^ suy ra ^ = +3. Ml Nam sd cdn vilt la 3, 9, 27, 81, 243 hoac -3, 9, -27, 81, Vdi (? = 3 ta cd 57 = ^’^Ij-}^ = Vdi 9 = -3 ta cd S^ = 547. b) Ta cd Ml = -2, Mg = 256. Mat khae, ^”^ = -^ = -^r- = -128 = {-2) suy raq = -2. Sdu sd edn vidt la 4, -8, 16, -32, 64, Ta cd Mi5 = -2. (-2)’^ =

118 Vidu 3 Day sd (M ) dugc cho nhu sau Ml = 2004,M2 = 2005 “n+l = a) Ldp day (v ) vdi v = M +I – u. Chiing minh day (v ) la cdp sd nhdn. b) Ldp cdng thflc tfnh M theo n. 2M_ + M _, ” ^ “^ vdn /2 > 2. a) Tfl gia thiit suy ra 3″n+l = 2w + ” -l Gidi <=> “n+l-“n= – g K – “n-l) <=> v 3V -i. 1 Vdy (v ) la cdp sd nhdn, ed ^ = va vi = 1. b) Dl tfnh u, ta vilt “n = (“n – “n-l) + (“n-l ” “n-2) (“2 ” “l) + “l = V _i + V _ Vl + Ml = r ly-^ -1 V 3y, = n-l = f n-l 116

119 Vidu 4 Cho cdp sd nhdn a, b, c, d. Chflng minh ring a) (b – cf + {c- af + {d- bf = {a-df-; h){a + b + c){a-b + c) = a^ + b^ + c^. Gidi 2 2 Ta cd i> =ac ; c =bd;ad = be. a) Bidn ddi vl trdi {b – cf + {c- af + {d- bf = b^ + c^ – 2bc + c^-2ac + a^+ (f – 2bd + b^ = (! – 2ad + / = (a – df. i){a + b + c){a-b + c) = {a + cf -if = cf + 2ac + c” -? Vi du.^ Tim cdp sd nhdn (M ) bidt = cl- + c^ + 2b^ -b^ = a^ + b^ + c^. TMI + M2 + M3 + M4 =15 [Mf + M^ + M + M4 = 85. (1) Tathd’yqr^ l.khidd, (1)«”i(g -1) = 15 q-l “‘Y-^)=85 «> Gidi ^^-1 Chia tflng vd cua hai phuong trinh, ta dugc I uf{q’ – If _ {q-lf “W-l)_ ^^-1 = (/-1)V-1) 225 {q + lf{q^ + 1) 45 (<?-1)2(^8-1) 85 /+1 17 <=> 14/ – 17<?^ – 17^^ – 17^ + 14 =

120 ., 2 1 Chia hai ve cua phuong trinh cho q vadatx- q +, ta cd 14/ ^ 17x – 45 = 0 o xi = – ; xo =-. ^ 2 ^ 7 Ta ed hai phuong trinh q + = – (vo nghiem) va 9 + = X- Giai phuong trinh nay tim duoc q = 2, q q 2 Tuong flng cd Ml = 1, Ml = 8. Vdy, ta cd hai cdp sd nhdn 1,2, 4, 8,…(Ml = 1,^ = 2) va 8,4,2, 1,…(MI = 8,<7=^). 2 Vi du 6 Mdt cdp sd cdng va mdt cdp sd nhdn diu la cdc day tdng. Cac sd’ hang thfl nhdt diu bing 3, cdc sd hang thfl hai bing nhau. Ti sd gifla cae sd hang ^ 9 thfl ba cua cdp sd nhdn va cdp sd cdng Id. Tim hai cdp sd d’y. Gidi Nlu ed cdp sd cdng 3, M2′ “3 thi cdp sd’ nhdn la 3, «2′ 9″3 Theo tfnh chdt cua cdc cd’p sd, ta cd M2 = 3 + M, o 9M3 – vamf = 3. ^ hay 3 + M3Y _ 27M3 Biln ddi dua vl phuong trinh 5M3-78M = 0 (M3 > 3). 118

121 Giai ra ta cdm3 = 15. Vdy cac cd’p sd cdn tim la : Cdp sd cdng 3,9, 15. Cdp sd nhdn 3, 9, 27. Vidu 7 Cho bd’n sd nguyen duong, trong dd ba sd ddu ldp thdnh mdt cdp sd cdng, ba sd sau lap thanh mdt cdp sd nhdn. Bilt ring tdng cua sd hang ddu va cudi Id 37, tdng cua hai sd hang gifla la 36, tim bd’n sd dd. Gidi Ggi bdn sd phai tim Id MI, M2, M3, M4, ta cd Cdp sd cdngm2 – d, U2, U2 + d vd cdp sd nhdn M2, M2^, U2q. Theo gia thil’t ta cd Tfl(l) suy ra r2m2 +d = U2{l + q) = 36 (1) u2-d + u^q^ = 37. (2) 36-d 36 _,., “2= ;^ =-^ ^^d=3b- (3) ^ 2 1+^ l+q Tfl (2) suy ra 37 + ^^^.37 + ^ 36 l + q^ l + <? 1 + ^ Thay d d (3) vao he thflc (4) va rut ggn, ta dugc phuong trinh 36^^ – 13q + 35 = 0. Giai ra duoc q = > q = 4 9. Vdy, vdi ^ = – thi 4 119

122 M2= T= 16, M3=16. -=20, M4 = 20. -= A 4 ^ 4 va Ml = 37 – M4 = = 12. Bd’n sd cdn tim la 12, 16, 20, Gid tri (7 = khdng thoa man, vi cac sd MI, M2, M3, M4 khdng nguyen. 9 C. BAI TAP 4.1. Trong cac day sd (M ) sau ddy, day sd ndo la cdp sd nhdn? 2n+l a) M = (-5) c) I Ml = 2 l”n+l – “n ‘ 4.2. Cd’p sd nhdn (M ) cd [MI +M5 =51 [M2 +Mg = 102. / 1 ^n ^Sn+l b)m n = (-l) -3 d)- Ml = 1 2 “n+l =”n+^”n a) lim sd hang ddu vd cdng bdi cua cdp sd nhdn ; b) Hdi tdng cua bao nhieu sd hang ddu tien se bing 3069? c) Sd la sd hang thfl md’y? 4.3. lim sd cac sd hang cua cdp sd nhdn (M ), bilt a)q = 2, u = 96, S = 189; ^s ^ 1 ^ 31 b)”i = 2, “n=g’ ^””^Y’ 4.4. lim sd hang ddu va cdng bdi cua cdp sd nhdn (M >, bidt a) IM5 – Ml = 15 IM4 M2 = 6 ; b) I 222 ~ “4 + “5 = 10 [M3 -M5 +Mg =

123 4.5. Bdn sd ldp thanh mdt cdp sd cdng. Ldn Iugt trfl mdi sd d’y cho 2, 6, 7, 2 ta nhdn dugc mdt cd’p sd nhdn. Tim cac sd dd Vie’t bdn sd xen gifla edc sd 5 va 160 dl dugc mdt cd’p sd nhdn Cho day sd (M ) : Ml =0 2″n ^1 “n+l = T VOI /2 > 1. “+^ M + 4 a) Ldp day sd {x ) vdi x = -r^ -. Chflng minh day sd (x ) la cdp sd nhdn. u^ + J b) Tim cdng thflc tfnh.r, M theo n Ba sd khdc nhau cd tdng bing 114 cd thi coi Id ba sd hang lien tidp eua mdt cdp sd nhdn, hodc coi la cdc sd hang thfl nhdt, thfl tu vd thfl hai muoi lam eua mdt cdp sd cdng. Tim cdc sd dd Cho cdp sd nhdn a, b, c, d. Chiing minh ring ^1 1 o = a^ +b^ +c^ ; a) a be a^ b^ c^j h) {ab + bc + cdf = {a^ +b^+ c^){b^ + c^ + d Mdt cd’p sd cdng vd mdt cdp sd nhdn cd cac sd hang diu duong. Bilt ring cac sd hang thfl nhdt vd thfl hai eua chflng trung nhau. Chflng minh mgi sd hang cua cd’p sd cdng khdng ldn hon sd hang tuong flng cua cd’p sd nhan. Bai tap on chirong III Gidi cdc bdi tap 1,2,3 bdng phuang phdp quy nap. 1. Chflng minh ring a) n^ – n chia hdt cho 5 vdi mgi sd tu nhien n ; b) Tdng cae ldp phuong eua ba sd tu nhien lien tilp chia hit cho

124 2. Chflng minh cac dang thflc sau vdi n e 1 1 I n{n + 3) a) A = -rrrr =nr-r /2(/2 + l)(/2 + 2) 4(/2 + l)(/2 + 2) ‘ n{n + 1) n{n + l){n + 2) b)b =l Chiing minh cdc bd’t dang thflc,n-l a) 3″”‘ > /2(/2 + 2) vdi /2 > 4 ; b) 2″”^ > 3/2-1 vdi /2 > Cho day sd (M ) : [MI = 1, t<2 = 2 l”n+l = 2Mn – “n-l + 1 vdi /2 > 2. a) Viet nam sd hang ddu cua day sd ; b) Ldp day sd (v ) vdi v = M +I – M. Chiing minh day sd (v ) la cdp sd cdng ; e) Tim cdng thflc tfnh M theo n. 5. Cho day sd (M ) : 1 “1=3 {n + 1)M_,. ^. “n+l= 3 ” V01/2>1. a) Vidt ndm sd hang ddu cua day sd. b) Ldp day sd (v ) vdi v = -^. Chiing minh day sd (v ) la cdp sd nhdn. c) lim cdng thflc tfnh M theo n. 6. Ba sd cd tdng la 217 cd thi coi la cdc sd hang lien tilp cua mdt cdp sd nhdn, hoac la cdc sd hang thfl 2, thfl 9 vd thfl 44 cua mdt cdp sd cdng. Hdi phai ldy bao nhieu sd hang ddu cua cdp sd cdng dl tdng cua chung Id 820? 122

125 7. Mdt cdp sd cdng vd mdt cdp sd nhdn cd sd hang thfl nhd’t bing 5, sd hang thfl hai cua cdp sd cdng ldn hon sd hang thfl hai cua cd’p sd nhdn la 10, cdn cdc sd hang thfl ba bing nhau. Tim cdc cd’p sd dy. 8. Chflng minh ring nlu ba sd ldp thanh mdt cdp sd nhdn, ddng thdi ldp thanh cdp sd cdng thi ba sd dy bing nhau. 9. Cho cdp sd nhan (M ) cd cdng bdi la q va sd cdc sd hang la chan. Ggi S^ la tdng cdc sd hang cd chi sd chin va 5/ la tdng cae sd hang cd chi sd le. Chflng minh ring q= 5 -^. 10. Cd thi cd mdt tam gidc vudng ma sd do cac canh cua nd ldp thdnh mdt cdp sd’cdng khdng? 11. Tfnh tdng :, /2-1 a) ; 2 2^ 2^ 2″ b) 1^ – 2^ + 3^ – 4^ (-1)””^. /2I 12. Hm m dl phuong trinh / – (3m + 5)x^ + (m + 1)^ = 0 cd bdn nghiem ldp thdnh cdp sd cdng. Bdi tap trdc nghiem (13-19) 13. Trong cdc day sd (M ) sau ddy, hay ehgn day sd giam : (A) u = sin/2 ; (B) M = n’+l n (C) u =yf^- yf^t^l ; (D) M = (-1)”(2″ + 1). 14. Trong cdc day sd (M ) sau ddy, hay chgn day sd bi chdn : (A) M = Vn^+1 ; (B) M =n + -; (C)M =2″+1; (D)M =-^- 15. Cho cdp sd nhdn (M ), bilt MI = 3, M2 = -6. Hay chgn kit qua dflng : (A) Mg = -24 ; (B)M5 = 48 ; (C)M5 = -48 ; (D)M5 =

126 16. Trong cdc day sd (M ) sau ddy, day sd nao la cdp sd cdng? f”i = 1 [MI = 2 (A) 3 (B) ‘ [M +1=M^-1; K+l=«n+«; TM, = -I f”i = 3 (c) r, (D> 9.1 l”n+l – “n = 2 ; l«n+l = 2M Cho cdp sd cdng 6, x, – 2, y. Kdt qua ndo sau ddy Id dflng? {A) x = 2,y = 5 ; {B) x = A, y = 6 ; {C) x = 2,y = -6; (D) x = A, y = Cho cdp sd nhdn Hay chgn kit qua dung : -2, X, – 18, y. (A)x = 6,>’ = -54; (B)JC = -10,>’ = -26; {C)x = -6,y = -54 ; {D)x = -6,y = Cho day sd (M ) vdi u = 3”. Hay chgn hi thflc dflng : /» X “l + “9 /ON “2”4 (A) ^-y^ = M5 ; (B) ^ = M3 ; (C) 1 + Ml + M Mioo = – ^ ; (D) M1M2… Mioo = M5050- Ldl GIAI – HUdNG DAN – DAP SO CHUONG III a) Ddt vl trai bing 5. Kilm tra vdi /2 = 1, hd thflc dung. 124 fe(3fe +1) Gia sfl da cd 5^^ = -^^5 – (fe + l)(3fe + 4) T^,.,^ vdi fe > 1. Ta phai chflng minh

127 c c o/,,x, fe(3fe + l),, H 3fe2+fe + 6fe + 4 5jt+i = 5^ + 3(fe + 1) – 1 = 2 + 3^ + 2 = 2 3fe2 + 7fe + 4 (fe + l)(3fe + 4),^ = ^ =.!^ ‘-^ (dpcm). b) Ddt vd trdi bing P, lam tuong tu nhu cdu a) a) Dat vd trdi bing 5. Vdi /2 = 1, vd trdi chi cd mdt sd hang bing 1, vd phai bing ‘ = 1. fe(4fe^ -1) Gia sfl da cd 5^^ = :r vdife > 1. Ta phai chflng minh Thdt vdy, tacd _(fe + l)[4(fe + l)2-l] 5^+1= 3 Sk+i =Sk+ [2(fe+l) -lf = Sk + (2fe + 1)^ _fe(4fe^- 1)… ^,2_(2^ + l)[^(2^-l) + 3(2fe + l)] 5 *” {^k +1) ^ _ (2fe + l)(2fe^ + 5fe + 3) ^ (fe + l)(2fe + 3)(2fe + 1) ^ (^ + 1)[4(^ + 1)^ ” IJ ~ b) Ddt vd trdi bing A. De thdy vdi /2 = 1, he thflc dflng. Gia sfl da cd Afc = ^^ ^. (fe > 1). Ta cd Afc+i =A^fc + (fe + 1)^ =-^-^^^^ + (fe + 1)^ (fe + lf{k^ + 4fe + 4) (fe + l)^(fe + 2f 1.3. a) Dat A = 11″^^ + 12^””*. Dl thd’y Aj = 133, chia hit cho 133. Gia sfl da cd A^ = 11*^^ + 12^*”^ chia hdt cho

128 TacdA^+1 = ir-^”+12″””^ = 11.11″”‘+ 12^’^- 12^ = *^^ + 122^-1( )= 11.A^ ^*^”^ ViA;t : 133nenA^+i : 133. b) HD : Dat B = 2/2^ – 3n^ + n, tfnh Bi. Gia sfl da cd B^ = 2fe^ – 3fe^ +fechia hdt cho 6. Ta phai chiing minh B^+i = 2(fe + 1)^ – 3(fe + 1)^ +fechia hit cho a) Vdi/2 = 1 thi 2^^^ = 8 > 7 = Gia sfl bd’t ding thflc dung vdi /2 =fe > 1, tflc la 2^^^ > 2fe + 5. (1) Ta phai chiing minh nd cung dung vdi n = k + 1, tflc la 2 > 2(fe + 1) + 5 hay 2^”S2fe + 7. (2) Thdt vdy, nhdn hai vl cua (1) vdi 2, ta dugc 2^^^ > 4fe + 10 = 2fe fe + 3. Vi 2fe + 3 > 0 nen 2^^^ > 2fe + 7 (dpcm). 2 2 ‘ b) Vdi /2 = 1 thi sin a + cos or = 1, bdt dang thflc dung. Gia sfl da cd sin a + cos a < 1 vdife > 1, ta phai chiing minh sin^^”^^ a + cos^*”^^ a <l. Thdt vdy, ta cd 2i-+2 2k+2. 2k 2 7k 9 sm^”^^ a + cos a = sm a. sin” a + cos^ a. cos a _ sin a + cos a<l Ddy thuc chdt la bai toan giai bdt phuong trinh tren N*. Phuang phdp. Cd thi dung phep thfl, sau dd du dodn kit qua va chflng minh. 126 a) Dung phep thfl vdi n = 1, 2, 3, 4 ta du doan : Vdi /2 > 3 thi bd’t dang thflc dung. Ta se ehiing minh dilu dd bing quy nap. Vdi /2 = 3, hiln nhidn da ed kdt qua dung, vi 2^ = 8 > = 7. Gia sfl bd’t dang thflc dflng vdi n = k, tflc la 2* > 2fe + 1,

129 ta se chflng minh bd’t ding thflc cung dflng v6in = k+ 1, tflc la 2^”‘^>2fe + 3. (2) That vay, nhan hai vl cua (1) vdi 2, ta dugc 2^””^ > 4fe + 2 = 2fe fe – 1 > 2fe + 3. b) HD : Dung phep thfl. Vdi /2 tfl 1 din 6, bdt ding thflc diu khdng dflng. Tuy nhiln khdng thi vdi vdng kit ludn bd’t phuong trinh vd nghiem. Nlu thfl tidp ta thdy ring bd’t phuong trinh dung Ichi /2 = 7. Ta cd thi lam tidp dl di tdi du dodn : Vdi /2 > 7 thi bd’t phuong trinh dugc nghiem dflng. Sau dd chflng minh tuong tu nhu cdu a). c) Ldm tuong tu nhu cdu a) va cdu b). bs : /2 > a)Tfnh5i = i, 52=, 53=^, 54=^ b) Vidtlai 5 = – =, 52 ==r = -rw^^ S. = T^f r’ ^4 = -rr-^ ^ ^ Ta cd thi du dodn 5 =- ” “4/2 + 1 Hgc sinh tir chflng minh cdng thflc tren Vdi /2 = 1, bdt ding thflc dflng. Gia sfl bdt ding thflc dflng vdi /2 =fe > 1, tflc la (1 + ai) (1 + 02) – (1 + fl;t) > 1 + «! + ^ «/t- (1) Nhdn hai vl cua (1) vdi 1 + a^t+i ta dugc (1 + ai) (1 + 02)… (1 +flfc) (1 + a^t+i) > (1 + fli + ^ «/t) (1 + «A:+l) = = l+ai+a ak + a^+i + ai^^t+l + «2«<:+i ^k-^^k+i- Vi aiofc+i + a2«<:+i ^k ^k+i > 0 nen (1 + fli) (1 + 02) – (1 + a/t) (1 + ^k+i) ^ 1 + fli + ^ «/t + Ok+i, J nghla la bd’t dang thflc cung dflng vdi /2 =fe

130 1.8. Vdi /2 = 1 thi Iflil = Iflil. Vdi /2 = 2 thi ai+a2<ai + 102!. Ddy la bdt dang thflc kha quen thude vd dd’u bing xay ra khi <3i, 02 cung ddu. Gia sfl bd’t dang thflc dung vdi /2 =fe > 2. Ddt oi a^ = A, ta ed lai < lail + 1^ aku (1) ma la + a^t+il < lai + la^+il < lai) + Ia2l + + lo/t’ + ‘”jt+i’ nen loi + ^2 + + ^/t + ^it+i’ – 1^1′ + ‘^2’ + + ‘^jfc’ + ‘^it+i” tflc la bd’t dang thflc dung vdi n = k a) > r-. -, -, – Du dodn day (M ) giam ^ 10^ 10^ 10^ y jql-2(n+l) J Dl chflng minh, ta xet ti sd -S±L = ; = – < 1. Vdy day sd giam. “n 10^”^” 10^ b) -A, 2, 20, 74, 236. Xlt dd’u cua hieu M +I – M e) 3,.. > Lam tuong tu cdu b) ‘ ‘, 3 9>/2 27V3 81^/4 243V5 ^..^ ^, ^,,.,, d)» >..- Phdn tiep theo co the lam tuong tu cdu a) ^ B. > ^ Chu y. Qua bdn bai tap tren, hgc sinh cd the rut ra nhan xet ve tfnh hdp If ccia viec xet hieu Un+i – a,, hay xet tl sd -^^, khi khao sat tfnh don dieu ccia d y sd a) Ta ed MI = Xet hieu M +i – M = (/2 + 1)^ – 4(/2 + 1) /2^ + 4/2-3 = 2/2-3. f “1 =0 Vdy cdng thflc truy hdi la “n+l =M + 2/2-3 vdi /2 > 1.

