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X b) — [5, (15^ X + bJ > 3" + " + " vai moi x HDDS a) Xet ham s6 f(x) = e" - x - 1, x > b) Dung bat dang ihuc Cosi Bai tap 7: T i m gia tri Ian nhat va gia tri nho nhat cua ham so a) f(x) = b) f(x) = x - Inx + tren khoang (0; + 0 ) tren doan [-1; ] HD-DS a) Kk qua f ( x ) = f ( ) = ; max f ( x ) = f ( ± l ) = xe[-l,l] b) Kel qua xe[-l,l] m i n f ( x ) = f (1) = , khong c6 gia tri ion nhat xe(0;-too) Bai tap 8: T i m m dd hk phuang trinh 4.2^°^"' + 2'"'°^'' > 2^'"' + m : a) CO nghiem b) c6 nghiem v o i moi x HDDS Dat t = 2'°'^\ < t < 2 CHU D E r x PHCrONG TRINH MU Vfi LOGflRiT I• - Phuang trinh mu ca ban: 0, a ^ 1) Neil h ^0 phuang trinh vd nghiem Neu h > 0, phuang trinh c6 nghiem nhat x = logah - Phuang trinh mii i/'"" (f'''^ (a > 0) "a = l a ^ l , f ( x ) = g(x) Phuffng phdp: - Dua ve ciing mot ca so - Dgt an phu - Logarit hod - Su dung tinh chat cua ham so ddnh gid hai ve Chu y: Ngoai phuang phdp chinh de gidi phuang trinh mii, ta c6 the ditng iinh nghia, hien doi thdnhphuang trinh tich so, diing hdt dang thirc, 217 Bai toan 1: Giai cac phuong trinh sau: a) 0,125.4^"-^ = (472)" b) (2 + Vs )^''= - Vs Giai 5x a) PT: 0,125.4'"-' = (4 V2 )" « T\' = 2'^" « 2'''-' = 2= « 4x - = — « 8x - 18 = 5x » b) PT:(2+ V3)''' = - X 5x = (2 + 73)'" = ( + Vs y ' x =-1 o x = Bai toan 2: Giai cac phuong trinh sau: a) ^ - ylogx _ ^Iogx+1 _ 2 = 2 -3'"-' ^logX-1 _ J2 ylOgx-1 G/di a)PT: 9^+-.9^ = ^9Y'^V2 v2 ^ -.9''=3.2 ^ x+- x+- ^+2.2 I ! I X -1 = log, — thi PT: 218 - 31 - + - 3t^ - 4t + 12 = (t - 2)(t + 2)(t - 3) = =0« y Chon nghiem t = hoac t = nen x = ln2 hoac x = ln3 b) Chia vk cho 8' > thi PT: 27 ^12Y + V y - = Dat t = 2) PT: t^ +1 - = « (t - l)(t^ +1 + 2) = « t = « Bai toan 5: Giai cac phuomg trinh: a) 2.25"+ 5.4'= 7.10' a) PT: (2^ 2\ -7 5j f2^ 5, PT: 5t^ - 7t + = X = ,t>0 b) " + " = " Giai = Dat t = ,t>0 t = hoac t = - (thoa man) , Suy nghiem x = hoac x = dat y = — va chia hai vi cho 4^, ta c6: b) Dieu kien x (3^ 2v — — , '3^ y 2, - =0 + V5 '3^ • i+Vs , 1+V5 2, + V5 , o - -X = log,2 — - — « -X = log, -1 « X = log^^_, - 2\ Bai toan 6: Giai cac phuong trinh: a) (V2-V3r+(V2+V3 =4 = / a) Ta CO V2-V3.V2 + V3 = 1, dat t = fV2 + V3l , t > PT: t + - ^ = c ^ t - - t + l = t o t = + V3 hoac t = - V3 « X = hoac x = -2 b) D a t t = 2^^^'^' - , t > thi PT: t^ - - t = « 2t^ - 5t - 12 = Chon