131 b) M = /2-4/2 + 3 = (/2-2)^ – 1 > -1. Vdy day sd (M ) bi chan dudi nhung khdng bi chan tren (Hgc sinh tu giai thfch dilu nay). c) 5 = 1 + 2^ + 3^ /2^ – 4 ( n) + 3n _ n{n + l){2n + 1) n{n + 1) 6 ^- 2 ‘^ _ n{n + l){2n + 1) – I2n{n + 1) + 18/2 _ n{n + l){2n – 11) + 18/2 6 ~ 6 ” 2.3. b) HD : Tim hieu M +I – M. K = i DS: <^ [M +I = M + (/2 + 1)2″ vdi /2 > 1. c) HD : Xet dd’u M +I – M a) Tfl M +i – u = n ta ed Ml = 1 “2 ~ “l = 1 o3 M3 – M2 = 2 “n-l – “n-2 = (” – -2)^ “n- “n-l =(«-l)^ Cdng tflng vl n ding thflc tren va rflt ggn, ta dugc M = 1 + 1^ + 2^ (/2-1)^ Sfl dung kit qua bdi tdp 1.2 b) – 1 ta cd i^ + 2^ (/2-i)^ = ^^i:^. 4, n^{n – if Vdy M = 1 + b) Mioo = BTBS&GT11-A 129

132 2.5. a) Tuong tu bai 2.4. DS b) Tuong tubal 2.1. M = 5 + (/2-l)(3/2-4) 2.6. a) HD : Vidt vai sd hang ddu dl du doan cdng thflc rdi chiing minh. DS:M =^^. n h) HD : Lam tuong tu bai 2.1. DS : u = 3 – n. c) Vdi chfl y ring ^n+l _ 3. Ldp tfch cua n-lti sd “n-l “3 “2 _ on-1 ^n-1 “n-2 M2 Ml Rflt ggn vd trai dugc -2- = 3 n-l = 3″ a) suy ra Ml = 1 M ” 2 =-3″-^ “2 = 3 M3 = 6 M4 = 10 b) Sd cdc tam giac u tao thanh tfl B vd n + 1 dilm chfnh la sd td hgp ehdp 2 eua n+l phdn tfl : r’2 “n- ‘-n+l- Ap dung cdng thflc C k _ f-‘k, /-^k-x n ” ^n-1 + *-n-l n-l ‘^n ta cd C +2 = C +i + C +i hay “n+l = u^ + n+ I BTBS&GT11-B

133 2.8. Vi 0 < M < 1 vdi mgi n nen 1 – M +I > 0. Ap dung bd’t dang thflc Cd-si ta cd Mat khae, tfl gia thidt So sdnh (1) vd (2) ta cd «+l(l-m +i)<- (1) 1 “n+l < 1 – XT ^”y ”^ tm “n+l “n < “n – ^ hay – < U (1 – M +i). (2) “n+l (1 – “n+l) < “n (1 ” “n+l) hay M +i < M a) M +i – M = 3(/2 + 1) – 1-3/2 + 1 = 3. Vi M +i = M + 3 nen day sd (M ) la cdp sd cdng vdi MI = 2, li = 3. b) M +i – M = 2″ + 1-2″ – 1 = 2”. Vi 2″ khdng la hing sd nen day sd (M ) khdng phai la cdp sd cdng. c) Ta cd M = 2/ Vi M +i – M = 2(/2+l) + 1-2/2-1 = 2, nen day da cho la cdp sd cdng vdi Ml = 3 ; <i = 2. d) Dl chiing td {u ) Ichdng phai la cd’p sd cdng, ta chi cdn chi ra, ching han “3 ~ “2 ‘^ “2 ~ “i la <lu a) Ml = 8, rf = -3. b) Ml = 1, rf = 3. c) Ml = 36, d = -13. d) Ui = 3,d = 2 hoac MI = -17, cf = a) DS : M = An {u^ = -l,d = A). b) 520 =

134 3.4. DS : /2 = 6. ) Mi +M2 +M3 = uf +u^ +u^ = 275. (1) (2) Ap dung cdng thflc 5 = ^r tim dugc MI +M3 = 18, suy ram2 = 9 (3) ThayM2 = 9 vao (1) va (2) ta dugc he imi = 18 ul +u^ = 194. Tfl day tim duge MI = 5,M3 = 13. vay ta ed cdp sd cdng 5, 9, 13. b) Ta ed b^ = M? + (MI + d)^ [u^ + {n – l)df = nui + 2MI (/2-1)] + d\^ + 2^ {n -if] 2,,, ^ n{n-l){2n-)d^ = /2Mi + n{n – )uid + -^ n{n – l)d Mat khdc, a = nu^ + (1) (2) Tfl (2) tim dugc MI, thay MI vao (1) dl tim d. Kdt qua d = ± ll2{nb^ – cf) n^{n^ – I) Ml = 1 a nijr-l), a 3.6. HD : Sfl dung cac cdng thflc M = “”^1! “”-^ vas (Ml + u )n di chiing minh Sn ^ S^ _ S2 n 3n n Tfl (1) suy ra he thflc cdn chflng minh. (1) 132

135 3.7. Ta cd 2MI + (m – l)d Sm = 2 “* ‘ Theo gia thidt _ 2MI +{n- l)d 5 = n. 5^ _ [2MI + (m – l)^]m _ nf S [2MI +{n-l)d]n rf Suy ra (2MI – d){m – /2) = 0 (vdi m ^ /2). Tfl dd Vdy ^ “1 = 2 “m _ «i + (/n – 1) J _ 2″ ^ ^^ ~ ^)^ _ 2m – 1 M M, + (/2 – l)rf 5n-n- Tfl dd tim dflge x = MIO = = 47. b) Xet cd’p sd cdng 1, 6, 11,…, 96. Ta ed 96 = 1 + {n- l)5=>n = = 0. Suyra 520= = 20(1 + 96) = 970 va Tfldd J:= 1. 2x = a) Cd thi ldp ti sd -^^. Cdp sd nhdn cd MI = -125, <? =

136 b) Cap sd nhdn cd MI = -81, <7 = -27. c) Day sd (M ) khdng phai la cdp sd nhdn. 7 d) Cdp sd nhdn vdi MI = 1, ^ = DS : a) Ml = 3, 9 = 2. b) /2 = 10. C)/2= DS : a)n = 6. h)n = a) Ta cd hd “i^ Ml =15 Mi^ – Mi^ = 6 hay Mi(/-l) = 15 “i(<? -^) = 6. (1) 15 Do (1) nen <7?t +1, suy ra q’-l _ci’+l 6 q{q^-l) ^ Bidn ddi vl phuong tnnh Giai ra duoc q = 2vaq = 2 Ne’u^ = 2 thi Ml = 1. Ndu o = thi Ml = ^ b)ds: Ml = 1,9 = 2. 2^^ – 5^ + 2 = HD : Ggi 4 sd cdn tim la x, y, z, t, ta ed : Cdp sd cdng x,y,z,t Cdp sd nhdn x-2,y -6,z -1,t-2. Ta cd he X + z = 2y y + t = 2z {y-6f ={x-2){z-l) [{z – If ={y- 6){t – 2). DS:x = 5,y=l2,z= 19,? =

137 4.6. DS : 10, 20, 40, Tfl gia thidt cd Ldp ti sd “n+l (“n + 4) = 2M + 3 hay M +I. M + 4M +I = 2M + 3. ^n+l _ “n+l – 1. “n+3 _ “n+l”n + 3Mn+i – «- 3 JC M, ‘*n+l +3 ” – 1 Wn+l”n ‘ “n+l + 3M – 3 (1) (2) Tfl (1) suy ra u^^-^.u = 2M + 3-4M +I, thay vao (2) ta dugc 2M + 3-4M ^, + 3M ^, – M ^n+l _ ^M -r J – ^u.n+ “^ -“*n+l “«-‘ _ “n M – M «n+l 1 2M + 3-4M +I – M ^i + 3M – 3 5(M – M +I) 5 Vdy x ^.i = x, ta cd cd’p sd nhdn (jc ) vdi q= vaxi= Ta cd /ja”-! ^J Tfl dd tim duoc M = ^ ^ I- x 1 + (1 A”-l v5y / J A«-l v5; _ f 1 A«-l + 1 ^J /^la”-l + 1 v5y 4.8. HD : Lam tuong tu Vf du 7. DS : Ba sd phai tim la 2, 14, a) Bie’n ddi vl trdi ^ b c a b c + a e a b a^ b^ c’ a b + acc^ {b^f a^ac a + b. + 3, 1,3, 3 = a + b + c. h) HD : Ap dung bdt dang thflc Cd-si cho cac sd a, b, c va b, e, d HD : Chflng td hai day sd diu la day tdng rdi chflng minh bing phuong phdp quy nap. 135

138 Bai tap on chirong III 1. a)hd:xemvfdu2, 1. b) HD : Ddt A = n^ + {n+ if + {n + 2)^ dl thd’y Ai ‘: 9. Gia sfl da ed A^^ : 9 vdi fe > 1. Ta phai chiing minh A^^.i : 9. Tfnh A^+i = A^ + 9fe^ + 27fe a) HD : Kilm tra vdi n=l, sau dd bilu diln ^k+x b) HD : Kilm tra vdi n = 1. ^., v^~ ‘R fe(fe + l)(fe + 2) Gia su da co Bt = * 2 Ta cdn chflng minh 1 ^fc + (^ + 1)(^ + 2)(fe + 3) (fe + l)(fe + 2)(fe + 3) ^, ^ ^. ^ ^, r. o (-t + 1)^^ + 2) 3. a) Vdi /2 = 4 thi 3^^”^ = 27 > 4(4 + 2) = 24. Gia sfl da cd 3^^^ >fe(fe + 2) vdi fe > 4. (1) Nhdn hai vd eua (1) vdi 3, ta ed 3 3<:-l ^ 3(/:+l)-l > 3^ (^ + 2) = (fe + 1) [(fe + 1) + 2] + 2fe^ + 2fe – 3. Do 2fe^ + 2fe – 3 > 0 nen 3^*^^^”‘ > (fe + 1) [(fe + 1) + 2], chflng td bdt ding thflc dflng vdi /2 =fe + 1. b) Giai tuong tu cdu a). a) Ndm sd hang ddu la 1, 2, 4, 7, 11. b) Tfl cdng thflc xdc dinh day sd ta cd 136 “n+l = 2M – M _i + 1 hay M +I -U = U – M _I +1. (1)

139 Vi v = M +i – M nen tfl (1), ta cd v = v _i + 1 vdi /2 > 2. (2) Vdy (v ) Id cdp sd cdng vdi Vi = M2 – MI = 1, cdng said= I. c) Dl tfnh u, ta vidt Vi = l V2 = “3 – “2 V3 = M4 – M3 V _2 = U _i – M _2 V _i = M – /2 _i Cdng tiing v6n-l hd thflc tren vd rut ggn, ta dugc Vl + V V _i = 1 – M2 + M = M = M – 1,., n{n -1) suy ra M = 1 + Vl + V v _i = a) Nam sd hang ddu la, -) 9 J J b) Ldp ti sd Vfl _ “n+l «_ “n+l /2 + 1 M M n + l Theo cdng thflc dinh nghia ta cd -^±^ = M 3/2 V Tfl (1) va (2) suy ra -^±^ = – hay v +i = -v. (1) (2) Vdy, day sd (v ) la cdp sd nhdn, cd Vi =, 9 = c) Dl tfnh M, ta vidt tfch cua n-lti sd bing V V “-n-l ^3 ^2 ^n *- V -l V _2 V2 Vl r I A”~^ v3y 137

140 hay n _ Kh Vdy M = 3″, suy ra v 3 J_ 3″ 6. HD : Ggi sd hang thfl hai cua cdp sd cdng la M2, ta cd M9 =M2 + Id, M44 =M2 + A2d. Sfl dung tfnh chdt cua cd’p sd nhdnm2.u^=i^ va tdng cac sd Id 217, ta co mdt he phuong trinh dl timm2 va d. DS : /2 = DS : Cdp sd cdng 5, 25, 45. Cdp sd nhdn 5, 15, HD : Ggi 3 s6 dola a – d, a, a + d rdi dp dung tfnh chdt cua cdp sd cdng va cdp sd nhdn. 9. Ggi sd hang thfl nhd’t cua cdp sd nhdn la MI vd cdng bdi la q. Ta cd 5/ = Ml + u^q + u^q ^. = u^q + u^q + u^q +… Nhdn hai vl cua (1) vdi q ta cd Vdy 3 5 qsi – u^q + u^q + u^q +… = 5^ Sc (1) (2) 10. Ggi sd do ba canh cua tam gidc vudng lax- d,x,x + d Theo gia thidt taco {x + d) = {x – d) + x. Tfl (1) tim dugc x = 0,x = Ad. Nhu vdy cd thi cd tam gidc vudng thoa man ddu bai, cdc canh cua nd la 3d, Ad, 5d. Ddc biet, ndu J = 1 thi tam gidc vudng cd cdc canh la 3, 4, 5 (tam giac Ai Cdp). (1)

141 11. a) HD : Ddt tdng la 5 va tfnh 25. DS:5 =3-^^^±^. 2″ ‘ b) HD:n^-{n+ if = -2/2-1. Ta cd 1^ – 2^ = -3 ; 3^ – 4^ = -7 ;… Ta cd Ml = -3, J = -4 vd tfnh 5 trong tflng trudng hgp n chan, le. 12. Ddt X = y, ta cd phuong trinh /-(3m+ 5)j + (m+1)^ = 0. (1) Dl phuong trinh cd 4 nghiem thi phuong trinh (1) phai cd 2 nghiem duong Ji. ^2 CVi < y2)- Bd’n nghiem dd la -yfy^, – ^, yjy[, ^. Dilu kien dl 4 nghiem tren ldp thanh cdp sd cdng la ^Jy2 – yjyi = 2yjy^ hay ^2 = 9^1, kdt hgp vdi dinh If Vi-lt tim dugc m = 5 vd m = 25 Dap an Bai tap trao nghiem * 13. (C); 14. (D); 15. (B) ; 16. (C); 17. (C); 18. (C); 19. (D). 139

142 %ieangiv. GI6IHAN 1. Gidi hqn cua day so A. KIEN THOC CAN NHd 1. Gidi han hum han lim M = 0 khi va chi khi IM I cd thi nhd hon mdt sd’ duong be tuy y, kl tfl mdt sd hang nao dd trd di. lim v = a <=> lim (v – a) = 0. n >+oo n->+oo 2. Gidi han vo circ* lim M = +00 khi va chi khi M cd thi ldn hon mdt sd duong ldn tuy y, n >+oo kl tfl mdt sd hang ndo dd trd di. lim M = -00 «. lim (-M ) = +c». n >+oo n >+<» ^ Luu y : Thay cho lim u = a, lim M = ±ao, ta viet tat lim2< = a, lmu = ±0 3. Cac gidi han dac biet 140 a) lim- = 0 ; lim j^ = 0 ; lim/2* = +00, vdife nguyen duong. «n” h) lim^” = 0 ndu 1^1 < 1 ; lim^” = +QO nlu q> I. c) lime = c (c la hing sd).

143 4. Djnh If ve gidi. han huru han a) Ndu limm = a vd limv = b,thi lim(m + v ) = a + 6; lim(u -v^) = a-b; limm v = ab ; lim-^ = -p (ndu b ^ 0). v b b) NduM > 0 vdi mgi n valimm = a, thi a > 0 va limyju^ = yfd. 5. Dinh li lien he gicira gidi han huru han va gidi han vo cifc a) Ndu limm = a va limv = +oo thi lim-^ = 0. b) Ndu limm = a > 0, limv = Ova v > 0 vdi mgi n thi lim-^ = +oo. ^n c) Nlu limm = +00 vd limv = a > 0 thi limm v = +oo. 6. Cap so nhan ICii vo han Cap sdnhdn IM vo hqn la cdp sd’ nhdn vd han cd cdng bdi q thoa man? < 1. Cdng thflc tfnh tdng 5 cua cdp sd nhdn lui vd han (M ) 5 = Ml + M2 + M M +… _ = “1 1-q B. VI DU Vidu 1 Cho day sd (M ) vdi lim M = 1. Chflng minh ring, kl tfl sd hang nao dd trd di, tdt ca eae sd hang cua (M ) diu nim trong khoang : a) (0,9 ; 1,1); b) (0,99 ; 1,01). Gidi limu = 1 <=> lim(m -1) = 0. Do dd, M – 1 cd thi nhd hon mdt sd dflong be tuy y, kl tfl mdt sd hang nao dd trd di. 141

144 a) Ld’y sd dflong nay la 0,1 (bing ‘ ), ta ed : IM – 11 < 0,1 <» -0,1 < M – 1 < 0,1 o 0,9 < M < 1,1 kl tfl mdt sd hang nao dd trd di. Ndi each khdc, tdt ca cae sd hang cua day (M ), kl tfl mdt sd hang nao dd trd di, diu ndm trong khoang (0,9 ; 1,1) b) Ld’y sd duong nay la 0,01 (bdng r ^ ), ta cd : IM – 11 < 0,01 -0,01 < M – 1 < 0,01 <» 0,99 <u < 1,01 kl tfl mdt sd hang nao dd trd di. Ndi each khae, tdt ca cac sd hang cua day (M ), kl tfl mdt sd hang ndo do trd di, diu nim trong khoang (0,99 ; 1,01). Vidi 1? Bilt day sd limm = 0. (“n) thoa man “«< n + I vdi mgi n. Chflng minh ring 142 Gidi l_ J_ T^- n + l ^,….n + l,. /2 n^ r> T>..»’ I I ‘ Datv =. Ta co limv = lim = lim ” = 0. Do do, v co thi nhd hon mdt sd duong be tuy y kl tfl mdt sd hang nao dd trd di. (1) Mat khdc, theo gia thidt ta cd M < V < V. (2) Tfl (1) vd (2) suy ra M cd thi nhd hon mdt sd dflong be tuy y kl tfl mdt sd hang nao dd trd di, nghia la limm = 0. Vidu3 Cho bie’t day sd (M ) thoa man u > n vdi mgi n. Chflng minh ring limm = +00. Gidi Vi lim/2^ = +00 (gidi han dac biet), nen n cd thi ldn hon mdt sd duong ldn tuy y, kl tfl mdt sd hang nao dd trd di.

145 Mat khdc, theo gia thidt M > n vdi mgi n, nen u cung cd thi ldn hon mdt sd duong ldn tuy y, kl tfl mdt sd hang ndo dd trd di. Vdy lim M = +OO. ^ Nhan xet : Trong cac vf du tren, ta da van dung true tiep cac djnh nghta ve gidi han ccia day sd. Vidu 4 Tinh lim An^ -n-l 3 + 2/2^ Ta cd lim An”- ViduS. -n-l 3 + 2/2^ Tfnh lim yj3n^ n 1-2/2^ Gidi 4-i-L = lim- n n 4- n = 2. Gidi.. Tl,, n3 + r^ + n 1-2/2^ l-2rf 1 / 3 n + 1 n’ + 1 n -2 Vidu 6 Tfnh lim 2 2 n – h + l Gidi n^ + n^ -2 lim n = lim n + l n + l = lim n 1 1 n n ^ =

146 Vidu7 -i_ Tfnh lim(-/2^ + n4n + 1). Gidi lim(-/2^ + n4n + 1) = lim{-n^) I- A ; 1 V/2 n’ 00. ViduS Tfnh lim //2 + n -p -) lim Gidi, ^ ^. (V/2^ +/2 – V/2^ -1 (v/2^ +/2 + V/2^ -1 V/2^ + /2 – V/2^ – 1 = lim-^^. ‘^ ‘- ^ ‘ yjn^ +n+ yfn^ – 1 n + l = lim y/n^+n + yln^ = lim- l+- n njl+ + njl – = lim 1.1..LX ‘ I ^ 144 Luu y : Khi giai bai toan 6 Vf du 7, ta da bien ddi ve dang cd thd ap dung hai tinh chat sau : limm = + 00 <:> lim(-m ) = -OO. (1) Neu limm = +oo v^ limv = a > 0 thi limm v = +oo. (2) Tuy nhien, nhong bien ddi tren Ichdng cdn thfch hgp vdi Vf du 8. Qu^ thirc, ndu lam tuong tu nhu vay ta se cd : limnn^ +n-^rf -11 = lim nal + nal – Vi lim -HH V = lim/2 r V n = 0, nen khdng the ap dung tinh chat (2) d tren.

147 ^ Nhan xet: De tim gidi han ccia mot day sd ta thudng dua ve cac gidi han dang dsc biet va ap dung cac dinh li ve gidi han hou han hoac cac djnh li ve gidi han vd cue. De cd the ap dung dudc cac djnh If ndi tren, thong thudng ta phai thuc hien mdt vai bien ddi bieu thflc xae dinh day sd da cho. Sau day la vai ggi >^ bien ddi, cd the van dung tuy theo tflng trudng hop : – Ndu bieu thflc cd dang phan thflc ma mau va tfl deu chfla cac luy thfla coa n, thi chia tfl va mau cho n, vdi k la sd mu cao nhat. – Neu bieu thflc da cho co chfla n dudi dau can, thi cd the nhan tfl sd va miu sd vdi cijng mot bieu thflc lien hgp. Vidu 9 Cho day sd (M ) xae dinh bdi Ml = V 2 “n+l = v2 ” “n Vdi/2>1. Bilt (M ) cd gidi han huu han 1 dii n > +00, hay tim gidi han dd. Gidi Ddt limm = a. Ta cd ”n+l.^2 + M => limm +i = lun.y/2 + M => a = yf2 + a =>a -a-2 = 0^^a = -l hoac a = 2. Vi M > 0 nen limm = a > 0. VdylimM^ = 2. Luu y : Trong Idi giai tren, ta da dp dung tfnh ehd’t sau ddy. “Nlu limm = a thi limm +i = a”. Ban dgc cd thi chiing minh tfnh chdt nay bing dinh nghia. Vidu 10 Cho day sd (M ) xdc dinh bdi cdng thflc truy hdi 1 “1 = 2 M ^i *n+l = 1 2-M vdi n >l. Day sd (M ) cd gidi han hay khdng khi n^> +(p’? Ndu cd, hay tim gidi han dd. 10. BTBS&GT11-A 145

148 Gidi Ta cd Ml = ;M2 = :r- ;M3 = ;M4 =. Tfl dd dir dodn u = -. (1) Chflng minh dfl dodn trln bing quy nap : – Vdi /2 = 1, ta cd MI = – – = – (dung). – Gia sfl dang thflc (1) dflng vdi /i =fe (fe > 1), nghia la MJ^ = fe + 1 Khi dd ta ed M^+I = = r = -r ^, nghia la dang thflc (1) ^”fett cung dflng vdi n-k+l. ^ – Vdy u = ^ V/2 e N*. ” n + l 1 Tfl dd ta cd limm_ = lim = lim = 1. ” n+l, n Nhan xet : De tim gidi han ciia day sd cho bang cdng thflc truy hdi ta cd the tim cdng thflc tdng quat, cho phep tfnh u theo n, bang each dfl doan cdng thflc nay, va chflng minh du doan bang quy nap. Sau dd, tim gidi han cua (i2 ) qua cdng thflc tdng quat. Vidu 11 Giai Day sd vd han 2,- V2,1, j=,,… la mdt cdp sd nhdn vdi cdng bdi -yf BTDS&GT11-B

149 Vi q = yfi = j= < 1 nen day sd nay la mdt cdp sd nhdn lui vd han. V2 2V2 Dodd,5=2-V2 + l-4= + 4 yfi J_”V2 + l’ Vidu 12 lim dang khai triln cua cdp sd nhdn lui vd han (v ), bilt tdng cua nd bing 32 va V2 = 8. Gidi 8 Tfl gia thidt suy ra, ^ = 32. Mat Idiac, V2 = Vi<7 = 8 ^> Vi = The vao dang thflc tren ta cd : – = 32 <» 4(7-4<7 + 1 = 0 <=> 9 = -. q{l -q) 2 Tfl dd v = V2^” = 8 1 2«-2 2″-5′ Vdy dang khai triln cua (v ) la : 16, 8, 4, 2, 1,,…, 2,…, 2 n-5 ‘ ” Vidu 13 Vidt sd thdp phdn vd han tudn hodn sau ddy dudi dang phdn sd hiru ti: a = 2, (ehukil3). Gidi 13 ^ ^^^r,..^ r, inn a = 2, = = 2 + ^^^^ ^ 100″ ^ = = 99 ^^’ Too’T55^’-‘I^’- ^^””^^ ”^P ‘^ “^^” ^”‘ ^^ ^^’ ”^”^ ^^’ “^ ^ 100^- 147

150 ^ Nhan xet: – Cach tinh tsng coa mot cap s6 nhiin lui v6 han : Nhan dang xem day sd da cho cd phai la mot cap sd nhan lui vd han khdng (ndu dieu nay chua dugc neu len trong gia thiet ccia bdi toan). Sau dd, ap dung cdng thflc tinh tdng da biet trong SGK. – Cach tim cap so nhan lui vo han khi biet mdt so diiu ki n : Dung cdng thflc tfnh tdng de tim cdng bdi va sd hang dau. – Cach viet mot sd’ thap phan v6 ban tuin hoan dudi dang phin sohou ti: Khai triln sd da cho dudi dang tdng coa mdt cap sd nhan lui vd han va tinh tdng nay. C. BAI TAP 1.1. Bidt ring day sd (M ) cd gidi han la 0. Giai thfch vi sao day sd (v ) vdi v = IM I cung ed gidi han la 0. Chiiu ngugc lai cd dflng Ichdng? 1.2. Vi sao day sd (M ) vdi M = (-1)” khdng thi cd gidi han la 0 khi n -> +QO? 1.3. Cho bilt day sd (M ) cd gidi han hiiu han, cdn day sd (v ) khdng ed gidi han hiiu han. Day sd {u + v ) ed thi cd gidi han hflli han khdng? 1.4. a) Cho hai day sd (M ) va (v ). Bilt limm = -oo vd v < u vdi mgi n. Cd kit ludn gi vl gidi han cua day (v ) khi n -^ +oo? b) Tim limv vdi v = -«! Tfnh gidi han cua cae day sd cd sd hang tdng quat sau ddy, khi n -^ +oo. 2/2-3/2^+1 a) a 3 2 n^ +n 2/IV/2 c) f n = -T^ n^ +2n-l e) «= 2″ + 1 3/2^ -5/2 + 1 b)&n = n^ +A d)rfn = I-An (2-3nf{n + if ‘ yf2^” f)v = + V “” J 4″ g) “n = 3″ – 4″ ” + 2″ h)v yjn^ +n-l- yjan^ – 2 n

151 1.6. Tfnh cdc gidi han sau : a) lim{n^ +2/2-5) ; e) lim [4″ +(-2)”l ; b) lim(-/2-3/2-2); d) limn\ln -1 ^/ Cho hai day sd (M ) va (v ). Chflng minh ring nlu limv =0 va M ^ v vdi mgi n thi IimM = Bilt M – 2 <. Cd kdt ludn gi vl gidi han eua day sd (M )? 1.9. Dung kdt qua cdu 1.7 dl tfnh gidi han cua cdc day sd ed sd hang tdng quat nhu sau : a)«n=;^; (-1)” d) M = (0,99)” eos/2; e) M = 5″ – cos vn n. C)”n = 2 -«(-!)” 1 + 2/2 2 ‘ Cho day sd (M ) xdc dinh bdi cdng thflc truy hdi Ml = 2 M +1 “n+l =-^vdi/2>l. Chflng minh ring (M ) cd gidi han hiiu han khi n -> +00. Tim gidi han dd. r’ i”-^ Tfnh tdng cua cdp sd nhdn lui vd han 1,- > >-,, V 2y n-l Tfnh tdng 5 = 1 + 0,9 + (0,9)^ + (0,9)^ (0,9)” ‘ lim sd hang tdng qudt cua cdp sd nhdn lui vd han cd tdng bdng 3 va cdng bdi^=f Cho day sd’ (6 ) cd sd hang tdng quat la b = sina + sin a sin”a n vdi a^ + kn. lim gidi han cua {b ). 149

152 1.15. Cho sd thdp phdn vd han tudn hoan a = 34, (chu ki la 12). Hay vie’t a dfldi dang mdt phdn sd Gia sfl ABC la tam giac vudng cdn tai A vdi dd ddi canh gde vudng bing 1. Ta tao ra cae hinh vudng theo cac bude sau ddy : – Bude 1 : Dung hinh vudng mdu xdm cd mdt dinh la A, ba dinh cdn lai la cac trung dilm cua ba canh AB, BC va AC (H.l). Kf hieu hinh vudng nay la(l). C A C A Hinh 1 Hinh 2 Hinh 3 – Bude 2 : Vdi 2 tam gidc vudng cdn rnau tring cdn lai nhu trong hinh 1, ta lai tao dugc 2 hinh vudng mau xam khdc theo each tren, kf hieu Id (2) (H.2). – Bude 3 : Vdi 4 tam giac vudng cdn mau tring nhu trong hinh 2, ta lai tao dugc 4 hinh vudng mdi mdu xdm theo each tren (H.3) Bude thd n : O bude nay ta cd 2″ hinh vudng mdi mau xam dugc tao thanh theo each tren, kf hieu la (n). a) Ggi u la tdng dien tfch cua tdt ea eae hinh vudng mdi dugc tao thanh d bude thfl n. Chiing minh ring M = -. b) Ggi 5 la tdng dien tfch cua tdt ca cac hinh vudng mdu xdm ed dugc sau n bude. Quan sat hinh ve dl du doan gidi han cua 5 khi n > +oo. Chiing minh du doan dd.

153 2. Gidi hqn cua ham s6 1. Gidi han huru han A. KIEN THCTC CAN NH6 Cho khoang K chfla dilm XQ va ham sd y = f{x) xdc dinh tren K hodc tren K{xo}. lim /(JC) = L khi va chi khi vdi day sd (jc ) bd’t Id, jc e KXIXQ} vd x -^ XQ, ta cd lim/(x ) = L. Cho ham sd y =f(x) xdc dinh tren khoang {XQ ; b). lim /(x) = Lkhi vd chi khi vdi day sd {x ) bd’t ki, XQ < x XQ ta cd lim/(a: ) = L. _ ‘ S;; Cho ham sd y =f(x) xde dinh tren khoang {a ; XQ). lim /(JC) = L khi vd chi khi vdi day sd (jc ) bdt ki, a < jc < XQ vd jc -^ JCQ, x->x^ ta cd Um/(x ) = L. Cho hdm s6y =f{x) xde dinh tren khoang {a ; +00). lim /(jc) = L khi va ehi khi vdi day sd {x ) bd’t ki, jc > a vd jc ^ +00 thi j:->+tx3 limf{x ) = L. Cho ham sd y =f{x) xde dinh tren Idioang (-00 ; a). lim f{x) = L khi va ehi Ichi vdi day sd (jc ) bd’t ki, x < a va x -> -00 thi limf{x ) = L. 2. Gidi han vo cue Sau ddy la hai trong sd nhilu loai gidi han vd cue khdc nhau : Cho ham sd y = f(x) xdc dinh tren khoang {a ; +00). lim /(JC) = – 00 khi vd chi khi vdi day sd (jc ) bd’t ki, jc > a va x > +00, X->-HO ta ed lim/(:«: ) =

154 Cho khoang K chiia diim XQ va ham sd y = f{x) xde dinh trln K hoae tren KWXQ]. Um /(JC) = +00 khi va chi khi vdi day sd (jc ) bd’t ki, jc e A’ {JCO} va jc -^ XQ, ta cd lim/(jc ) = + co. ^ Nhan xet :/(x) cd gidi han +00 khi va chi khi -f{x) cd gidi han Cac gidi han dac biet a) lim X = jcg. b) lim c = c ; c) lim c = c ; d) lim = 0 (c la hing sd^. X^>XQ -x^±t» jc^+oo X e) lim X* = +00, vdifenguyen duong. X^-HC f) lim X* = -00, nlufela sd le ; g) lim x* = + 00, ndufela sd chdn. 4. Djnh li ve gidi han huru han Dinh li 1 a) Neu lim /(x) = L vd lim g{x) = M, thi X->X() X->XQ lim [/(x) + g{x)] = L + M X^XQ lim [/(x) – g{x)] = L-M ; X^XQ lim [/(x).g(x)] = L.M ;. X^XQ lim 44 = ^ (nlu M ^ 0) ; h) Ne’u/(x) > 0 va lim /(x) = L, thi L > 0 va lun V7W = V^. A” Chu y : Djnh If 1 vin dung khi x ^ +00 ho&c x -^

155 Dinh li2 lim f{x) = L khi va ehi khi lim /(x) = lim /(x) = L. X->XQ X^XQ X^XQ 5. Quy tac ve gidi han vo circ a) Quy tdc tim gidi hqn eua tichf{x).g{x) lim /(x) X-^XQ L>0 L<0 lim g{x) X-^XQ lim f{x)g{x). X-^XQ b) Quy tdc tim gidi hqn cua thuang f{x) 8{x) lim /(x) X^XQ lim g{x) X^XQ Dd’u cua g{x) X^XQ g{x) L L>0 L<0 ± Tuyy (Ddu eua ^(x) xet tren mdt khoang K nao dd dang tfnh gidi han, vdi x ^ XQ). B. VI DU Vidul Cho ham sd f{x) = 2x^ + X – 3 x^n Dung dinh nghia ehiing minh ring lim /(x) = 5. x->l 153

156 Gidi Ham sd da cho xdc dinh trdn R {1}. Gia sfl (x ) la day sd bdt ki, x 9^ 1 vd x -^ x^+x -3 2(x – l)(x +-) lim /(x ) = lim ±52J3L_^ = lim 2_ n->+oo n->+oo X 1 n >+co X^ 1 = Um 2(x +1) = 5. Do dd, Um/(x) = 5. Vidu 2 Cho ham sd fix) fx, nlu X > 0 [l – X, ndu X < 0. Diing dinh nghia chiing minh ring ham sd fix) khdng edgidi han khi x-> 0. Ham sd da cho xdc dinh tren R. Ldy day sd (x ) vdi x =. Gidi Ta ed x -> 0 va lim /(x ) = lim x = lim = 0. (1) n»+oo – n-»+oo n >+oo /2 Ldy day sd {y ) vdi y =. Ta cd >’ ^ 0 vd lim f{y ) = lun (1 – y ) = Um (1 + -) = 1. (2) n->+oo n->+oo n->+oo /2 Tfl (1) va (2) suy ra ham sd/(x) khdng cd gidi han khi x -> 0. ^ Nhan xet 154 De dung djnh nghta chflng minh hdm sd y =fix) khdng cd gidi han khi x -» XQ, ta thudng lam nhu sau : Chgn hai day sd khae nhau (a ) v^ (i ) thoa man : a v^ b thugc tdp xae djnh coa ham sd y =fix) va khae XQ ; a -> XQ ; i -> XQ ;

157 Chflng minh rang lim /(a ) ^ lim f{b ) han nay khdng tdn tai. n-»+oo n-»+oo hosc chflng minh mdt trong cac gidi ^ Luu y : Trudng hgp x ^ Xg, x -> XQ hay x -> ±00 chflng minh tuong tu. Vi du 3 Tfnh a) lim (Vx^+5-1) ; b) lim ^ ^ ; c) lim (-x^ + x^ – x + 1); x^-2 J x^3~ X – 2 A:->-CO,.,. 1-X.,. 2x-l d) hm ; e) lim -. ^-^’^ (x – 4) x^3- x-3 Gidi a) lim (yfx^ lj = yl{-2f = 2; b) lim ^^ = = 4 ; x^r X-1 i-1 c) lim (-x^ + x^ – X + 1) = lim x^(-l + r- + -r-) = -H». d)taed lim (l – x) =-3 < 0. (1) lim(x-4f =0va(x-4f >0 vdimgix^4. (2) x^a f{x) Ap dung qui tic vl gidi han vd cue ddi vdi thuong ^^-rr, tfl (1) va (2) suy ^,. 1-x ra lim = -00. ^^^x-af e) Ta cd lim (2x – l) = 5 > 0, lim (x – 3) = 0 va (x – 3) < 0 vdi x-^3 x^3 2x-l moi X < 3. Do dd, lim = -00. x^3~ X-3 Nhan xet Trong cac vf du tren ta da dung true tidp cac dinh If v^ gidi han coa tdng, hieu, tfch, thuong va can coa cac ham sd hosc cac quy tac vi gidi han vd cue. 155

158 Vi du 4 Tfnh cdc gidi han sau : x’^ + 2x – 3 a) lim ^^1 2x^ – X – 1 2x^ + 3x – 4 c) lim J:->+«) X x^ +1 e) Um – – ^ – l x-^o’-^v-^ + l J b) lim x^2 Vx ^,,. Vx^ – X – ylax^ + 1 d) lim r 5 AT >-00 ZX + J f) Um {1AX^ – X + 2x). Gidi,,. x’^+2x-3,. (x-l)(x + 3),. x a) Um = lun = Um = x2-x-1 -i2(;,_i)(^ + ) ->i2x + l X (2 – x)(vx ) ^ b) lim -^^ J-^ = lim ^-^; ^ = lim- (Vx ] = J:^2 VX A:^2 X-2 ;c->2 J 4_,,. 2x^+3x-4,. “^r2 v3 c) lim = lim ^ ^ = -2. x-^+00 _;c – X +1 x^-^ i 1 1. yfx^-yl^ Ax^ +1 d) lim 2x+3 ^ x^ xjl IxL 4 + r = lim – jr->-oo 2x+3 -xjl xj4 + r = lim – 2x+3 r e) lim ;c->0” X x Jl-i..4.J. = lim X = lim l-(x + l) -1 = lim ;t^o- x{x + 1) ;,^o- {x + 1) =

159 .2 ^ A Jl f) lim (V4x2-x+2x)= lim ^^^ “”^ “^”^ ^^^ ^^-^ yjax^ – X – 2x = lim, = lim, = lim, =. X-^^x> j 1 ;c->-oo / 1 At->-oo / 1 4 ^ Nhan xet Khi tfnh gidi han md khdng the ap dung true tidp djnh If ve gidi han trong sach giao khoa, ta phai bien ddi bieu thflc xae djnh h^m sd ve dang ap dung dugc cac djnh If nay. Sau day la mgt sd each bien ddi thudng dugc dtjng. Tinh lim – khi lim u{x) = lim v(x) = 0 x-*xf) v(x) X-^XQ X-*XQ – Phdn tfch tfl vd mdu thdnh tfch cdc nhdn tfl vd gian udc. Cu thi, ta biln ddi nhu sau :,. M(X),. (X-XA)A(X),. A(x).,,,. A(x) lim -7-^ = lim “; ; [ = lim ^ va tfnh Irni 7^. X-^XQ V(X) X-^XQ (X Xo)B(x) X->XQ B(x) X-^XQ B(x) – Ne’u M(X) hay v(x) ed chfla bidn sd dudi dd’u cdn thi cd thi nhdn tfl va mdu vdi bilu thflc lien hgp, trudc khi phdn tfch chflng thdnh tfch dl gian udc. Tinh lim khi lim u{x) = ±00 va lim v(x) = ±00 J:->±CO V(X) X^XQ X^X^ – Chia tfl vd mdu cho x” vdi n la sd mu bdc cao nhd’t eua bidn sd x (hay phdn tfch tfl vd mdu thanh tfch chfla nhdn tfl x” rdi gian udc). – Ndu M(X) hay v(x) cd chfla bidn x trong dd’u can thflc, thi dua x ra ngoai ddu cdn (vdifela sd mu bdc cao nhdt cua x trong dd’u cdn), trudc Ichi chia tfl va mdu cho luy thfla eua x. Tinh lim [M(X) – v(x)] khi lim M(X) = +00 va lim v(x) = +00 X->XQ X->XQ X^XQ hoac lim M(X).V(X) khi lim M(X) = 0 va lim v(x) = +00. X^>XQ X^XQ X->.XQ Nhan vd chia vdi bieu thflc lien hgp (neu cd bieu thflc chfla bien sd dudi ddu cdn thflc) hosc quy dong mau de dua ve cung mot phan thflc (neu chfla nhieu phan thflc). 157

160 2.1. Dung dinh nghia tim cdc gidi han C. BAI TAP.,. x + 3 a) lim:r ; x^5 3- X 2.2. Chohamsd’/(x)=< x2, ndu X > 0 x^. -1, ndu X <0. b) lim x^+co x’^ +1 a) Ve dd thi eua ham sd/(x). Tfl dd du dodn vl gidi han cua/(x) khi x -> 0. b) Dung dinh nghia chiing minh du dodn trdn a) Chiing minh ring ham sd y = sinx khdng ed gidi han khi x > +oo. b) Giai thich bing dd thi kit ludn d cdu a) Cho hai hdm s6 y = fix) va y = g{x) cung xdc dinh trdn khoang (-oo ; a). Dung dinh nghia chiing minh ring, ndu lim /(x) = L vd lim g{x) = M x->-ao.)c->-oo thi lim /(x).g(x) = L.M Um gidi han eua edc ham sd sau : ^)fix) = ; khi X ^ 3 ; b) h{x) = khi x -> -2 ; ^ -1 {x + 2f c)fe(x) = sax^ -x + 1 khi X ^ – 00 ; d)fix) =x + x^ + 1 khi x -> -oo ; X 15 – e) h{x) = khi x ^^ -2 va khi x > Tfnh eae gidi han sau : 158,, ^ + 3,,,. (1 + x)^ -1 a) lun -^ ; b) lim-!^ ; -3×2+2x-3.^^0 X e) lim -^ ; d) lim ^->+ o X^ – 1 ‘.>:^5 Vx – >/5 ‘.,. JC-5 ^,. lx^+5-3 e) Iim -j= 7= ; f) lim ;c->+oo Vx + V5 x^-2 X + 2