nghiem t = nen x + Vx^ - = Vx'^ - = - x x^ (sin ~ )^ va (cos — > (sin ~ 5 f V T > (loai) Neu X = thi PT nghiem dung, la nghiem nhat b) P T: ( — )^ + ( — ) " va ta CO X = thoa man PT V i ve trai la ham so nghich 4 bien tren R nen c6 nghiem nhat x = Bai toan 11: Giai cac phuong trinh: a)x.2' = x ( - x ) + 2(2''-1) b) 2"^'- 4" = x - Giai a) PT: X.2'' - x(3 - x) - 2.2" + = 2"(x - 2) + x^ - 3x + = o 2"(x - 2) + (x - l ) ( x - 2) = « (x - 2X2" + x - 1) = o X - = hoac 2" + x = l x = hoac x = (Vi f(x) = 2" + X d6ng bign tren R va f(0) - 1) b) PT:2""' + ( x + l ) = 2'" + 2x Xet ham s6 f(t) = 2' + t, t e R thi f ' ( t ) = 2'.ln2 + Vi f'(t) > 0, V t nen f d6ng b i l n tren R PT f(x + 1) = f(2x) < ^ x + l = x o x = l 221 Bai toan 12: Giai cac phuong trinh: a) 3^'^'^'+2|x| = 3^^' b) ^ ^ V ^ ( x + l) = ^(2^^"^"^'^'+V2^) Giai a) Phuong trinh da cho xac dinh voi mpi x Xet x < Khi ta c6 3''"'^' + | x | > > 3"^', nen phuong trinh da cho khong C O nghiem khoang (0; + t » ) Xet x ^ Phuong trinh tro \ = 3''"'-(x+l)' Ta CO Vx" + , X + >1 Xet ham s6 f (t) = 3' - ^ voi t e [1; +oo) f (t) = 3' ln3 - 2t, f "(t) = 3' (ln3)^ - Vi 3'(ln3)^ - ^ 3(ln3)' - > 0, Vt > 1, nen f "(t) > 0, Vt ^ Suy f '(t) la ham so dong bien tren [1; +oo) Do f'(t) > f (t) = 31n3 - > 0, Vt ^ nen f(t) la ham so dong bien tren [ 1; +co) Phuong trinh: f ( V x ' + l ) = f ( x + 1) Vx'+l =x +l fx'+l = x'+2x + l « x =0 x^O x>0 Vay phuong trinh c6 nghiem nhat x = b) Dieu kien 3x4-1^0 2x-l>0 Phuong trinh tro 2>/2x- -.1 ) / x + l+l + A/2X-1 ,/3x +l +l ,2 '^^/3x + l+l 2' 2^^^ 2' 222 Taco V x - + l , V x + l > Xet ham so f(t) = - ^ f (t) = voi t e [1; +00) ln2 - ; f "(t) = 2'^'(ln2)^ - V i t ^ nenf"(t)>(21n2)^ - > Suy f '(t) la ham s6 d6ng bi^n tren [ ; +00) Nen f (t) > f ' ( t ) = 41n2 - > 0, Vt > Do f(t) la ham so d6ng bi^n tren [ ; +00) Phuong trinh f ( V x - +1) = f(V3x + ) V x - + = V3x + o 2x + V x - l =V3x + l o X +1 ^ 2V2JC-I = X + • ( x - l ) = x^ + x + l X = va X = Vay phuong trinh da cho c6 nghiem x = 1, x = Bai toan 13: Giai cac phuong trinh: a) 5' + 4" + 3" + 2" = ^ ^ + — - x ' + x ' - x + 16 Y y 6' b) 4" - 2^"' + 2(2" - l)sin(2'' + y - 1) + = Giai a) Xet ham s6: fi;x) - 5" + 4" + 3" + ' - I I P -+—+— 2" 3^ +4x^-2x^+x-16,xeR Ta c6: f ' ( x ) = 5'ln5 + 4''ln4 + " ^ + 2''ln2 ln2 V 2^ -+ ln3 l n -+ 3" 6" y + 12x^-4x + > Suy ham so dong bien va phucrng trinh