161 g) lim Vx-1 x-^i Vx i) lim-t- ;c->ojc^ U^ + l 1-2x + 3x’* h) lim x-^-w X -9 1 ; j) lim (x^ – 1)(1-2x)^ ‘ x’^ + X Tfnh gidi han eua cdc ham sd sau khi x -> +oo va khi x -> -oo Vx^ – 3x a)ax)= ^^,, b)/(x) = X + Vx^ – X + 1 ; c)/(x) = Vx^-x – Vx^ Cho hdm sd y fix) = cdddthinhuhinh4. 2x^ – 15x + 12 X – 5x + 4 M a) Dua vao dd thi, du doan gidi han eua ham s6fix) khi x -> 1″^; x-^l ;x->’4’^;x->4 ; x->+qovakhi x->-oo / ^ ” ” ^ / ”’ b) Chflng minh du dodn trdn Cho ham sd fix) = _1, ne’ux>l ^-1 x^-l mx + 2, ne’ux<l. Hinh 4 Vdi gia tri nao eua tham sd m thi ham ^6 fix) cd gidi han khi x > 1? lim gidi han ndy Cho khoang K,XQeK va ham s6y =fix) xdc dinh tren K {XQ}. Chflng minh ring ndu lim /(x) = +QO thi ludn tdn tai ft nhd’t mdt sd c thude X^XQ. Ar{xo} sao cho/(c) >

162 2.11. Cho ham sd y =/(x) xdc dinh trdn khoang {a ; +00). Chiing minh ring ndu lim J /(x) = -00 thi ludn tdn tai ft nhdt mdt sd c J:^+CC thude {a ; +00) sao cho/(c) < 0 3. Ham so iien tuc A. KIEN THQC CAN NH6 1. Ham so lien tuc Cho ham sd y =/(x) xdc dinh tren khoang ^ vd XQ e K. y =fix) lien tuc tai XQ khi vd chi khi lim /(x) = /(JCQ). X^XQ y =fix) Uen tuc tren mdt khoang ndu nd Udn tuc tai mgi dilm cua khoang dd. y = fix) lien tuc tren doan [a ; b] nlu nd lien tuc tren khoang (a ; b) va Um /(x) = f{a), Um /(x) = f{b). x^>-a* x^^b ^ Nhan xet : Dd thj ciia ham sd lien tuc tren mdt khoang la mdt “dfldng lien” tren khoang dd. 2. Cac djnh li 160 Dinh HI a) Ham sd da thflc lien tuc tren todn bd tdp sd thuc R. b) Ham sd phdn thflc hiiu ti va ham sd lugng gidc lien tue tren tflng khoang eua tdp xdc dinh cua chflng. Dinhli2 Gia sfl 3′ =/(x) vay = g{x) la hai ham sd lien tuc tai dilm XQ. Khi dd : a) Cdc ham s6fix) + g{x), fix) – g{x) vafix).g{x) cung lidn tuc tai dilm XQ ;

163 f{x) b) Ham sd ^^ -^ lien tuc tai JCO, ndu g{xq) * 0. g{x) Dinh li 3 Nlu hdm sd y = fix) lien tuc tren doan a ; b] va fia)fib) < 0 thi tdn tai it nhd’t mdt dilm c e {a;b) sao cho/(c) = 0. Menh de tuang duang : Cho ham sd y = fix) lien tuc tren doan [a ; b] va fia)fib) < 0. Khi dd phuong trinh/(x) = 0 ed ft nhdt mdt nghiem trong khoang {a;b). B. VI DU Vi du 1 ‘x + 3 Xlt tfnh lien tue cua ham sd fix) = tai dilm x = -l. x-r [ 2, ndu x^ -I nlu x= -1 Gidi tdp xdc dinh eua hdm sd da cho la D = R, chfla x = -1. Ta cd,/(-l) = 2 vd lim x + 3 = -l^/(-l). x^- X – I bo dd, hdm sd khdng lien tue tai x = – 1. Vidu2 X – 2x – 3, neu X ^3 Xet tfnh lien tue eua hdm sd /(x) = < x-3 5, ndu X = 3 tren tdp xdc dinh cua nd. 11.BTBS&GT11-A 161

164 Gidi Tdp xdc dinh cua/(x) la D = R. x^ – 2x Ndu x^3 thi fix) = Id hdm sd phdn thflc hiiu ti, ndn liln tuc tren cae khoang (-00 ; 3) vd (3 ; +00). ‘ – Tai X =3, ta ed/(3) = 5,,. x^-2x-3,. (x + l)(x-3) o.-. lim = Imi-^ -^ – = lmi(x + 1) = 4 5t /(3). x~^3 x-3 x-^3 x-3 x^3 Do dd/(x) khdng lien tuc tai x = 3. – Kdt ludn : Ham sd/(x) lien tuc tren cae khoang (-00 ; 3), (3 ; +00), nhung gian doan tai x = 3. Vidu 3 Chflng minh ring phuong tiinh sau ed ft nhd’t hai nghiem : 2x^-10x-7 = 0. Gidi Xet ham sd fix) = 2x – lox – 7. Ham sd nay la ham da thflc ndn lien tue tren R. Do dd nd lidn tue tren cac doan [-1; 0] vd [0 ; 3]. (1) Mat khae, ta ed : fi-l) = l;fio) = -l va fi3)=ll. Do dd /(-l).f(o) < 0 va /(0)./(3) < 0. (2) Tfl (1), (2) suy ra phuong trinh 2x^ – lox -7 = 0 cd ft nhdt hai nghiem, mdt nghiem thude khoang (-1 ; 0), edn nghiem kia thude khoang (0 ; 3) BTBS&GT11-B

165 Vi du 4. Chiing minh ring phuong trinh sau ludn cd nghiem vdi mgi gia tri ciia tham sd /n : {l-m^-3x-l=0. Gidi Xet ham sd’/(x) = {l-m)x -3x-l. Vi /(O) = – 1 < 0 va/(-l) = m^+l>0 nen/(-l)/(0) < 0 vdi mgi m. (1) Mat khae, /(x) la hdm da thflc, lien tuc tren R, nen lien tuc tren doan ^ [-1; 0]. (2) Tfl (1) vd (2) suy ra phuong trinh fix) = 0 cd ft nhd’t mdt nghiem trong 9 S khoang (-1; 0), nghia la phuong trinh {I – m )x -3x-l=0 ludn cd nghiem vdi mgi m. Nhan xet De chiing minh phuong tnnh fix) = 0 cd ft nhat mot nghiem, chi can tim duoc hai sd ava b sao cho : fia).fib) < 0 va ham sd/(x) lien tuc tren doan [a ; b]. Chu y. Neu phuong trinh chfla tham sd, thi chgn ava b sao cho : fia) vafib) khdng cdn chfla tham sd hay chfla tham sd nhung cd dau khong ddi ; hosicfia).fib)chfla tham sd nhung tchfia).fib) ludn am. C. BAI TAP (x – 1)1x Cho hdm sd f{x) = ‘ ^. X Ve dd thi eua ham sd nay. Tfl dd thi du dodn cac khoang tren dd ham sd lien tuc va chiing minh du doan dd Cho vf du vl mdt ham sd lien tuc trdn (a ; b] va tren (b ; c) nhung khdng lien tue tren {a ; c). 163

166 3.3. Chflng minh ring nlu mdt ham sd lien tue trdn (a ; b] vd tren [b ; c) thi nd lien tue tren {a ; c) Cho ham sd y =fix) xae dinh tren Ichoang {a ; b) chfla dilm XQ. f(x) f{x ) Chflng minh ring nlu Um ”-^ ‘ = L thi ham sd fix) lien tuc tai dilm XQ. X-^XQ X – XQ Hudng ddn : Ddt g(x) = /(^) ~ /^^o) _ ^ yd bilu diln/(x) qua g{x). X XQ 3.5. Xet tfnh lien tue eua cdc hdm sd’ sau : 3) fix) = Vx + 5 tai X = 4 ; b)^(x) = ‘_x-l I. ndu X < 1 V2-X-1-2x, ndu X > 1 taix=l Xlt tfnh lien tue cua cdc ham sd sau trdn tdp xae dinh eua chflng a)/(x) = x^-2 f^, ndu x^ y]2 X – V2 b) g{x) = 2^, ndu x=y/2; 1-x -, nduxti 2 {x-2f 3, ndu X = 2. x^ – X lim gia tri cua tham sd m di hdm sd f{x) = >, ndu X 9t 2 x-2 m, ndu X = 2 lien tuc tai x = 2. Vx Tim gia tri cua tham sd m di hdm sd /(x) = x^ -I, ndu X ^ 1 lien tuc tren (0 ; +oo) Chiing minh ring phuong trinh a) X – 3x – 7 = 0 ludn ed nghidm ; m, ndu X = 1 b).cos2x = 2sinx – 2 cd ft nhd’t hai nghidm trong khoang 6′”” 164 😉 Vx + 6x = 0 cd nghiem ducmg.

167 3.10. Phuong trinh x”* – 3x^ + 1 = 0 ed nghidm hay khdng trong khoang (-1 ; 3)? Chiing minh cdc phuong trinh sau ludn ed nghidm vdi mgi gia tri eua tham s6 m : a){l-m^){x+lf+x^-x-3 = 0; b) /72(2eosx – V2 ) = 2sin5x Chflng minh phflong trinh x” + flix” + a2x” a -ix + a = 0 ludn cd nghiem vdi n la sd tu nhien le Cho ham sd y = fix) lien tuc trdn doan [a ; b]. Ndu fia).fib) > 0 thi phuong trinh fix) = 0 cd nghidm hay khdng trong khoang (a ; b) 1 Cho vi du minh hoa Ndu ham sd y = ^x) khdng Udn toe tren doan [a ; b] nhung/;a)/(6) < 0, thi phuong trinh y(x) = 0 cd nghiem hay khdng trong khoang {a;b)l Hay giai thfch cdu tra Idi bing minh hoa hinh hgc. Bai tap on chitdng IV 1. Tfnh cae gidi han sau {n -^ +oo) :,,. (-3)” + 2.5″ ‘ ‘ux, /I a) lim^^ ; b) lun ; 1-5″ rf +n + e) liml V/2^ + 2/ ^rf +n-\. 2. Tim gidi han cua day sd (M ) vdi ^ (-1)” u^ 2″-/2 ^^ “«= ~2~~T ‘ b)m = -. /2^ + 1 3” Vilt sd thdp phdn vd han tudn hodn 2, (chu ki 131) dudi dang phdn sd. 165

168 4. Cho day sd (M ) xae dinh bdi Ml = 1 u ^, “n+l = a) Chflng minh ring M > 0 vdi mgi n. 2M +3 w +2 vdi/2 > 1. b) Bie’t (M ) cd gidi han hiiu han. Tim gidi han dd. 5. Cho day sd (M ) thoa man M < M vdi mgi n. Chiing minh ring nlu lim u = a thi a<m. 6. Tfl dd cao 63m cua thap nghieng PISA d Italia (H.5) ngudi ta tha mdt qua bdng cao su xudng ddt. Gia sfl mdi ldn cham dd’t qua PISA ITALIA bdng lai nay len mdt do cao bing dd cao ma qua bdng dat dugc ngay trudc dd. Tfnh dd dai hanh trinh eua qua bdng tfl thdi -[j^ dilm ban ddu cho ddn Ichi nd nim ydn tren mat ddt. Hinh 5 1. Chiing minh ring ham sd/(x) = cos khdng cd gidi han Ichi x ^ Tim cac gidi han sau : x + 5 a) lim x-^-2 x”-, + X-3 c) lim (x^ +2x’^y[x – I) 9. Tim cac gidi han sau : b) lim Vx^ + 8x + 3 ; x^3 d) lim J C ^ – l 2x^ – 5x – 4 {X + if 166 a) lim I x^ ^-^ 4-Vx2+16 c) lim 2x* + 5x – 1.X->-HX) – x^ + X^ e) lim x(vx l x.v->+x..,. X-y[x h) lim p: ; ^^ 1 Vx -1 X + V4x^ -X + 1 d) lim ;c->-oo l-2x f) Um JC->2+Vx 1 1 X 2

169 10. Xde dinh mdt hdm sd y =fix) thoa man ddng thdi cae dilu kidn sau : a) fix) xae dinh trdn R {1}, b) lim /(x) = +00 ; Um /(x) = 2 vd lim /(x) = 2. x^ x-^+oo x-»-oo 11. Xet tfnh liln tue cua ham sd/(x) = trdn tdp xae dinh cua nd. x^ + 5x + 4 x^ +1 ndu X # – 1 1, nlu X = Xae dinh mdt ham sd y =fix) thoa man ddng thdi cac dilu kidn sau : a)/(x) xdc dinh tren R, b) y =/(x) lien tuc tren (-oo ; 0) vd tren [0 ; +oo), nhung gian doan tai x = Chflng minh ring phuong trinh : a) x^ – 5x-1 = 0 cd ft nhdt ba nghiem ; b) m{x – l)^(x -4)+x -3 = 0 ludn ed ft nhd’t hai nghiem vdi mgi gid tri cua tham sd /n ; c)x -3x = mc6ii nha’t hai nghism vdi moi gia tri cua m e (-2 ; 2). V I C y Cho ham s6fix) = r = 0. Phuong trinh/(x) = 0 ed nghidm hay khdng Ji ^ ^ a) Trong khoang (1 ; 3)? b) Trong khoang (-3 ; 1)? 15. Gia sfl hai ham sd y =/(x) va y =/(x + -) diu lidn tue tren doan [0 ; 1] va /(O) = /(I). Chiing minh ring phuong trinh fix) – fix +-) = 0 ludn ed nghidm trong doan [0 ; ]. 167

170 Bai tap trao nghiem 16. Chgn menh dl dung trong edc mdnh dl sau : (A) Nlu lim M = +00, thi limm = +00 ; (B) Nlu lim M = +00, thi limm = -00 ; (C) Neu limm = 0 thi lim M = 0 ; (D) Nlu limm = -a thi lim M = a. nn _ nn 17. lim bing 2″ +1 (A)l; (B)-a); (C) 0; (D) lim V/2^ -n + l – n bing (A) 0 ; (B) 1 ; (C) -^ ; (D) ‘ 19. lun (x – x^ + 1) bing (A) ; 1 (B) -o) ; (C) 0 ; (D) x lim bing x^2~ X-2 (A)-<»; (B) ; (Q1; (D) + co. 2x Cho ham sd /(x) = – -. lim /(x) bing (A)+a); (B) ; 3 + 3x x^-l^ (Q 1 ; (D)-oo. 168

171 – 22.,. X -6 lim bang x^~ x (A) ; (B)-oo; (C) ^ ; (D) + oo.,,, ^/,4x^-x + l lim bang.ic >-oo x + 1 (A) 2; (B)-2; (C) 1 ; (D) Cho ham sd/(x) xdc dinh trln doan [a ; b]. Trong edc menh dl sau, minh dl nao dflng? (A) Nlu hdm sd/(x) lien tue tren doan [a ; b] vafia)fib) > 0 thi phuong trinh/(x) = 0 khdng cd nghidm trong khoang {a ; b). (B) Nin fia) fib) < 0 thi phuong trinh/(x) = 0 cd ft nhd’t mdt nghidm trong khodng (a ; b). (C) Nlu phuong trinh/(x) = 0 cd nghidm trong khoang {a ; b), thi hdm sd fix) phai lien tue trdn khoang {a ; b). (D) Ndu ham sd fix) lien tue, tdng trdn doan [a ; b] va fia)fib) > 0 thi phuong trinh/(x) = 0 khdng thi ed nghidm trong khoang {a;b). 25. Cho phuong trinh 2x’* – 5x^ + x + 1 = 0. (1) Trong cae menh dl sau, minh dl ndo dflng? (A) Phuong trinh (1) khdng cd nghidm trong khoang (-1 ; 1) ; (B) Phuong trinh (1) khdng cd nghidm trong khoang (-2 ; 0); (C) Phuong trinh (1) chi ed mdt nghidm trong khoang (-2 ; 1); (D) Phuong trinh (1) cd ft nhd’t hai nghidm trong khoang (0 ; 2). 169

172 LOI GIAI – HUONG DAN – DAP SO CHUONG IV Vi (M ) cd gidi han la 0 ndn M cd thi nhd hon mdt sd duong be tuy y, kl tfl mdt sd hang ndo dd trd di. Mat khae, v = M = u . Do dd, v cung ed thi nhd hon mdt sd duong be tuy y, kl tfl mdt sd hang nao dd trd di. Vdy, (v ) cd gidi han la 0. (Chflng minh tuong tu, ta ed ehilu ngugc lai cung dflng) Vi M = (-1)” = 1, ndn M khdng thd nhd hon mdt sd duong be tuy y, kl tfl mdt sd hang ndo dd trd di. Ching han, u khdng thi nhd hom 0,5 vdi mgi n. Do dd, day sd (M ) khdng thi cd gidi han la Day (M + v ) khdng cd gidi han hiiu han. Thdt vdy, gia sfl ngugc lai, (M + v ) cd gidi han hiiu han. Khi dd, cae day sd {u + v ) vd (M ) cung cd gidi han hflu han, ndn hieu cua chung cung la mdt day cd gidi han hiiu han, nghia la day sd cd sd hang tdng qudt la M + v – M = v cd gidi han hiiu han. Dilu ndy trdi vdi gia thidt (v ) khdng cd gidi han hiiu han a) Vi lim u = -oo ndn lim(-m ) = +oo. Do dd, (-M ) cd thi ldn hon mdt sd duong ldn tuy y, kl tfl mdt sd hang ndo dd trd di. (1) 170 Mat khae, vi v (-M ) vdi mgi n. (2) Tfl (1) va (2) suy ra (-v ) ed thi ldn hon mdt sd duong ldn tuy y, kl tfl mdt sd hang ndo dd trd di. Do dd, lim(-v ) = +oo hay lim v = -oo. b) Xlt day sd (M ) = -n. Ta CO -nl < -n hay v < M vdi mgi n. Mat khdc lim u^ = lim{-n) = -oo. Tfl kdt qua cdu a) suy ra limv = lim(-/2!) = -oo.

173 1.5. a)-3; e) lim f)0; ( 2″ + b)+oo; 1 «1 “^'”^” g)- ; + 1 n c)0; – = +00 ; 2″j h)-l a) +00 b)-c»; c) +00 ; d) limv = 0=> v cd thi nhd hon mdt sd duong be tuy y, kl tfl mdt sd hang naoddtrddi. (1) Vi M ^ v vd v < v vdi mgi n, ndn M < v vdi mgi n. Tfl (1) va (2) suy ra M cung ed thi nhd hon mdt sd duong be tuy y, kl tfl mdt sd hang ndo dd trd di, nghia la lim M = limm = 2. m 1.9. a) Vi n < vdi moi n va lim = 0 ndn lim- = 0. n ‘ n n b)0; cosyfnn c)0; d)0; cosyfnn e) Ta cd M = 5″ – cos yfnn = 5″ 1-5″ V / Vi 5″ r Do dd, lim 1 ^ 1 V 1-1 n A 1- cosyfnn. < va lim = 0 ndn lim = 0. 5″ 5″ 5″ yfn: cosy/nn 5″ Mat khae, lim 5″ = +oo. = 1>0. Tfl (1), (2) va (3) suy ra lim(5″ – cosyfnn) = lim5″ yfnv COSV/271 5″ = +00. (1) (2) (3) 171

174 1.10. Ml = 2 M ^, = -‘^ voi /2 > *«+l Tacd, Ml = 2, “2=2” “3 ” 4’ “4=g-. “s = j^- Du dodn, M = 2″-^ + 1 ; vdi n e N*. /^n-l Chflng minh du dodn trdn bing quy nap (ban dgc tu chiing minh). 2″”^ +1 Tfldd, IimM_ = lim ; r IY”! = lim 1 + = lim Y = 1. ” 2″”^ f Nn-l 1.13.M = I Day sd : sina, sin a,…, sin”a,… vdi a^ n + fen, la mdt cdp sd nhdn vd han, cdng bdi ^ = sin a. Vi sina < 1 vdi a5t + fejt ndn (sin” a Id mdt cdp sd nhdn lui vd han. 9 n Hon nfla, 6 = sina + sin a sin a = 5. -^.,,.,.2. n sina Do do, hm b = sma + sin a sin a +… = :. 1-sma Giai tuong tu Vf du 13, ta cd a = 34, = i^ a) Chflng minh bing quy nap M = >n+l (1) Vdi /2 = 1, mdt hinh vudng dflge tao thdnh ed didn tfch Id MJ = -.

175 Vdy (1) dung. – Gia sfl cdng thflc (1) dung vdi n =fe (fe > 1), nghia la M^ = ; ;- Ta cdn 2 chflng minh (1) dflng vdi /2 =fe+1,tfle la chflng minh M^+I = 2^+2 Thdt vdy, d bude thflfeta cd 2*~^ hinh vudng mdi mdu xdm duge tao thanh. Ong vdi mdi hinh vudng nay ta lai tao dugc hai hinh vudng mdi trong bude thfl fe+1. Tdng didn tfch cua hai hinh vudng mdi ndy trong bude thfl fe+1 bing nfla didn tfch eua hinh vudng tuong flng trong bude thfl fe. Do dd, tdng didn tfch tdt ca cdc hinh vudng mdi cd dugc trong bude thflfe + 1 ^^ “^+1 = 2 – ^ ^ ~^- ^^y ^^^ ^^””^ vdi22 =fe Ket ludn : Vdi mgi n nguydn duong ta ludn cd M = 2n+l b) Dudodn : S -> :r-5^gc khi «^ +, bay lim5 =. C/22ing/n2/2/2.-5 = Mi +M M =^ + ^ ^ = i-i^. Tfl dd, lim5 = a) – 4 ; b) +oo. x^, nlu X > /(x) = x^ – 1, nlu X < 0. a) (H.6) Du dodn : Hdm sd/(x) khdng ed gidi han khi x -> 0. b) Ld’y hai day sd ed sd hang, tdng qudt la a = vab =. Ta ed, a -> 0 vd & -> 0 khi/2->+00. (1) ^- 173

176 Vi->0 nen /(a )= ^ 2 Do dd, lim /(<3 ) = Um = 0 (2) n >+oo n-^+<x> n Vi,-<0 ndn f{b ) = -L-i. Dodd, lim f{b^)= lim f4-i = -1. (3) n-^+<x>n ) Tfl (1), (2) va (3) suy ra/(x) khdng cd gidi han khi x > 0. n 2.3. a) Xlt hai day so («) vdi a = 2nn va {b ) vdi b^ = + 2nn {n e Ta cd, lima = lim2nn = +oo ; limz/ = lim n + 2nn ‘ lim sin a = limsin2/27i = limo = 0 ; = lim/2 ^ + 22i = +<x>; Umsin6 = lim sin + 2nn = liml = 1. V2 y Nhu vdy,a -^ +oo, b -^ +oo, nhung lim sin a ^ limsin&. Do dd, theo dinh nghia, ham sd y = sinx Ichdng cd gidi han khi x -> +oo Gia sfl (x ) la day sd’ bd’t ki thoa man x < a vd x -> -oo. Vi lim /(x) = L ndn lim /(x ) = L. Vi lim ^(x) = M nen lim g(x ) = M. x-^-co n->+oo Do dd, lim f{x ).g{x ) = L.M. n >+oo Tfl dinh nghia suy ra lim f{x).g{x) = L.M. 11A

177 2.5. a) 0 ; b) – oo ; c) lim y/ax^ – x + 1 = lim Ixlj4 – x- = lim A:->-OO V -\’-*^ = +00. d) lim (x^ + x^ + 1) = lim x^(l ) = -oo. A: >-oo ;c»-oo X x e) – 00 vd x + 3 x + 3, a) lim I ;:, = lim – – = Um r = r. X^f 3 jc^ + 2x – 3 x-^-3 {x – l)(x + 3) x-^-3 X (i+xf-i b) lim. (i+^-i)r(i+^)^+(i+^)+ii = lim _ lim L =L Jc->0 X = limx^q c) Um r(i+x)2+(i+x)+ii r.. J= ^ = Um (1 + xf + (1 + x) + 1=3. X jc^ol -I i_ J_ ^-1 = ii, l4i.o. x->+oo ;c2 1 ;t->+oo -!_ (VI – >/5)(>/I + Vs) X – 5 d) lim p: ^ = lim ^.- j- x-^5 Vx – V5 x^5 Vx – V5 = lim(>/x + V5) = 2>/ e) iim lim j=r- =?=r- j= = = Um um -^ jr-*+ao Vx + V5 X-^-KX> 1 V5 Vx X (Vi ^ + > 0 vdi moi x > 0). Vx ^ ^ =

178 -.,. Vx^ = lim f) lim x^-2, x->-2 x + 2 x^ (x + 2)fVx J = lim (-^X-^^) =Uni -^ =±. ^^-2 (X + 2) Vx^ ^^-2 Vx^ ^ VI-1,. (VI – l)(vit3 + 2) g) lun = lim -^ ,, x^yjx x^ X (Vx – l)(vx ) (Vx – l)(vx ) ~ ;r’^l X-1 ~ ^”^1 (VI-l)(VI + l),. Vx = lim 7= x-^ Vx + 1 ^,,. l-2x + 3x^ h) lun J:->+OO X 9 = = lim ^^ ^ JC->+oo i) lim ^5- ‘^ ‘ -1^ = lim^;- x-^qx^x^ +1 ‘-7 ( -x’ ^ x^ +1 = 3. = lim -1 x-^q x^ +1 O sfl -2.f X J) lin.<^’-“<‘-^^” = lim ^ f^ x’ +X + 3 = -1. = limi ^ ar->-oo = (-2)^ = -32. X X 2.7. a) lun – = lun J -^ = lun!! -^ = Um -!! i-. r-i-9 r 4. 9 Y-I-9 9 x->+oo X + 2 jc->+oo X + 2 x~*+a2 X + 2 J:->-+OO, X = 1; 176

179 lim 47^ 3x = lim ‘1-1 X x > oo X + Z ;c_^_oo, X + 2 = lim – ^ ^ X + 2 lim – X X = -1. b) lim (x + Vx^ – X + 1) = lim jr->+ao J:^+OO lim imlx + Vx -x + l) = x->-<» = lim.x->-<» = lim x-1 X – X Jl 1 – ^ x2 i-i X i +.i-i+4 = Um X JC >+oo X + Xjl + ^r ^ l+jl-1+4 = +00 x^ – (x^ – X + 1) x-1 lim ‘, =^ = hm ^-^- X – Vx^ – X + 1 ^^- X – Vx – X + = = lim 1 1 A:->-OO 2′ X X + xjl + -7r e) lim J:->+OO = lim (V7r;_777T)= lim (f-^)-(f ^’) / x-^+<^y/x^ -x + yjx^ +1 ;c->+oo -X-1 = Um -i-i X xjl- + XJl + -r “” Jl- + jl + ^r r //I n :,. (x2-x)-(x2 + i) Um Vx” – X – Vx’^ + 1 = lim , X-^-ooV / ^ ^ ^ V x ^ – X + V x ^ + l = Um 1 1 = lim _ 2 2″ 12. BT0S&GT11 -A 177

180 2.8. a) Du dodn : lim /(x) = +00 ; Um /(x) = -00 ; Um /(x) = -00 ; x^^ x^r x^^* b) Ta ed, lim /(x) = +00 ; Um /(x) = 2 ; lim /(x) = 2. lim (2×2 _ jg^ ^ j2) = -1 < 0, lim fx^ – 5x + 4) = 0 va X – 5x + 4 < 0 vdi mgi x G (1 ; 4) ndn 2×2 _ jgj^. ^ J2 lim r = +00. ;c->l”‘ X – 5x + 4 Vi lim (2×2 _ j^^ ^ ^2) = -1 < 0, lim (x2-5x + 4) = 0 vi- V / v_».i- ‘ JC->1 ^ ‘ x-> va X – 5x + 4 > 0 vdi moi x < 1 nen 2x^- 15x +12 Um = -00. x^r jt^r X x2-5x + 4 Vi Um (2×2 _ j5^ ^ ^2) = -16 < 0, Um (x2-5x + 4] = v_,.4+v / v_^/l+v / va x2-5x + 4 > 0 vdi mgi x > 4 nen,. 2×2 _ j5^ ^ j2 Um = -CO. x-*a^ X – 5x + 4 Vi Um (2×2 _ j5^ ^ ^2] = -16 < 0, lim (x2-5x + 4) =.r->4 ^ ‘ x->a va X – 5x + 4 < 0 vdi mgi x G (1 ; 4) nen 2×2 _ j5^ ^ j2 lim = +00. J:->4″ X – 5x ^ + 12,. 2×2-15x + 12,. -x v2, lun = lun ^ = 2; j._>4<x) X – 5x + 4 ^^-w 1 _, _1_ X ^2 2 X ^,. 2×2-15x + 12,. X y^ ^ lim = lun ^ = 2. Jc->^» X – 5x + 4 jc->-^ 1 4. _ ^ x BTBS&GT

181 2.9. Um /(x) = lim x-1 3 ] = lim x^ + X – 2 x^l^ (X – 1)(X + X + 1),. (x-l)(x + 2),. x + 2 = lim ^^ -^ = lim z =1. x^t (x – 1)(X + X + 1) x^l* X + X + 1 Um /(x) = lim (mx + 2) = m + 2. x->r x->r fix) ed gidi han khix->l<=>/n + 2=l<=>/?2 = -l. Khi dd lim /(x) = 1. x->l Vi lim /(x) = +oo nen vdi day sd (x ) bd’t ki, x e ^ {XQ} va x -^ XQ ta X^XQ ludn cd lim /(x ) = +oo. n->+oo Tfl dinh nghia suy ra/(x ) cd thi ldn hon mdt sd duong bd’t ki, kl tfl mdt sd hang ndo dd trd di. Ndu sd duong nay la 1 thi /(x ) > 1 kl tfl mdt sd hang ndo dd trd di. Ndi each khdc, ludn tdn tai ft nhd’t mdt sdx^ e K {XQ} sao cho/(x^) > 1. Ddt c = x^, ta COfie) > Vi lim /(x) = -oo ndn vdi day sd (x ) bd’t ki, x > a va x ^ +oo ta ludn jr->+oo cd lim /(x ) = -00. n-*+oo Do dd lim [-/(x )] = +oo. n->+oo Theo dinh nghia suy ra -fix ) cd thi ldn hon mdt sd duong bdt ki, kl tfl mdt sd hang ndo dd trd di. Ndu sd duong ndy la 2 thi -/(x ) > 2 kl tfl mdt sd hang nao dd trd di. Ndi cdch khdc, ludn tdn tai ft nhd’t mdt sd x^ e {a ; +QO) sao cho -/(x^) > 2 hay/(x^) < – 2 < 0. Ddt c = x^, ta CO/(c) <

182 3 5 1 ^^/^ (-^-1)^ [x-1, neux> a)/(x) = ‘-> = < X [1-x, ndu x<0. Ham sd nay cd tdp xdc dinh la R {0}. b) Tfl dd thi (H.7) du dodn fix) Udn tue tren edc khoang (-oo ; 0), (0 ; +oo), nhung khdng lidn tuc tren R. Thdt vdy, – Vdi X > 0,fix) = X – 1 la ham da thflc nen uen tue tren R, do dd Udn tue tren (0 ; +oo). – Vdi X < 0, fix) = 1 – X cung la ham da thflc nen uen tuc tren R, do dd Udn tuc tren (^;0). Hinh 7 Di thdy ham sd gian doan tai x = 0, vi lim /(x) = -1, lim /(x) = 1. x^o* x^0~ x + 2, nlu x< Xet ham s6fix) = – -^ 1, neu x>0. Ix Trudng hap x < 0. fix) = X + 2 Id hdm da thflc, lidn tue tren R, nen nd lidn tue tren (-2 ; 0]. Trudng hap x > 0. /(x) = -7 la ham sd phdn thflc hiiu ti ndn lidn tuc trdn (0 ; 2) thude tdp X xdc dinh cua nd. Nhu vkyfix) lidn tuc tren (-2 ; 0] vd tren (0 ; 2). Tuy nhidn, vi lim /(x) = lim -r- = +00 nen ham sd fix) khdng cd gidi x^o* x^o” X han hiiu han tai x = 0. Do dd, nd khdng lidn tuc tai x = 0. Nghla la khdng lidn tuc tren (-2 ; 2). 180

183 33. Vi ham sd Uen hie tren {a; b] nen lien mc tren {a; b) va lim /(x) = f{b). (1) x->b Vi hdm sd Uen tue tren [b; c) nen Uen tuc tren {b; c) vd lim /(x) = /(6). (2) Tfl (1) vd (2) suy ra/(x) lien tue tren cdc khoang {a ; b), {b ; c) vd lien tuc tai X = 6 (vi lim /(x) = f{b)). NghTa la nd lidn tuc tren {a; c). x-^b 3.4.Ddtg(x)=^^^^^^/^-L. X – XQ Suy ra g{x) xdc dinh tren {a;b) {XQ} vd Um g{x) = 0. Mat khdc,/(x) =/(xo) + L{x – XQ) + (X – XQ) g{x) ntn lim /(X)= lim [/(xo) + L(x – XQ) + (x – XQ) g{x) X~*XQ X->XQ = Um /(XQ) + lim L(x – XQ) + lim (x – XQ). lim g{x) = /(XQ). X->.»^ ^->^ X^XQ X-^XQ Vdy hdm sd y =fix) lien tuc tai XQ a) Ham sd/(x) = Vx + 5 cd tdp xdc dinh Id [-5 ; +oo). Do dd, nd xae dinh trdn khoang (-5 ; +oo) chfla x = 4. Vi lim /(x) = lim Vx + 5 = 3 = /(4) nen/(x) lien tuc tai x = 4. x->4 A:->4 x-1., nlu x< 1 b) Hdm sd g{x) = V2 – X – 1 CO tdp xdc dinh la R. -2x, nlu x> 1 Ta cd, g(l) = -2. (1),., ^ X-l.. (X – 1)(V2 -X + 1) hm g{x) = lim, = lim-^ -z x^r x^r V2 – X -1 x^r > – x = lim (-V2 – X – 1) = -2. x^r lim g{x) = lim (-2x) = -2. ;t->l'” ;c-*r (2) (3) Tfl (1), (2) vd (3) suy ra lim ^(x) = -2 = g{l). Vdy g{x) lidn tuc tai x = 1. X->1 181

184 3.6.a)/(x) = x2-2 r-, nlu X^ yf2 X-V2 2 V2, nlu x= V2. Tdp xae dinh cua ham sd la D = R. Neux^ V2 thi/(x)= X-V2″ Ddy Id hdm phdn thflc hiiu ti nen lien tuc tren cac khoang (-00 ; V2 ) va (>^;+oo). Tai X = V2 : x2-2 lim /(x) = lim,- x^sl2 x^^ X – V2 = Um x^y/2 (X – V2)(x + V2) X-V2 = lun (x + V2) = 2yf2 = /(V2). Vdy ham sd lien tuc tai x = V2. x-^^j2 Ket ludn : y =fix) lien tuc tren R. 1-x b) g{x) = {x-2f -, ne’u X ^ 2 ed tdp xdc dinh la D = R. Neu X 5!: 2 thi g{x) =, ndu X = 2 1-x (x-2)^ tren eae khoang (-oo ; 2) va (2 ; +00). 1 -X Tai X = 2 : lim g{x) = lim ; x^2 lien tue tai x = 2. X-^2(Y- {x-2)’ la ham phdn thflc hiiu ti, nen nd lidn tuc -00. Vdy ham sd y = g{x) khdng Ket ludn : y = g{x) lien tue tren cdc khoang (-00 ; 2) va (2 ; +00), nhung gian doan tai x = m = /w=±

185 3.9. a) Xet/(x) = x^ – 3x – 7 va hai sd 0 ; 2. b) Xet/(x) = cos2x – 2sinx + 2 tren cac khodng n n ^’2″ /” n c) Ta cd, Vx^ + 6x = 0 o x^ + 6x + 1 = 4 o x^ + 6x – 3 = 0. Ham sd/(x) = x^ + 6x -3 lidn tue tren R nen lien tuc tren doan [0 ; 1]. (1) Tacd/(0)/(l) = -3.4<0. (2) Tfl (1) vd (2) suy ra phuong trinh x^ + 6x – 3 = 0 ed ft nhd’t mdt nghidm thude (0; 1). Do dd, phuong trinh Vx + 6x = 0 ed ft nhd’t mdt nghidm duong Hudng ddn : Xet fix) = x”^ – 3x^ + 1 = 0 tren doan [-1 ; 1]. Trd Idi: Cd a) (1 – mx + 1)^ + x2 – X – 3 = 0. fix) = (1 – mx + 1)^ + x2 – X – 3 la ham da thflc ndn lien tuc tren R. Do dd nd lidn tue tren [-2 ; -1]. Ta c6fi-l) = -1 < 0 vafi-2) = m^ + 2>0 n6nfi-l)fi-2) < 0 vdi mgi m. Do dd, phuong trinh/(x) = 0 ludn ed ft nhdt mdt nghidm trong khoang (-2 ; -1) vdi mgi m. Nghia Id, phuong trinh (1 – /n )(x + 1) + x – x – 3 = 0 ludn cd nghidm vdi mgi m. b) /?2(2eosx -42) = 2sin5x + 1. HD : Xet ham sd/(x) = /n(2cosx – V2 ) – 2sin5x – 1 tren doan n n 4 ‘ Hdm sd/(x) = x” + a^x^ ^ + a2x” ^ a _ix + fl xdc dinh tren R. Ta cd Um /(x) = lim (x” + ^ix””^ + 02-^””^ + + «n-i^ ^ ‘^n) J:->+OO A:->+C» = lim x”(l + ^ + ^ ^ + 5L) = +00. ^->+<x, X x^ X””^ X” Vi lim /(x) = +00 ndn vdi day sd (x ) bdt ki ma x -^ +00, ta ludn ed lim/(x ) =

186 Do dd, /(x ) ed thi ldn hon mdt sd duong bdt ki, kl tfl mdt sd hang ndo dd trd di. Nlu sd duong nay la 1 thi /(x ) > 1 kl tfl mdt sd hang ndo dd trd di. Ndi each khae, ludn tdn tai sd a sao cho fia) > 1″. (1) lim /(x) = lim {x’^+a^x”‘^ + a2x”~^ a _ix+a ) = lim x”(l + -^+ % ^ + -^)^-00 (do/2le). ;c->-<x> X x^ X””^ x” Vi lim /(x) = -00 ndn vdi day sd (x ) bd’t ki ma x -^ -oo, ta ludn cd x->-<» lim/(x ) = -00, hay lim[-/(x )] = 4-oo. Do dd, -/(x ) cd thi ldn hon mdt sd duong bd’t ki, kl tfl mdt sd hang ndo dd trd di. Nlu sd duong nay la 1 thi -fix ) > 1 kl tfl sd hang nao dd trd di. Ndi each khdc, ludn tdn tai b sao cho -fib) > 1 hay/(&) < -1. (2) Tfl (1) va (2) suy va fia) fib) < 0. Mat Ichae, ham da thflc/(x) lidn tuc tren R, nen lien tuc tren [a ; b]. Do dd, phuong trinh/(x) = 0 ludn ed nghidm Ndu ham sd y =fix) lien tue tren doan [a ; b] vafia)fib) > 0 thi phuong ttinh fix) = 0 cd thi cd nghidm hodc vd nghidm trong khoang {a ; b). Vi du minh hoa fix) = x2-1 uen tue tren doan [-2 ; 2],/(-2)/(2) = 9 > 0. Phuong trinh 2 1 X – 1 = 0 cd nghiem x = ± 1 trong, khoang (-2 ; 2). fix) = x2 + 1 lien tucfl-endoan [-1 ; 1] vafi-l)fil) = 4 > 0. Cdn phuong trinh X + 1 = 0 lai vd nghidm trong khoang (-1 ; 1) Ndu ham sd v =/(x) khdng lien tue tren doan [a ; b] rihmgfia)fib) < 0 thi phuong trinh/(x) = 0 cd thi cd nghiem hodc vd nghidm trong khoang {a;b). 184

187 Minh hoa hinh hgc (H.8) 3’i O h X a) Hinh 8 a)/(jc) = 0 vd nghidm trong {a; b); b)/(x) = 0 cd nghidm trong {a; b). Bai tap on chudng IV I. a)-2;,b)- ; 2. a) Tacd, uj = (-1)” n^ i^- 1 Dat v = -^ n^ +1 n” +1 (1) Tacd limv = lim 1 2 -^ = lim “, = 0. Do dd, v cd thi nhd hon mdt n’+l 1+1 sd duong be tuy y, kl tfl mdt sd hangnao dd trd di. Tfl(l)suyra, M = v = v. Vdy, JM cung cd thi nhd hon mdt sd duong be tuy y, kl tfl mdt sd hang ndo dd trd di, nghia Id limm = 0. b) Hudng ddn : M = 2″-/2 3″+l 2″ 3″ + l 3. 2, = 2 + ill + ^ ^ iooo2 1000” 131 ^2 + ^M_ = 2 + Hl = ^i^

188 (Vi, -,…,,… la mdt cdp sd nhdn lui vd han vdi cdng bdi 1000 iooo2 1000″ 1 s ^ 1000^’ 4. a) Chiing minh bing quy nap : M > 0 vdi mgi n. (1) – Vdi /2 = 1, ta cd Ml = 1 > 0. – Gia sfl (1) dung v6i n = k> I, nghia Id M^ > 0, ta cdn chflng minh (1) dung’ ^’di n = :fe+l. Tacd “A+1 _2M, Ket ludn ; M > 0 vdi mgi n. h) Ddt limm = a. 2M^ +3 ViM^>OnenM;t+i=-^^^>0 2M +3,.,. 2M + 3 2a + 3 _, /r u j.i ‘*n+l – = ” r-,,, 9 => -^ limm,, “‘””n+l M +2 “+^ = lim – M +2 => a = a+ 2 => a = ±V3. “n ^ ^ Vi M > 0 vdi mgi n, ndn limm = a > 0. Tfl dd suy ra limm = Vs. 5. Xet day sd {y ) vdi v^ = M – u. > M < M ydi mgi/2 => v > 0 vdi mgi/2. (1) Mat khae, limv = Um(M – M ) = M – a. (2) Tfl (1) va (2) suy ra M – a > 0 hay a < M. 6. Mdi khi cham ddt qua bdng lai nay ldn mdt dd cao bing dd cao eua ldn roi ngay trudc dd vd sau dd lai roi xudng tfl dd cao thfl hai nay. Do dd, dd ddi hanh trinh cua qua bdng kl tfl thdi dilm roi ban ddu dln_ – thdi dilm cham ddt ldn thfl nhd’t la ^i = 63 ; – thdi dilm cham dd’t ldn thfl hai la ^2 = ; thdi dilm cham dd’t ldn thfl ba la rfj = ‘102 ‘