f(x) = CO khong qua mot nghiem va f ( l ) = ( Vay phuong trinh da cho c6 nghiem nh^t la x = b) Phuong trinh da cho tuong duong v o l (2^' - 2.2' + 1) + 2(2"- l)sin(2" + y - 1) + = o (2" - \f + 2(2" - l)sin(2" + y - 1) + s i n ^ + y - 1) + cos^(2" + y - 1) = o [2" - + sin(2" + y - 1)]^ + cos^(2" + y - 1) = 2' + sin(2^ + y - l ) = l cos(2' + y - l ) = Vi cos(2" + y - l ) = = > s i n ( " + y - 1) = ± Ta CO hai truomg hop sau: - NSu sinCZ" + y - 1) = thi 2" = 0, v6 nghiem - NSu sin(2' + y - 1) = -1 thi 2" = » X = TC Suy sin(y + l ) = -ly = - — - + k2n Vay phuong trinh da cho c6 nghiem la: x = 1, y = - — - + kn, k e Z Bai toan 14: Tim dieu kien de phuong trinh: a) 3^'"'' +3*^°^'^ =m c6 nghiem b) ( V s + 1)" + 2m( Vs - 1)" = 2' CO nghiem nhdt Gidi a) Dat t = 3"""', vi < s i n \ nen < t < 9 _ t BBT: X m Xet f(t) = t + ^ , < t < ; f ' ( t ) f t^-9 ; f'(t) = Okhit + 10 f 10 " ^ ^ ^ Vay diSu kien f(t) = m c6 nghiem thoa l < t ^ a < m < X b) p r « + 2m =1 , V5+1 Vs - ' = , datt = laco: 2 V PT: t + — = » t^ - + 2m = t Xet t = ^ m = thi PT: t^ - = « t = hay t = 1: thoa man Xet t ^ 0, diSu kien c6 nghiem t > 0: ti < < t2 hoac < ti < t2 « P < hoac (A > , P > , S > ) < = > m < hoac m = Vay: m < hoac m = Cach khac: Xet ham so va lap bang bien thien 224 DANG TOAN PHlTOfNG TRINH LOGARIT - Phuang trinh logarit ca ban: logaX = b (a > 0, a ^ 1) Phuang trinh logarit ca ban luon co nghiem nhdt x = a* - Phuang trinh logarit log/fx) = logagfx), (a>0, a^l) ^ f(x)>Ohayg(x)>0 l f ( x ) = g(x) Phirang phdp: - Dua ve ciing mot ca so - Dot dn phu - Ma hod - Sir dung tinh chdt cua ham so, ddnh gid hai ve Chu y: Ngodi phuang phdp chinh de gidi phuang trinh Idgrarit, ta co the dimg dinh nghia, bien doi phuang trinh tich so, diing hat dang thicc, Bai toan 1: Giai cac phucmg trinh sau: a) log2|x(x - 1)] = b) log2(9 - 2'') = o'"^*^-''* Gidi a) PT: log2[x(x - 1)| = « x(x - 1) = » x^ - x - = x = -1 hoac x = b) Dieu kien X < PT: log2(9 - 2") = lO'"^''-"* « - 2' = 2^"^ « 2^" - 9.2'* + = 2" = hoac 2" = Chon nghiem x = Bai toan 2: Giai cac phucmg trinh sau: a) 5-41ogx = b) S^log.C-x) = logj V x ^ + logx Giai a) Vai x > dat t = logx thi F T : j — ^ ^ +— = t ^ ^ ,t ^ - » 2t - 3t + = t = hoac t = - (chon) Suy nghiem x = 10 hoac x = VTo b) DK: X < 0, PT: V l o g , ( - x ) = l o g , ( - x ) « V l o g , ( - x ) ( - Vlog^C-"^) = o ^log2(-x) = hoac ^log2(-x) = 5 x = -1 hoac x = -2^"\ 225 Bai toan 3: Giai cac pliuong trinh: a) log2X + log2(x - 1) = b) log2X + log3X + log4X Giai a) D K : x > 1, PT < ^ log2x(x - 1) = o x(x - 1) = x^ - X - = Chon nghiem x = b) D K : X > 0, PT: (1 + log32 + log42).log2X = 1 (3 + log32)log2X = - » log2X = + log, Vay nghiem x = 2^''^'"^'" Bai toan 4: Giai cac phuang trinh: a) log3(3^-l) log3(3"^'-3) - 12 b) log,.,4 = + iog2(x - Giai a) D K : x > 0: PT: log3(3' - 1)[1 + Iog3(3'' - 1)] = 12 Dat t = log3(3^ -1) thi PT: t ( l +1) = 12 log3(3' - 1) = -4 hoac log3(3' - 1) = « t ' + t - 12 = « « o t = - hoac t = 3' - = ~ hoac 3' - = 27 81 8? 3' = — hoac 3' - 28 X = log382 - hoac x = log328 81 b) D K : x > 1.x ^ , PT: + l o g ( x - 1) log^Cx-l) Dat t = log2(x - 1) thi PT: - = 1+ t « t^ + t - = t = hoac t = -2 Giai nghiem x = — hoac x = 3 • Bai toan 5: Giai cac phuong trinh: a) log4L(x + 2)(x + 3)1 + ~ log2 - 2 x+3 b) l o g ( x + ) I o g x = l Giai (x + 2)(x + ) > a) D K : x-2 x2 >0 x +3 226 Phuang trinh mat cau la: (S): x + ?1V f 298 + (y-20)^4 = 49 V Cau 9.b Ta c6 z] + = Z|Z2 = > Z|(z2 - zi) = Zj => Zi Z2-Z1 Z2 ZJ' + Z^ = Z1Z2 = > Z2(Zi - Z2) = Z^ =5> I Z2 i I Z i - Z2 Do nen Zi - Z2 Zi - Z2 — Zl Zl Z2 Vay - z, 3_ ^ Z2 Zl Z2 -Z-, + 2' D E L U Y E N THI T O N G HOP SO I Phan chung cho tat ca thi sinh: (7,0 di6m) X -3 Cau Cho ham so y = (1) x+1 1) Khao sat su bien thien va ve thi (C) cua ham so (1) 2) Viet phuang trinh tilp tuyen cua d6 thi (C), biet khoang each tu giao diSm I cua hai duofng tiem can cua (C) d6n tiSp tuy^n b ^ g V2 Cau Giai phuang trinh: (1 + 2sinx) cosx(2x + — ) = — x'+2x'y = Cau Giai he phuang trinh , ( X , y e R) X " +y" + y = Cau Tinh tich phan I = In" X rrdx x(l + lnx)- Cau Cho hinh chop S.ABC c6 mat phang (SAC) vuong goc vai mat phSng (ABC) va CO SA = SB = SC = 2a, AB = 3a, BC = a V3 (a > 0) Tinh dien tich cua mat cau ngoai tiep hinh chop S.ABC theo a Cau Tim tham so m de phuang trinh sa'u c6 nghiem thuc: (Vx + Vx - l | mVx + J-— + l / x ( x - l ) = I Vx-1 II Phan rieng: (3,0 diem) Thi sinh chi dtu-gc lam mot hai phan: A hoac B A Theo chuoTig trinh chuan Cau 7a Trong mat phang vai he toa Oxy, cho cac dilm P ( l ; 1), Q(4; 2) Lap phuang trinh duang thang d cho khoang each tu P va Q dSn d l i n luot bing 2va3 403 Cau 8a Trong khong gian vod he toa Oxyz, cho mat phang (P): x - 2y + 2z + = ck mat c4u (S): (x - if + (y + 3)^ + (z + 