189 #^1 ^”3 ^’2 – thdi dilm cham ddt ldn thfl tu la t/4 = ^ + 2. r- ; ^ thdi dilm cham ddt ldn thvtn{n>l) la ^,,, 63 ^ 63 ^ 63 <i = ^ ‘ “-^’ (Cd thi chiing minh khing dinh nay bing quy nap). Do dd, dd ddi hdnh trinh eua qua bdng kl tfl thdi dilm roi ban ddu ddn Ichi nim yen tren mat dd’t la : flf = ^ + 2.-^ ^ +… (met). 10 io2 io«-i /TO rfto /TO 1 Vi 2. -,2. :r,…,2. -,… la mdt cdp sd nhdn lui vd han, cdng bdi o = -, 10 io2 io”-i *^ ” ^ 10 2^ nen ta ed 2.^ + 2.-% ^ +… = -If = io2 io«-i 1 ± Vdy, rf= ^ + 2.-^ ^+…= = 77 (mit). 10 io2 io”-i 7. Hudng ddn : Chgn hai day sd ed sd hang tdng quat Id a = – vd b =. Tinh vd so sdnh lim/(a ) vd lim/(6 ) dl kit ludn vl gidi han eua/(x) khi x -> a) -3 ; b) 6 ; c) +00 ; d) a)4 ; b)l ; c) 2 ; d) ; x(x2 +l-x2j / r~2 ” +1 ~-“c I X e) lim X Vx +1-X = lim ^. = lim, x-^+co ) ^^+<» V;c2+l+x ^^^;j. 1 + i.-*” 10 = lim, = -. JC->+<o / 1 Z 187

190 f) lim jc-^2″‘ VX X-2 = lim 1 – (x + 2) JC-^2^ X^ – 4 = lim – x^t” x2-4 = 00. 2x Chang han/(x) = -. Di dang kilm tra duoc ring/(x) thoa man cae (x -1)^ dilu kidn da neu. 11. Ham sd lidn tuc tren R. 12. Hudng ddn : Ching han xlt /(x) = X, ndu x>0 x-1, ndu X< Hudng ddn: a) Xlt ham sd fix) = x – 5x -1 tren cdc doan [-2; -1], [-1; 0] va [0; 3]. b) Xet ham sd fix) = m{x – lf{x^ – 4) + x”^ – 3 ttln eae doan [-2; 1], [1; 2]. c) Xet ham sd fix) = x -3x r- m tren edc doan [-1; 1], [1; 2]. 14. a) Vdi X 9i 2 ta cd ^ + ^^ + ^ = 0 o x^ + 8x + 1 = 0. x-2 Vi x^ + 8x + 1 > 0 vdi mgi x G (1 ; 3) ndn phuong trinh x^ + 8x + 1 =0 khdng cd nghidm trong khoang (1 ; 3). Do dd, phuong trinh/(x) = 0 khdng cd nghidm trong khoang ndy. b)/(x) Id hdm phdn thflc hiiu ti, nen lidn tue tren (-oo ; 2). Do dd, nd liln tue trdn [-3 ; 1]. Mat khdc,/(-3)/(l) = -100 < 0. Do dd, phuong tnnh fix) = 0 ed nghidm trong khoang (-3 ; 1). 15. Xet ham sd g{x) = fix) -/(x + ^ ). Ta ed g{0) =/(0) -/(O + ) =/(0) -/(i), 188 ^( )=/( )-/( + )=/( )-/(l)=/( )-/(0) (vi theo gia thidt/(o) =/(l)).

191 Do dd, g{0) g(l) = [fio) -fi^)]fi^) -fio)] = – [fio) -fi^ )f < 0. – Nlu g{0) g( ) = 0 thi X = 0 hay x = Id nghiem cua phuong trinh g{x) = 0. -Nlug(0)g( )<0. (1) Viy =fix) vay = fix + ) diu lien tue tren doan [0 ; 1], nen ham sdy = g{x) cung lidn tuc trdn [0 ; 1] vd do dd nd lidn tuc trdn 0; (2) Tfl (1) vd (2) suy ra phuong trinh g{x) = 0 ed ft nhd’t mdt nghidm trong khoang (0 ; -). Kit ludn : Phflong trinh g{x) = 0 hay/(x) -fix +-r-) = 0 ludn ed nghidm trong doan [0 ; ]. :,’h Dap an Bai tap trac nghiem 16. (C). 17. (B). 18. (C). 19. (D). 20. (A) 21. (D). 22. (B). 23. (B). 24. (D). 25. (D) 189

192 huang V. DAO HAM 1. Djnh nghla va y nghla cua dqo ham A. KIEN THUC CAN NHO 1. Djnh nghta Cho ham sd y =fix) xde dinh trdn khoang {a ; b), XQ e {a ; b), XQ + Ax e {a; b). Ndu tdn tai, gidi han (hflu han) lim Ax->0 /(XQ + AX) – /(XQ) duge ggi la dqo hdm cua fix) tai XQ, kf hidu la/'(xo) hay y'(^o) Ax f'{xq)= Um /(XQ + AX)- /(XQ) _ /(X) – /(XQ) Um Ax X->XQ X X( 0 2. Quy tac tfnh dao ham bing djnh nghta Bude 1 : Vdi Ax la sd gia cua ddi sd tai XQ, tinh Av Bude 2 : Ldp ti sd -^ ; Ax, Bude 3 : Tfnh Um 4^. Ax^oAx Ay =fixo + Ax) -/(XQ) ; ^ Chu y : Trong djnh nghta va quy tac tren day, thay Xo bdi x ta se cd djnh nghta va quy tac tinh dao ham ccia ham soy =fix) tai diem x e (a; b). 190

193 3. Quan he giura tinh lien tuc va sir co dao ham fix) cd dao hdm taixfl < ^ fix) lien tuc taixg 4. Y nghta hinh hoc cua dao ham Ndu tdn tai,/(xo) la he sd gde eua tidp tuyin cua dd thi ham sd y = /(x) tai MQ (XQ ;fixf^). Khi dd phuong trinh tidp tuyd’n eua dd thi ham sd tai MQ Id >’-3’o=/'(^o)(^-^o)- 5. Y nghta vat li cua dao ham v{t) = s'{t) Id vdn tdc tflc thdi cua chuyin ddng s = 5(0 tai thdi dilm t. B. VI DU %Vidu1 Bing dinh nghia, hay tfnh dao hdm eua ham sd y = V2x -1 taixo = 5. Gidi Tdp xae dinh cua hdm sd da cho la D = -j x x > Vdi Ax la sd gia eua ddi sd tai XQ = 5 sao cho 5 + Ax G D, thi Tacd Khidd Ay = V2(5 + Ac)-1 – VlO – 1. Ay Ax V9 + 2Ax – V9 Ax y(5) = um 1^ = lim (V?T2A^-3X^^^^ Ax^o Ax ^x^o Ax(V9 + 2Ax + 3) 9 + 2Ax – 9,. ^ = lim, = hm AA:->0 AC(V9 + 2AX + 3) Ax-^o V9 + 2Ax + 3 3′ 191

194 Vidul Cho hdm sd ^'” 2 X + X X-2- a) Hay tfnh (bing dinh nghia) dao hdm cua ham sd da cho tai x = 1. (^) b) Vidt phuong trinh tie’p tuydn eua (^) tai dilmi4(l; -2). ‘ Gidi a) Vdi Ax la sd gia cua dd’i sd tai x = 1, ta ed Ay = Ay Ax (1 + Ax)^ + (1 + Ax) l + Ax Ax + Ax Ax + 2 Ax Ax + Ax2 + 2Ax – 2 5Ax + Ax2 Ax-1 Ax Ax Ax-1 ‘ lim = lim = -5. Ax-*0 Ax Ax->0 Ax-1 Vdy y(l) = -5. b) Phuong trinh tilp tuydn eua C^) tai A(l ; -2) la y + 2 = -5(x-l) hay y = -5x + 3. Vi du 3 (x – 1)2, ndu X > 0 Chflng minh ring ham s6fix) = (x + 1), ndu x<0 khdng ed dao hdm tai x = 0, nhung lidn tue tai dd. 192

195 a)tacd/(0) = 1. Gidi Trudc hit, ta tfnh gidi han ben phai cua ti sd i. Ta cd x-0 /(x)-/(0) ^ (X-1)2-1 dodd x-0 X X – 2x X = x-2(vdi x^o), Um^<^>-{‘ >=lim(.-2) = -2. JC->0″^ X-0 x^o* Sau dd ta tfnh gidi han bdn trdi: x)-f lim ^W-f””^ lim ‘”‘>’-‘ x^o’ X-0 x-^0 = Um (x + 2) = 2. x^0~ Vi gidi han hai ben khdc nhau nen khdng tdn tai gidi han cua ti sd khi X -> 0. Dilu dd chflng td hdm sd y = fix) Ichdng cd dao ham tai x = 0. b) Vi lim /(x) = lim (x – 1)2 = 1, Um /(x) = lim (x + 1)2 = 1 x->0″ x->0 vd /(O) = 1 ndn hdm sd/(x) lien tuc tai x = 0. Vi du 4 feosx, neu x > 0 Chflng minh rang hdm sd y = g{x) = < [-sinx, neu x <0 khdng cd dao ham tai x = BTBS&GT 11-A 193

196 Gidi Vi Um ^(x) = Um cosx = 1, x-^o* x-^0* Um g{x) = lim (-sinx) = 0, x-*0~ x->0″ g{0) = coso = 1 ndn ham sd y = g{x) gidn doan tai x = 0. Do dd ham sd ndy khdng ed dao hdm tai dilm x = 0. C. BAI TAP 1.1. Sfl dung dinh nghia, hay tim dao ham eua edc ham sd sau : a) y = 3x – 5 ; h)y = x’-9; 2 d)y=v3x + l ; c) y = 4x – X ;, Cho/(x) = 3x^ – 4x + 9. Tfnh/'(I) Cho/(x) = sin2x. Tfnh /’ v4y 1.4. Cho/(x) = Vx-1. Tfnh/'(0),/'(l). g 1.5. Cho ^x) =. Chflng minh ring ^(-2) = ^(2) Chflng minh ring hdm sd y = x – l khdng ed dao hdm tai x = 1, nhflng lidn tue tai dilm dd Chflng minh ring ham sd 1, ndu X > 0 y = signx = < 0, ndu X = 0 -^Lnlux <0 khdng cd dao ham tai x = BTDS&GT11-B

197 1.8. Vie’t phuong trinh tidp tuydn cua dd thi cua cdc ham sd X + 4x + 5 a) y = – tai dilm cd hoanh dd x = 0 ; ^ x + 2 b) y = x^ – 3×2 + 2 ^^. ^.^j^ ^_ J. _2^. c) y = V2x + 1, bidt he sd gde cua tidp tuydn la Cac quy tac tfnh dqo ham 1. Cong thurc A. KIEN THUC CAN NHO (c)’ = 0 / nx, n-l (X ) = /2X (VI)’ = -^ 2Vx (c = const); {n G N*, X G R); (x > 0). 2. Phep toan {U + V-W)’ = U’ + V’-W ; {uvy = irv + uv’; {ku)’ = ku’ ^U_y V U’V-UV’ {k = const); {V^O); V XL y2 3. Dao ham cua ham hop yx = yu-u. 195

198 Vidu 1 Tim dao ham cua hdm sd B. VI Dg y = (4x^ – 2×2-5x) (x2-7x). Gidi Ap dung cae cdng thflc vd phep todn dao hdm ta duge : y = (4x^ – 2×2 – g^y ^^2 _ ^^y ^ ^4^3 _2x^_ 5^) (^2 _ ^^y = (12×2 _ 4^ _ 5) (^2 _ ^^^ ^ ^4^3 _ 2^2 _ 5^^ ^2x – 7) = llx”^ – 84x^ – 4x^ + 28x^ – 5×2 ^ ^^^ + gjc”^ _ 28x^ – 4x^ + 14x^ – IQx^ + 35x = 20x’* – 120x^ + 27×2 + 70x. Vidu 2 Gidi Vdy y = y = V V + 3x I (Vx – l) + [ – + 3x j(vx – l)’ – MH^x- X X + 3]{4’x-l) 1 3x xvx 2Vx + 7= + – Vidu 3 Hm dao ham cua ham sd -x^ + 2x + 3 x

199 Gidi Ap dung quy tic tfnh dao hdm eua mdt thuong, ta dugc y Vdy y’ = (-x2 + 2x + 3)'{x^ – 2) – (-x2 + 2x + 3)(x^ -2)’ (x^ – 2)2 (-2x + 2)(x^ – 2) – (-x2 + 2x + 3).3×2 (x3-2)2-2x’^ + 2x^ + 4x (-3x’^ + 6x^ + 9×2) (x3-2)2 x^ – Ax^ – 9×2 + 4x – 4 (x3-2)2 V//7H4 Tim dao ham eua ham sd y-= {x- 2)7x Gidi (x-2)(x2+l)’ y = (X- 2)’>/?7i + (x-2)fv77tl’ = V77I + ^^”^^^^ -^ ^ ^ 2Vx2+l o ‘ n 7 -^U – 2) 2×2 _ 2JC +1 Suy ra y’ = Vx^ ^ 1 =, Vx2 +1 Vx2 + 1 C. BAI TAP Tim dqo hdm cua cdc hdm sd sau ( ) : It 5 ^ 3 2 X 2.1. y = x -4x -X + -;z y = -9x^ + 0,2x^-0,14x , 2 4^ y= T + r r- Jf x2 x^ 7x’* 2.4. y = -6Vx

200 2.5. y = (9-2x) (2x^ – 9×2 _^ ^^ 2.6. y = 2x-3 x + 4 ‘ 2.7. y = 5-3x – x^ x y = (x2 + 1) (x^ + 1)^ (x”* + if. y = x^-1 ytx y = b c a + + ^r X r^ {a, b,c la cdc hing sd). IL y = Vx^ – 2x Rflt ggn V x-1 /(^) = + 1 2(Vx + l) ‘ ^J’Vx + 1 ‘ va tim/'(x). Vx-2 Vx Vx Cho fix) = x^ + x^ – 2x – 3. Chflng minh ring /'(I) +/'(-l) = -4/(0) Cho/(x) = 2x^ + X – V2, g{x) = 3x^ + x+ yfi. Giai bdt phuong trinh /'(x) > g ‘(x) Cho/(x) = 2x^ – x2 + V3, g(x) = x^ + ^ – V x-2./ x2-4 – X + 2 j Giai bd’t phuong trinh /'(x) > gx) Cho/(x) =-, g{x)=^-^. Giai bd’t phuong trinh/(x) < gx) Tfnh/'(-l), biet ring/(x) =- + + ^. X X X Tfnh g'{), biet rang g{x) = – + -^ + x^. X y/x 198

201 2.19. Tfnh h'{0), bidt ring h{x) = ^ V 4 ^ (x-2)(8-x) Tinh <p'{2), bilt ring (p{x) = Chiing minh ring nlu S{r) Id dien tfch hinh trdn ban Icfnh r thi S'{r) Id chu vi dfldng trdn dd Chflng minh ring nlu V{R) Id thi tfch hinh cdu ban kfnh R thi V'{R) Id dien tfch mdt cdu dd Gia sfl V Id thi tfch hinh tru trdn xoay vdi ehilu cao h va bdn kfnh ddy r. Chflng minh ring vdi r Id hing sd thi dao ham V'{h) bing didn tfch day hinh tru vd vdi h la hing sd thi dao hdm V'{r) bing didn tfch xung quanh cua hinh tru Dqo ham cua cac ham so luong giac A. KIEN THUC CAN NHO sinx lim = 1. x^o X (sinx)’ = cosx ; (cosx)’ = -sinx. (tanx)’ = r ; 1 (cotx)’… = 1 2 ‘ ^ ‘. 2 COS X sm X B. VI DU Vidul Tim dao hdm cua cdc hdm sd a) y = sin3x + cos + tan Vx ; 2 a b) y = sin(x – 5x + 1) + tan. X 199

202 I X Gidi a) Ta cd y’ = 3eos3x – sin + ^ 5 2Vxeos Vx ^a^’ h) Ta cd y’ = (x^ – 5x + 1)’ cos (x2-5x + 1) + Vf du 7 Hm dao ham eua eae 2fl COS X = (2x – 5) eos(x – 5x + 1) a 2 2 «X cos X a)y b)y c)y ham sd. 1 = sm r- ; x^ = Vxeot2x ; 2 2 = 3sin X cosx + cos x. Gidi a) y’ = sm V x’-) f 1 A cos 7- = r-cos-y x^ x”^ x” Vx2y b) y’ = (Vx)’cot2x + Vx(cot2x)’ = ;=cot2x – 2Vx c) y’ = 3(sin x)’cosx + 3sin x (cosx)’ + (cos x)’ 2 3 = dsinxcos x – 3sin x – 2cosxsinx = sinx (6cos x – 3sin x – 2eosx). 2Vx sin 2x C. BAI TAP Tim dqo hdm ciia cdc hdm sd sau ( ) : 3.1. y=vtan^x y = eosi T” – 5x 200

203 3.3. y = sinx y = cos x y = tan X – cotx. cost. n 3.6. f{t) = ^-tait= -. ‘^ l-smt y=vx+^ + 0,lx^. Vx J.O. y 2x + X + 1 x2 – X cos 0 + sin g{(p)= -^ ^. I – COS(p y-= (l+3x + 5xV y = (3 – sinx)^ y = sin 3x + r cos X 3.13.y=Vl + 2tanx y = COtVl+x y = yjx + yjx + y/x Cho/(x) = 5×2 _ jg^ ^ -, Tfnh/'(l),/'(4),/’ v4y Giai phuong trinh/'(x) = 0, bidt ring ^ x/ ^ a ^ a)/(x)=3x + r- + 5; ^ X sin3x f smjx /r sm X + cos3x o)fix) = – + cosx – yl3 V Giai cdc bdt phuong trinh ^1} a)/'(x) > 0 vdi fix) = l-x’ – ^x”^ + 8x – 3 ; b)g'(x)<0 vdi g(x) = c) ^'(x) < 0 vdi ^x) = X – 5x + 4 x-2 2x-l x2+r 201

204 3.19. Xdc dinh m di bd’t phuong trinh sau nghidm dflng vdi mgi x e R a)/'(x)>0 vdi f{x) = ^ 3x^+mx-5; 3 2 b)^'(x)<0 vdi g{x) = ^ – ^ + {m + l)x-l Chfltig minh ring/'(x) = 0 Vx G R, nlu : a)/(x) = 3(sin x + cos x) – 2 (sin x + cos x); b)/(x) = cos X + 2sin x cos x + 3sin x cos x + sin x ; n C) fix) = COS X – – leos 7t 1 ( n ;, + _ +COS x n 1 2 2n – d)/(x) = cos X + cos I ‘f I + cos V Tim/'(l),/'(2),/'(3) ndu Tim/'(2) ndu /(x) = (x-l)(x-2)2(x-3)l fix) = x2sin(x – 2). X eos X Cho y = ^ + ^ – 2x. ^3 2 Vdi nhiing gid tri ndo eua x thi: a)y'(x) = 0; b)y'(x) = -2; c)y'(x) = 10? Tim dqo hdm cua cdc hdm sd sau ( ) : y = a + 5a x – x y = {x – a) {x – b) y = ax + b y = (x + 1) (x + 2)2 (x + 3)^ a + b ‘ y = (xsina + cosa) (xeosa – sina) y = (1 + nx'”) (1 + mx”) 202

205 3.30. y = (1 – X) (1 – x2)2 (1 – x^f y = 1 + X – X 1 – X + X 2′ ^ y = (l-x)2(l + x) y = (2-x2)(3-x^) (1-X) y=xvl + x y = Va2 – x y = (2 – X ) cosx + 2xsinx y = sinx – xcosx cosx + xsinx’ y = sin(eos x). cos(sin x) y = tan – cot. ^ 2 2 X X y = tanx – ^tan x + -tan x. 4. VI phan A. KIEN THUC CAN NHd Djnh nghta Cho ham sd y =fix) xde dinh tren {a ; b) va ed dao ham tai x G {a ; b). Gia sfl Ax la sd gia cua x sao cho x + Ax e {a ; b). Tfch/'(x)Ax (hay y’.ax) dugc ggi la vi phdn eua ham sd/(x) tai x, ling vdi sd gia Ax, kf hidu la dfix) hay dy. ^ Chu y; Vi dx = Ax ndn dy = dfix)=f{x)dx. 203

206 B. VI DU Vi du 1 Tim vi phdn eua cdc hdm sd a) y = sinx – xcosx; b) y = 4. Gidi a) Ta cd y’ = cosx – cosx + xsinx = xsinx, do dd dy = {xsinx)dx. 3 b) Vi y’ = – ndn ta cd x^ 3dx Vidu 2 Gidi ^, d(svnx) (sinx)'(ix cosx (, n, Ta ed -7 7 = 9 ^7^ = : = -cotx x ^ k-,k a(cosx) (cosx) ox -sinx v 2 C. BAI TAP 4.1. Cho ham sd fix) = x – 2x+ 1. Hay tfnh A/(l), dfil) vd so sdnh chung, ndu a) Ax = 1 ; b) Ax = 0,1; e) Ax = 0,

207 Tim vi phdn cua cdc hdm sd sau ( ) : 4.2. y = 4-. x’^ 4.3. y = x + 2 x-r 4.4. y = sin x Tim <i(tanx) d{cot x)’ 4.5. y tan Vx Vx 4.7. Chflng minh ring vi phdn dy va sd gia Ay cua ham s6y = ax + b trung nhau Chflng minh ring vdi Ixl rd’t bl so vdi a > 0 (Ixl < a) ta ed 4~a’ + X» a + – (a > 0). 2a Ap dung cdng thflc trdn, hay tfnh gdn dflng edc sd sau : a) Vl46 ; b) V34 ; e) Vl Dqo ham cdp hai LDinh nghta A. KIEN THCC CAN NHd Gia sfl ham sd’/(x) ed dao ham/'(x). Ne’u/'(x) cung cd dao hdm thi ta ggi dao ham cua nd Id dqo hdm cdp hai cua fix) vd kf hidu la/”(x): Tflong tfl (f'{x)y =/”(x). (3). {f”{x)y=r{x)hoacf”{x) {/” %)y =/”x) ;nen* d ddy kf hidu/ x) =/(x);/%) la dao ham cdp n eua ham sd/(x). 205

208 2. Y nghta co hoc cua dao ham cap hai Dao ham cd’p hai f”{t) la gia tdc tdc thdi eua chuyin ddng s = fit) tai thdi dilm t. Vidu 1 Tfnhy”, bilt ring a)y = b)y = B. VI DU xvl + x2 ; tanx. Gidi a) y = vr77+-^ l + 2x^ y = Vl + x ?-^(i±l!l vr77 l + x2 4x(l + x2)-x(l + 2×2) (1 + x2)7 1 + X^ x(3 + 2×2) (1 + x2)7 1 + X^ 1 b) / = 2″‘ ^”y ^^ cos X (cos x)’ 2 cos xsinx y = 4 COS X 2sinx ‘^ cos^x V 4 cos X Tt X 9^ + A:7t, ^ G Vl + x2 ; suy ra yirfm2 Cho/(x) = (2x – 3)’. Tfnh/”(3),/”‘(3). 206

209 Gidi Tacd /'(x) = 5.2 (2x- 3)”^ = 10 (2x- 3)^ /”(x) = 80 (2x – 3)^ ; /” (x) = (2x – 3f = 480 (2x – 3)2. Tfl do /”(3) = 80.3^ = 2160 ; /”‘(3) = = C. BAI TAP Tim dqo hdm cdp hai ciia cdc hdm sd sau ( ) : 5.1. y = sin5xeos2x y = x2-l 5.4. y – y- 2x + l x2 + X – 2 x + 1 x-2″ 5.5. y = X sinx y = xvl + x y = (1 -x2)cosx y = sinx sin2x sin3x y = ytx. y- x2 1-x’ y = xcos2x y = “yf^- Bai tap on chudng V 1. Tim dao ham eua cae ham sd sau :, /2,v sinvx a) y = xcot X ; b) y = ; cos3x c) y = (sin2x + 8)^ ; d) y = (2x^ – 5)tanx. 207

210 2. Tim dao hdm cua hdm sd tai dilm da chi ra : a)/(x)=, ‘ /'(0) =? Vx b)y = (4x + 5)^ y'(0) =? c) g{x) = sin4x cos4x, g f- V 3. Chflng minh ring/'(x) > 0 Vx G R, ndu a)/(x) =^x^ – x^ + 2x^ – 3×2 + 6x – 1; b)/(x) = 2x + sinx. 4. Xdc dinh a dl/'(x) > 0 Vx G R, bilt ring fix) =x +{a- )x- + 2x Xdc dinh a di g'(x) > 0 Vx G R, bilt ring g{x) = sinx – asin2x – sin3x + 2ax. 6. Tim he sd gde eua tilp tuyin cua dd thi ham sd y = tanx tai dilm ed hoanh ddxo=. 7. Trdn dudng cong y = 4x – 6x + 3, hay tim dilm tai dd tilp tuydn song song vdi dudng thing y = 2x. 8. Dd thi ham sd y = 7=sin3x eit true hoanh tai gde toa dd dudi mdt gde bao V3 6 nhidu dd (gde gifla true hodnh vd tidp tuyin cua dd thi tai giao dilm)? 9. Cho edc ham sd fix) =x + bx- + cx + d; {^) 208 g{x) = X – 3x – 1.

211 a) Xde dinh b, c, d sao cho dd thi (“g) di qua eae dilm (1 ; 3), (-1 ; -3) va /’ r^ v3y 3′ b) Vilt phuong trinh tilp tuyin cua C^ tai dilm cd hoanh dd XQ = 1 ; e) Giai phuong trinh /'(sinf) = 3 ; d) Giai phuong trinh f”{cost) = g'(sino ; ^^T…-u, /”(sin5z) + 2 e) um gioi han lim,,. ^,. ^ z^og'(sm3z) Chflng minh ring tidp tuydn cua hypebol y = ldp thanh vdi eae true toa dd mdt tam gidc cd didn tfch Ichdng ddi. 11. Chiing minh ring ndu hdm sd’/(z) ed dao ham din cdp n thi [f{ax + b)f”^ =a”f,^”hax + b). ^2 Ldl GIAI – HUdNG DAN – DAP SO CHUONG V a)y’ = 3; d)y’ = 2V3x +1 ‘ 1.2. /'(I) = /’ ^n} v4y = /'(O) = ^ ; khdng ed/'(l). b) y’ = 2x ; e) y = -1 {x-2f e) y’ = 4-2x ; 14. BTDS&GTif-A 209

212 1.5. (px) = -4-nIn ^'(-2) = <p'{2) = -2. x^ 1.6. HD. XemVfdu HD.XemVfdu a) y = -x + – ; b) y = 9x + 7 ; X 5 c)y= y 5x – 12x – 2x y’ = -27x^ + 0,4x-0, y x2^x x^”‘7x y’ VI 2′ 2.5. y’ = -16x^ + 108x^ – 162x y’ 11 2 (x + 4> 2.7. y -x’^ + 4x + 1 {x-2f 2 / y’ = 2x (x” + 1)” (x”* + 1)^ + 6x^ (x^ + 1) (x^ + 1) (x’* + 1)’ + 12x^ (x2 + 1) (x^ + 1)2 (x^* + 1)^ y A2 = 3 x^ Vx ^ 4 3 5x + V? 14. BTBS&GT 11-B 210

213 f- b cf(b 2c ^ y’ = -4 a ^ 3, Vx x J V ^ x y = 3x – 4x l^jl 2x^ /(x)=^^; /'(x)=- ^”^ x^ -4 {x’ -4) Ban dgc tu ehiing minh {-oo ; 0) u (1 ; +oo) (-00 ; 0) u (1 ; +oo) [-1 ; 0) i f Vi S{r) = nr ndn 5′(r) = 2nr Id chu vi dudng trdn Vi V{R) =^nr^ ndn V'{R) = 47t/?2 Id dien tfch mat cdu Vi V = 7:r2;2 ndn Vh) = nr la didn tfch day hinh tru ; Vr) = 2nrh la dien tfch xung quanh cua hinh tru y’ 3tan X 2 cos xv tan x 3.2. y’ 10 sin fi-5.1. V” J cos – 5x.6 ; 211

214 , 2x cosx – sinx y’ = sm 3.4. y’ = x + 1 {X +1) 2 2sinx 2x 3.5. y = cos X sm X – sin t{l – sin 0 + cos t 3.6. /'(0 = {l-sintf 1 ^;Dod6 /’ l = sinr 3.7. y’ = 2Vx X. 2xVx 3.8. y’ = -3×2 + 2x + 2 (x2 – X + 1) g(p) = cos^ – sin^ – 1 (1 – cos^)2 2x y = 4(1 + 3x + 5x^)^ (3 + lox) y = -3(3 – sinx) cosx y = 3sin6x + 2sinx cos X y = y = 1 Vl + 2 tanx.cos x -X Vl + x2 sin2 Vl + x y = 2Vx + yjx + yjx Vx + Vx V 2Vx 3.16./'(1) = 2, /'(4) = 36, 212 /’ 27 2

215 3.17. a) {+2, ±4} a) x< l,x>2. b) Vd nghidm. 1-V^ c) 00 u i + Vs a)m>3. 4 h)m< Cdch 1. Chutig minh cdc bilu thflc da cho khdng phu thude vdo x. Tfl dd suy ra/'(x) = 0. a)/(x)=l^/'(x) = 0; b)/(x)=l=>/'(x) = 0; c)/(x)= j(v2-v6)^/'(x) = 0; d)/(x)= 3^/'(x) = 0. Cdch 2. Ldy dao ham cua/(x) rdi chflng minh ring/'(x) = ; 0 ; y’ = x2 + X – 2 a)-2;l. b) -1 ; 0. c) -4 ; y = lod^x – 5x’^ y = 2x – (a + b). 213

216 3.26. y = a + b’ y = 2{x + 2) (x + 3f (3x^ + 1 Ix + 9) y = xsin2o; + eos2a y = mn[x”-^ + x'”-^ + (m + n)x'”^”-^’ y = -(1 – xf (1 – x2) (1 – x^)2 (1 + 6x + 15×2 + 14x^) y = y = 2(1-2x) (1 – X + x2)2 l-x + Ax (1 – x)^(l + x)^ (Ixl ^ 1) y = 12-6x – 6×2 _^ 2x^ + Sx”^ – 3x^ {x^iy {l-xf j.j-t. y l + 2x’ Vl + x2 a” y’ {cf – x2)va2 – x y 2. = X sinx y = -sin2x. cos(eos2x). (lxl< y x2 / s2 (cosx + xsmx) y sin X X^ Kll, ke Z) y’ = 1 + tan X x ^ {2k + l)-,k e 214

217 A/(l) = Ax + 3(Ax)2 + {Axf, dfil) = Ax. a) A/(l) = 5>rf/(l) = l. b)a/(l) = 0,131 >#(!) = 0,1. e) Afil) = 0, > dfil) = 0, dy = :rdx. X 4.3. dy = -dx. {x-lf 4.4. dy = (sin2x)d’x. 2Vx – sin(2vx) 4.5. dy = ^ ),-^dx. 4xVxeos Vx tan x X ^ k,k G V 4.7. y = ax + b =>y = av^dy – adx = aax ; Ay = a{x + Ax) + b- [ax + b] = aax. Vdy dy = Ay Ddt y(x) = Vfl^ + x, ta cd y'(x) = p= 2Va^ +. Tfldd Ap dung : Ay = y(x)-y(0)«y'(0) X =^ Va + X «a a) 12,08 ; b) 5,83 ; e) 10, r-X. 2a 215

218 y = sin5x cos2x = [sin7x + sin3x] 4* y” = -(49sin7x + 9sin3x) x , do dd ^~x2 + x-2″ ^-1 ‘ ^ y” = 2 (x-lf + {x + 2f 5.3. ^ x2-l x+1 x (X + 1)2 (X-1)2 y = (x + 1)^ (x – 1)^ 5.4 x + 1 y = = 1 + x-2 x-2 y = N2 ‘ y = (X – 2)’ (X – 2y 5.5. y” = (2 – x”)sinx + 4xeosx y = 2x^ + 3x (1 + x2)vl + x y” = (x – 3)cosx + 4xsinx. y 1 4xVx 1. ^ 1., 1., y = sin2x + -sm4x – -sin6x; y” = -sin2x – 4sm4x + 9sin6x.

219 5.10. y = -X x ‘ >’ 3- {l-xf y” = -4sin2x – 4xeos2x y” = ^ 4Vx Bai tap on chitdng V.. 7 2xcosx 1. a) cot X sin X b) cosvxeos3x + 6VxsinVxsin3x 2Vxcos 3x c) 6cos2x (sin2x + 8) ; ^, ^ 2 2x^-5 d) 6x tanx +. cos X 2. a)-; b)40; e) a)/'(x) = 6(x^ – x^ + x2 – X + 1) = 6xM ;c^ – x^ x^+6 ^ 2 ; -7-X + 1 f 1 = 6x’ x^- 2, r + 3x^+6 > 0, Vx G R. b)/'(x) = 2 + cosx > 0, Vx G R. 4. / (x) = 3×2 + 2(a-l)x + 2. A’ = (a – 1)^ – 6 = a2 _ 2a – 5. Ta phai ed 217

220 A'<0oa2_2a-5<0<=>1-V6<a<l + V6. Vdy/'(x) > 0 vdi mgi x G R nlu l-yfe <a<l + yf6. 5. g'{x) = cosx – 2aeos2x – eos3x + 2a 2 = Aasin x + 2sinxsin2x 2 2 = 4asin x + 4sin xcosx 2 = 4sin x{a + cosx). 2 Rd rang vdi a > 1 thi a + cosx > 0 va sin x > 0 vdi mgi x G R ndn vdi a > 1 thi g'{x) > 0, Vx G R (1;1) a)c = 2,b = -l,d=l 3 2 =>fix) = X – X + 2x + 1 ; b)/'(x) = 3×2-2x + 2 _>y^-(i) = 3 Phuong trinh tidp tuydn tai M(l ; 3) Id y – 3 = 3(x – 1) hay y = 3x. c)/'(sino = 3sin2f – 2sinf + 2. fxsint) = 3 o 3sin2r – 2sinr -1=0 f = + k2n sint = 1 1 ^ smr = – t = arcsm + k2n (-1) t = n-arcsin – + ^27r {k 218

221 d) /”(x) = 6x – 2 =>/”(coso = 6eosr – 2 ; g'{x) = 2x – 3 ^ g'{sint) = 2sinr – 3. Vdy 6eosf – 2 = 2sim – 3 <:^ 2sin/ – 6cosf = 1 «> sinr – 3eosr =. Ddt tan^ = 3, ta duge sin{t – <p) = cos0 sm3z 3z a 10. y = :i-^y(xo)=-^ Phuong trinh tidp tuydn tai M{XQ ; y^) Id y- = 7^{X-XQ) cfx 2cf + y = Suy ra dien tfch tam gidc OAB Id -f 2a^ I I 9 XQ = 2a = const. 11. HD : Chiing minh bing quy nap. 219

222 ON TAP CUOI NAM Chflng minh cdc he thflc sau : a) sina + sin a + TT V 3 J + sm 8 ^ a n 3 J f sin 4a cos 2a b) 1 + eos4a”l + cos2a = cot 2^-«0; 2 2 2a b c) (cosa – cos6) – (sina – svnb) = -4sin cos(a + b) ; d) sin2(45 + a) – sin2 (30 – a) – sinl5 cos(15 + 2a) = sin2a. 2. Bidn ddi thdnh tfch a) 1 + cos n f- ^5, ^ + 3a sm n -3a + cot 7t + 3a ‘ 2 2 cosla – cos8a – cos9a + cos 10a b) sin 7a – sin 8a – sin 9a + sin 10a c) -eos5a cos4a – cos4a eos3a + 2eos 2a cosa. 3. Gia sfl A, B, C la ba gde eua tam giac ABC, chflng minh ring : sinc a) cosacosb = tana + tanb ; u^ >. z> ^ A ^ B C b) sina + sinfi + sinc = 4eos -cos cos – ; ^ sina + sin5 + sinc A B c) -: : : ;; : ;7 = COt COt. sma + smb-smc Cho hdm sd y = sin4x a) Chflng minh ring sin4(x + k ) = sin Ax vdiyfc G Z. Tfl dd ve dd thi eua cdc ham sd y = sin4x ; y = sin4x + l. b) Xdc dinh gia tri eua m dl phuong trinh – Cd nghidm ; – Vd nghidm. 2^ (Cj) (C2) sin4x + l = /n (1)

223 n c) Vie’t phuong trinh tidp tuyin eua (C2) tai dilm ed hoanh dd XQ = 5. Tim gid tri ldn nhd’t vd gid tri nhd nhdt cua ham sd 6. Cho ham sd y = sin2x + 4sinxcosx-3cos2x + l. f{x) = tanx + smx cotx a) Tim tdp xde dinh cua ham sd da cho. b) Xet tfnh chdn, le eua ham sd. (C) NT,…’ ^i-i-g.^ tanx + sinx,,,,, c) Bien ddi bilu thuc thanh tich. cotx d) Chiing td ring dilm 7. Giai cdc phuong trinh ^n 9} 3’2j,. _ 4 X. 4 X a) sin2x = cos – sm ; b) 3sin5x – 2eos5x = 3 ; thude (C). e) cos ^n + 5x ^ ^ v2 y + sinx = 2cos3x ; d) sin2z + eos2z = V2 sin3z. 8. Giai edc phuong trinh a) cos X + cos 2x – cos 3x – cos 4x = 0 ; ^n, ^ yf2. b) cos4x cos(7i + 2x) – sin2x cos -Ax sin4x; V’ y c) tan( x) – tan(140 – x) = 2sin(80 + 2x); 221

224 J N. 2 – ^ 2 X X 2 X X 2 X.. d) tan + sm ;rtan + cos.cot + cot + sinx = 4 ; sin2r + 2cos2f-1 p _ 2= COS ^ COS? – cos3^ + sin3f – sinf 9. Giai eae phuong trinh a) cos(22 – t) eos(82 -t) + cos(l 12-0 cos(172 -t)= -(sinr + coso ; b) sin2(r + 45 ) – sin^(r – 30 ) – sinl5 cos(2? + 15 ) = – sindf; c) sin 2x + cos 2x = 128 d) V4eos2 X V4sin2 x + 3 = 4 ; e) tan(7icoso = eot(7isino. 10. Cd bao nhidu sd gdm tam ehfl sd, trong dd cd dflng hai ehfl sd 2? 11. Mdt td ed 10 hgc sinh trong dd ed An, Binh, Chi, Dung vd Huong. Cd bao nhidu cdch xdp 10 ban dd vdo 10 ghd sip thanh hang ngang sao cho An, Binh ngdi canh nhau vd Chi, Dung, Huong cung ngdi canh nhau? 12. Mdt trdm tdm the nhu nhau dugc ddnh sd tfl 1 ddn 100. Ld’y ngdu nhien mdt the. Kf hieu A vd B Id cdc bidn ed A : “The duge ld’y ghi sd chia hd’t cho 3″, B : ” The dugc ldy ghi sd chia hdt cho 5″. a) Tfnh P(A), P(fi) ; b) A va fi ed ddc ldp khdng, vi sao? c) Cung hdi nhu trdn nhung sd the la 105 vd duge ddnh sd tfl 1 din Cd hai hdp chfla bi. Hdp thfl nhdt chfla 1 bi dd vd 2 bi xanh, hdp thfl hai chfla 2 bi dd vd 1 bi xanh. Tfl mdi hdp ldy ngdu nhidn 1 bi. Tfnh xdc xud’t sao cho 2 bi ld’y ra cung mdu. 14. Tim cdp sd cdng aj, a2, 03, 04, a^, bilt ring 222 aj = -12 vd a^a-iflc^ = 80.

225 15. Vidt ba sd hang ddu eua mdt cdp sd cdng, bid’t ring tdng n sd hang ddu tidn eua cdp sd nay Id 16. Giai phuong trinh S = An – 3/ n 7 + X + X X +… = -, X 2 trong dd Ixl < Tim sd hang thfl nhdt aj va cdng bdi q cua mdt cdp sd nhdn (a ), bid’t ring, «4-«2 = -l va a6-a4 = 32 ^ Chflng minh ring ba sd hang ddu eua tdng V V3-1 3-V3 6 ldp thdnh mdt cd’p sd nhdn va tinh tdng trdn vdi gia thidt ring eae sd hang tie’p theo duge tao thdnh theo quy ludt cua cdp sd nhdn dd. Trong cdc bdi tap 19, 20, hay tfnh gidi han lim x. /I-» a) x =, j= ; b) x = Vl + /2^ – /2 ; y/n + l +y/n c) x =n^in- yfn^ + 1 j ; d) x = yjn^ – n^ + n ;, yln^+l+yf^., ( lyl-4/2 20. a) x = I ;.b) x = n – – ^ 2/22 V/2 +n-n 21. Xet tfnh bi chdn cua edc day sd vdi sd hang tdng qudt sau : V 5/22 2n n^ +3 ” + 1 e) z = ncosnn. 223

226 22. Chflng minh ring day sd sau ddy tdng va bi chdn trdn : X^ = I 1 1 X-, =»-?7T’ ^2-5TT-77T’ ^’ :: ^ Tfnh cdc gidi han 1 1 x = “+l 4x^ + 9x + 7 a) lim x^i 3x^ + x^ + 1 b) lim x^2 x^ + 3×2-9x – 2 x-6 x + 1 c) lim -iv6x x,,. ^10 – X – 2 e) lun ; x^2 x Tfnh cdc gidi han a) lim JC->+«) r.2 ^3×2-4 3x + 2 ) Um [V2x x] ; ^,,. V9 + 5x + 4×2 _ 3 d) lun ;c^0 X Vx V8x + 1 f) Um x^i V5-X – V7x – 3 b) Um is V9x^+l-3x ; A,. 4: -.,.,2x^+3 d) Um r x->+<x> Ax + 2 V2x2 +: e) Um x^-ao Ax Tfnh edc gidi han ^,. sinx-sina a) lun At->a X a._,. 2sin x + sinx-1 e) lun r.]l 2sin x-3sinx + l CX b) lim(l-x)tan-x- ; x-^l 2 d) lim tanx – smx x-^o sin X

227 26. Tfnh dao ham eua cdc hdm sd sau : a) y = 1 + X – X,.2 ‘ 1 – X + X e) y = cos2x – 2sinx ; b)y = d)y = (2-jc2)(3-x^) (1-;^)’ cosx 2 ‘ 2 sin X 2X X e) y = cos -o-tan- ; f) y = sin 2x 27. Cho ham sd fix) = ‘ /-> X Sin, neu x ^0 X A, nlu X = 0. Xde dinh A dl/(x) lien tuc tai x = 0. Vdi gid tri A tim duge, hdm sd cd dao ham tai X = 0 Ichdng? LOI GIAI – HUONG DAN – DAP SO ON TAP CUOI NAM 1. a) sina + sin ^ a + 7t 14 ^ = sina + { 8 + sin sin a 2n n ) + 2n 2n = sina+ 2sinaeos = 0. 3 b) sin 4a cos 2a _ 2 sin 2a cos 2a 1 +cos 4a 1 +cos 2a 2cos2 2a.2cos2a 2sinacosa 2 cos a fsn = tana = cot BTBS&GT 11 -A 225

228 2 2 c) (cosa – cosfo) – (sina – sinb) =.. 2^ + b. 2^ -b, 2a + b. 2<^-b = 4sin – sm 4cos – sin 2 2 = 4 sin’ a-b^ 2 a + b 2 a + b^ sm 2 cos 2 V J.. 2 O ~ bl. – 2 a + b^ = 4sin”-^ 2 V 1-2COS” 2 J d) sin2(45 + a) – sin2(30 – a) – sinl5 eos(15 + 2a) =.. 2(i-b,,, = -4 sin r eos(a + b). = [sin (45 + a) + sin (30 – a)] [sin (45 + a) – sin (30 – a)] – sin 15 cos (15 + 2a) 75 = 2sin ; cos, a 75.2 cos r- sin a -sinl5 cos(15 + 2a) = sin 75 sin(15 + 2a) – cos75 cos(15 + 2a) = -cos(90 + 2a) = cos (90-2a) = sin2a. 2. a) 1 + cos 7t – + 3a 2 sin n – 3a /, + cot n + 3a = 1 – sin3a + cos3a-tan3a b) cos3a-sin3acos3a + cos 3a-sin3a cos 3a (cos 3a – sin3a)(l + cos 3a) cos 3a V^ sin -3a 2 3a,2cos r- – /r 2 3a. 2v2cos -T-sm -3a A cos3a V cos3a cos 7a – cos 8a – cos 9a + cos 1 Oa 2 sin 8a sin a – 2 sin 9a sin a sin 7a – sin 8a – sin 9a + sin 10a 2 cos 8a sin a + 2 cos 9a sin a 17a. a sin 8a – sin 9a -^cos^-smy 17^ – cot cos 9a – cos 8a ^. 17a. a 2-2sin – sin BTBS&GT 11-B

229 e) -cos5a cos4a – cos4a cos3a + 2cos 2a cosa 2 = – cos4a (cos5a + cos3a) + 2cos 2a cosa = – 2cos4a eos4a cosa + 2cos 2a cosa = 2cosa (cos 2a – cos 4a) = 2cosa (eos2a + cos4a) (cos2a – cos4a) = 2cosa. 2cos3a cosa. 2sin3a sina = 2cosa sin2a sin6a. a) HD : Thay sinc = sin {A + B)..,…. ^ ^. A + B A-B ^. C C h) sina + sinb + smc = 2sin – cos – + 2sin cos ^ C = 2eos – 2 eos- A-B A + B + cos- A B C = 4cos cos cos c) Chiing minh tuong tu cdu b) ta cd sina + sinb – smc = 4sin sm cos Tfl (1) vd (2) suy ra dilu phai chiing minh. n a) Ta ed. sin4(x + k ) = sin(4x + 2kn) = sin4x vdi k e (1) (2) Tfl dd suy ra hdm sd y = sin4x la ham sd tudn hodn vdi chu ki 71 Vi ham sd y = sin4x la ham sd le ndn dd thi cua nd cd tam dd’i xiing la gd’c toa dd O. Cac ham sd y = sin4x (Cj) va y = sin4x +1 (C2) cd dd thi nhu trdn hinh 1 vd hinh 2. Hinh 1. Do thi hdm so y = sin 4x 227

230 Hinh 2. Do thi hdm so y = sin4x + 1 b) Vi sinax + l = m» sin4x = m-l va nen l<sin4x<l -l<m-l<l O 0</22<2. Tfl dd, phuong trinh (1) cd nghidm khi 0 < //2 < 2 vd vd nghidm khi m > 2 hoac m<0. e) Phuong trinh tilp tuyin cua (C2) cd dang y-yo=y'(xq){x-xq). ^^ Jfo=^ tacd yo=sin5 + l = y'(x) = 4eos4x => y'(xq) = 4cos = 2yf3. Vdy phuong trinh tilp tuyin la.4=2^/3′ X- “”^» y 2V5x-^^-^ 24 – rr cos 2X.. _ 3 ^, ^ ^, 5. Ta CO y = + 2sin2x- (l + cos2x) ^ 2 ‘ = 2sin2x-2cos2x = 2V2si sin 2x- n Do dd GTLN cua ham sd la 2V2, dat duge khi sin 2x – I = 1 hay 228 2x – = + ^271, tflc la khi X = + ^7t ; ^ G Z.

231 GTNN cua ham sd Id -2^^, dat dugc khi sin(2x-^) = -l hay 2x – = – + k2n, tflc la khi x = – + kn ; k e Z. n 6. a) Vi tanx xdc dinh vdi x^ + kn va cotx xdc dinh vdi x^ kn {k e Z) ntn tdp xdc dinh eua ham sd da cho la D = R\k^;ke b) Vi X G Z? <=>-X G D vd.. tan(-x) + sin(-x) _ -tanx-sinx _. / X)./ r 7 / X) eot(-x) -cotx nen ham sd la chdn tren D = R { k,k e. n c) Ta ed, vdi x ^ k. tan X + sm X… 2,-, ^ ^ 2 2 x = tanx(tanx +sinx) = tan x(l + cosx) = 2tan x.eos. cotx 2 d)taed / = 2 r 7c^^ tan- Do dd dilm fn 9} v3’2. thude (C). eos^t = 2(V3)2f:^” Trong cdc cong thdc nghiem dudi day (BT 7, 8, 9), k la sd nguyen. n 7. a) X = + ^7t ; X = + ^27t ; ^71 ^^” = To^ 2 ^ 2kn X = aretan5 – 5 5 c) X = {2k + 1) x={ak- 1)- n 6 4 n d) z = (8/t + 1)- 4 z={sk + 3) 20 9_ 2 57C X = – + k2n

232 _ ^71 _ kn o. a) X ; X. 2 5 b)x=(2k:+l) X=(-l) c) X = ^60. d) X = (Ak + 1)- ; X = (-1)^^^ arcsin – + kn. e)t = {Ak+l)-. A 9. a)r = B60 ; r = (4yt + 1)90. h)t = A:90 ; r = ±15 + A;90. ‘ c)x = (3^+l). 12 d)’x = + + kn. :.’ 6 ^. ‘^ ^ ^/2, e) t = ± arceos 1- kn. A A 10. a) Gia sfl chfl sd 2 diing ddu. Khi dd, chfl sd 2 kia se dugc xd’p vao mdt trong bay chd edn lai. Cd 7 each. Khi da xd’p xong hai chfl sd 2, edn 6 chd, ta xd’p 9 ehfl sd khae 2 vao 6 chd dd. Ta cd 9 cdch. Theo quy tic nhdn, cd 7.9 sd gdm 8 chfl sd ma chfl sd 2 dflng ddu. b) Chfl sd 2 khdng dflng ddu. Khi dd, trong 8 chfl sd khdc khdng va khdc 2, ta chgn mdt chu sd dl xd’p vdo vi trf ddu. Cd 8 each. Chgn hai chd trong bay chd dl xip chfl sd 2. Cd Cj each. 230 Xdp chfn chfl sd (khdc 2) vdo nam vi trf cdn lai, cd 9 cdch. 2 5 Theo quy tac nhdn, cd 8.C7.9 sd ma ehfl sd’ 2 khdng dflng ddu. Theo quy tie cdng, sd cac sd cd 8 chfl sd ma cd dflng hai ehfl sd 2 la 7.9S8.C7.9^ =

233 11. Ddu tidn ta ehi dung 7 ghd vd xdp An, Chi vd 5 ban khdng thude nhdm An, Chi vdo 7 ghd. Ta ed 7! each xip. Sau dd xl’p Binh ngdi canh An. Cd 2! each. Cudi cung xdp Chi, Huong ngdi cung nhdm vdi Dung. Ta cd 3! each. Theo quy tic nhdn, cd 7! 2! 3! = each. 12. Khdng gian mdu Q = {1, 2,…, 100}. A ={3, 6, 9,…,99},/2(A) = 33. B= {5, 10,…, 100], n{b) = a) P{A) = 100′ i^,p{b)=: ^ ‘ 100′ ^”^ b) A nfi= {15, 30,45, 60,75, 90} ; P(A n 5) = -^ ^ P{A). P{B). 100 Vdy A va B khdng ddc ldp. – Nlu cd 105 the thi xet tuong tu, ta cd : «W = 35,/'(A)= = ; n{anb) = l; Vdy A va B ddc ldp. 13. Kf hidu A, la biln cd “Bi ldy tfl hdp thfl / mdu dd”, 2 = 1,2. Bidn cd cdn tim xae sudt Id A = AjA2 u Aj Aj. Do Al va A2 ddc ldp nen P{A) =P{A^A2) + P{A,A2) = P{A^)P{A^) + P{A^)P{A2) -kl 1i-1 ” 3’3 ^ 3″3 ~ 9’ 231

234 14. Kf hieu cdng sai la d, ta cd aj aj = -12 l«ia3«5 = 80 «> I ai +2d = -A [ai(ai+2rf)(ai+4fif) = 80 [aj =-2d-A [16(J + 2){d – 2) = 80. Giai ra ta dugc d = ±3. Cdc cd’p sd cdng phai tim la 2,-1,-4,-7,… va 15. Ta ed -10,-7,-4,-1,… Sj = Mi = = 1 vd [2MI +{n- 2 l)d]n ^ ^^2 _ ^^ ^[2 + {n-l)d] = 2(4/2-3)=>d=S. Tfl dd Ui = 1,222 = 9,223 = Vi Ixl < 1 nen vdi 221 = x, 9 = x ta cd S= “1 l-q = X + X X +… = 1-x Dodd + X + x x” +… = + 5 = <s> 1 X X 1-x 7 2 <=> x2- x(l x + 1 -X) 7 2 Giairataduccx.a,.,4, 232

235 17. Ta cd he phuong trinh aiq – a^q = 45 ‘ «.^ -«i^ = Tfl dd rflt ra o =- =:>«= + ^ 16 ^ 4 Vdi o = thi a, = 6 ; 4 ^ 18. Ta cd q = thi ai = ^ 2 1 1,^ /r, V a! r^ T=^ = (2 + v3) = p = a,. a3..3-v3j 6(2-V3) 6’ V3-1 6 ^ ^ Vdy 3 sd hang ddu ldp thdnh cdp sd nhdn vdi cdng bdi la 1, V3 + 1 _ 1 V3-1 _ 1 ‘^~3->^’>^-l~ V3(V3-1) V3 + 1 ~ 3 + V3 Rd rdng 1^1 < 1, ndn tdng vd han trdn la V3+l.r 1_^ V3-1 A 3 + V3 S=3 + yf a) yfn 1,. 1 V/ V/2 n + l ^ + 1 b) = yfl + n^ -n = limx = 0. ^(1 + n^f + nyll + n^ + n^ 233

236 c) n^in-yln^ + 1 = -n. 1 + J1 + n -n n + V/2^ + 1 > -00 khi n > +00. d) = sn – n + /2 = 1.3;ii-i -3(1-,., 3fP2 3T2″ 3r’2 3, 2 ^(/2 – n ) – nsn – n + n > khi n > a) x = V/2^ V«3/ 3^ ~~ /2 + n – n A f ^-i] f4-> > liinx = b),.,_4ufj-2uf,-llfj-2,2 j’/2v2/2 => limx =l.(-2) = a) Day (x ) bi chan vi 234 n 5/22 0 < -^; < 5 voi moi n ; rf +3 b) Day (y ) bi chan vi yn = (-1)” e) Day (z ) khdng bi chdn vi z = l/2cosrttcl = /2. 2 K2n 2/2,., 2/2. I sia/21 < 7 < 2 ; n+l n+l

237 22. Day sd {x } tang vi x +i = x + ^ ^i ^ ^ > ^. Mat khdc, day sd nay bi chdn tren vi *:«+i x ” r ” + 1 < ^ ‘ -T ‘ 53 -T +- + = 5″,_1 23. a) 4. 1-i- < VOI moi n. 5«>’ A ^,,. x^ + 3×2-9x – 2 ^^ _ 2)(jj.2 + 5;^. + 1) b) lim = hm-^^ -^ x^2 x^ -X-6 x^2 {x – 2)(x2 + 2x + 3) x2 + 5x = lim ^->2jc2+2x ‘ _^ J (x + l)(v6x x] = lim J L c) Um. = lim ^^-1 V6x x ^^-1 3-3x^,. V6x x, = lim ^TT. r = 1. x^-i 3(1 – x) ^,,. V9 + 5x + 4×2 _ 3 5x + 4×2 d) lim = lim.. r ^^ ^ -^Ox V9 + 5x + 4x x = lim ^^ V9 + 5x + 4×2+3 0,,. ^10 – X – 2,. 2-x e) lim = lim., = ^^ r x^2 X-2 x^2 (^ _ 2) 3^(10 – xf + 2Vl0^ = -lim ^-^2 3^(10 – x)2 + 2VIO – X + 4 ^^ 235

238 VTTs-VsTTT,. 7(1 -x)(v5^ + VTT:^) = lim f) lim x^i V5-X – V7x – 3 x^i 8(1 – x)(vx V8x + l) 24. a) lim X >+oo r,.2 ^ 3×2-4 3x + 2 7,. V5 – X + V7x = hm, = r. ^x^ly/x + S+ y/sx + I 12 = Um x^^9x^ = lim 2x^ + 4×2 +6×2 _i2x X->+oo 9 + ^-i^-a ^ x2 X^ 2 9″ b) c) lim I V9x^ + 1-3x) = Um -; L x^^ I ^^^V9x2+l + 3x Um (V2x x) = Um V2x2-3 + (-5x) = +00. d) ^^J2 Um = lun ” * j:-»-l-ao 4x + 2 jr->+oo / e) lim i 2.’. 3, “T”? = lun – ‘ “^ 4x + 2 jr->-oo [ 2 x a) lim x-ya smx – sma X-a X + a. x-a cos – sm – Um 2 2 x-a = cosa. b) Ddt 1 – X = r (r ^ 0 khi X -) 1), ta cd 236 lim(l – x)tan- = lun rtan(l – i) x-^ 2 x-> = Um/cot f = Um ^ 2->0 2 f->0 ^ 71 tan-r 2^ Tl”

239 ^ Chu y. lim tanx lim sinx 1 = 1. J c->0 X Jf->P X COSX c) d) 1 + V3 5-3V3′,. tanx-sinx,. 1-cosx Um = lim I x-*o sin X x^o sin x cos x a) 2(1-2x) (l-x + x2)2 2sin^- 1 = lim 2 1. x^o. 2X 2X 2 4sm cos cosx 2 2 b) c) d) e) f) 12-6x – 6×2 + 2x^ + 5x^ – 3x^ (1-^)^ -2eosx(l + sinx). 1 + cos X 2sin^x 1. 2x X – sin-^i-tan cosi 2x – 6 sm 2x- n 27. A = 0. Khi dd/(x) ed dao hdm tai x =

240 MUC LUC Trang Chumg /. HAM SO LUDNG GIAC – PHUGNG TRINH LlTONG GIAC 3 1. Hdm sd Iflgng gidc 3 2. Phuong trinh lugng gidc CO ban Mdt sd phuong trinh lugng gidc thudng gap 24 Bdi tdp dn chuong I 35 Ldi giai – Hudng ddn – Dap sd chuong I 36 Chuang II. TO HOP – XAC SUAT Quy tic ddm Hodn vi, chinh hgp, td hgp Nhi thflc Niu-ton Phep thfl va bidn cd Xdc sudt cua bidn cd 69 Bdi tdp dn chuong II 73 Ldi giai – Hudng ddn – Dap sd chuong II 74 Chuang IIL DAY SO – C^ SO CONG VA CAP SO NHAN Phuong phdp quy nap toan hgc Day sd

241 3. Cdp sd cdng Cdp sd nhdn 114 Bai tdp dn chuong III 121 Ldi giai – Hudng ddn – Dap sd chuong III 124 Chumg IV. Gldl HAN Gidi han cua day sd Gidi han cua ham sd Ham sd lidn tuc 160 Bdi tdp dn chuong IV 165 Ldi giai – Hudng ddn – Ddp sd chuong IV 170 Chumg V. DAO HAM Dinh nghia vd y nghia eua dao ham Cdc quy tic tinh dao ham Dao ham cua cae ham sd lugng giac Vi phdn Dao hdm cdp hai 205 Bdi tdp dn chuong V 207 Ldi giai – Hudng ddn – Dap sd chuong V 209 On tdp cud’i ndm

242 Chiu trach nhiem xudt bdn : Chu tich HDQT kiem T6ng Giam d6c NGO TRAN AI Pho Tdng Giam d6’c kiem T6ng bidn tap NGUYfeN QUt THAO Bien tap ldn ddu: NGUYfiN XUAN BINH – NGUYfiN NGOC TU Bien tap tdi bdn : NGUYfeN XUAN BINH Bien tap Id thuat: NGUYfeN THANH THUt – TRAN THANH HANG Trinh bdy bia : TRAN THUt HANH Sih bdn in : Lfe THI THANH HANG Che bdn : C6NG TY CF THifeT KE VA PHAT HANH SACH GIAO DUC BAI TAP DAI SO VA GIAI TICH 11 Ma so : CB103T1 In cudn, khd 17 x 24 cm. In tai Cdng ty TNHH MTV in Quang Ninh. Sd in: So xuat ban: /CXB/ /GD. In xong va nop luu chieu thang 4 nam

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244 u< HUAN CHKONG HO CHI MINH VI/ONG MIEN KIM CUdNG CHAT LUONG QUOC TE SACH BAIlAPLCfP BAI TAP DAI so VA GIAI TICH BAI TAP HiNH HOC BAITAPVATLIU 4. BAI TAP HOA HOC BAI TAP SINH HOC BAITAPOIAU’H 7. BAI TAP TIN HOC BAI TAP NGO” VAN 11 (tap mot, tap hai) 9. BAlTAPUCHSCfll 10. BAI TAP TIENG ANH BAITAPTIENGPHAP BAI TAP TIENG NGA 11 SACH BAI TAP LCfP 11 – NANG CAO. BAI TAP OAI so VA GIAI TiCH 11, BAI TAP HOA HOC 11. BAI TAP HiNH HOC 11, BAI TAP NGUVAN 11 (tap mot, tap hai). BAI TAP VAT LI 11, BAI TAP TIENG ANH 11 Ban doc co the mua sach tai: Cac Cong ty Sach – Thiet bi truong hoc a cac dia phucmg. Cong ty CP Dau tu va Phat trisn Giao due Ha Noi, 187B Giang Vo, TP. Ha Noi. Cong ty CP Dau tu va Phat trien Giao due Phuang Nam, 231 Nguyen Van Cu, Quan 5, TP. HCM. Cong ty CP Dau tu va Phat trien Giao due Da Nang, 15 Nguyen Chi Thanh, TP. Da Nang. hoac cac cua hang sach cua Nha xuat ban Giao due Viet Nam : – Tai TP. Ha Noi : 187 Giang V6 ; 232 Tay Son ; 23 Trang Tien ; 25 Han Thuyen : 32E Kim Ma ; 14 3 Nguyen Khanh Toan ; 67B Cua Bae. – Tai TP. Da Nang : 78 Pasteur ; 247 Hai Phong. – Tai TP H6 Chi Minh 104 Mai Thi Luu ; 2A Dinh Tien Hoang, Quan 1 ; 240 Tran Binh Trong ; 231 Nguyin Van Cir, Quan 5. – Tai TP can Tho : 5 5 Duong 30’4. Tai Website ban sach true tuyen : Website: ” Gia: d

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