3)^ = theo giao tuy§n la duong tron (C) V i l t phuong trinh mat cku (S') c6 tam thuoc (a): x + y + z + v a chua duong tron (C) C a u 9a Giai phuong trinh: - log2(x + 3)^ - log2(4 - x ) ' = 3[1 + log2(x + 6)] B Theo chu"(mg trinh nang cao C a u 7b Trong mat phSng v a i he toa Oxy, cho d i l m K(3; 2) va duong tron (C): x^ + y^ - 2x - 4y + = vai tam I Tim toa d i l m M e (C) cho I M K = 60° C a u 8b Trong khong gian v a i he toa Oxyz, cho hinh hop chu nhat A B C D A , B , C , D , v a i A , ( ; 0; 0), , ( ; 0; 0), Di(0; 2; 0), A(0; 0; 3) Goi M , N, P, Q lan lugt la trung d i l m cac canh A B , B i C i , C i D i , D i D Chiing minh rang cac dikm M , N , P, Q cung thuoc mot mat phang (a) Xac dinh thiet dien ciia hinh hop k h i cat bai mat phang ( a ) , tinh the tich cua khoi chop c6 dinh C va day la thiet dien = 16 C a u 9b Giai he phuong trinh log4(x' + y^) = ^ + log4(xy) Lai Giai C a u 1.1) Su bien thien: y' = • Tap xac dinh D = R \} • y > V x ^ - , suy ham so dong bien tren timg (x + \y khoang (-co; -1), (-1; +cc) Giai han l i m y = ; l i m y = => Tiem can ngang: y = x->-cc X—»-f-co l i m y = +oo; lim v-+(-i) = - o o ^ Tiem can dung: x = - x^(-\y Bang bien thien X -1 -co +00 + y' + +00 y • -00 • ^ -1 Dothi Giao v a i Ox: (3; 0), giao v a i Oy: (0; -3) Do thi nhan I ( - l ; 1) lam tam doi xung / 404 X —3 2) Gia su M(xo; yo) thuoc (C) thi yo = — , XQ -1 Xo+1 Khi phuong trinh tidp tuyen A tai M la: (Xo+1) o Xo + (xo + 1)^ - 8(xo + 1)^ + 16 = (xo + 1)^ = o xo = hoac xo = -3 V o i Xo = 1, phuong trinh A: y = x - 2; V o i Xo = -3, phuong trinh A: y = x + Cau Phuong trinh tuong duong V3 (1 + s i n x ) ( - cos2x - - ~ - sin2x) = - o (1 + 2sinx)(cos2x - V3 sin2x) = cos2x - V3 sin2x + 2sinxcos2x - V3 sinxsin2x = - 2sin^x - V3 sinxcosx + 2sinxcos2x - V3 sinxsin2x = ' = - Vay he c6 nghiem (1; 1), (-1; 1),(V3;-1), ( - V ; - l ) Cau Dat t = + Inx =>dt = - d x , t - = Inx => \ n \ (t - 1)^ X Doi can: x = l = : > t = l ; x = e=>t = Ta c6: I = In" x (t-iy dx = x(l + l n x r 2/ dt = V t 1\ t-r t-21i^t-^ t = 21n2 Cau Ke S H ± A C Do S A = S C nen H la trung dikm A C (1) Vi ( S A C ) ( A B C ) nen S H ± ( A B C ) =i> H A = H B = H C (2) Tu (1) va (2) suy A A B C vuong tai B c6 H la tarn ducmg tron ngoai tiep Do A C = V B A ' + B C ' = 2V3a ^ SH = V A S ' - A H ' = a S H la true ducmg tron ngoai tiep tarn giac A B C , mat phang ( S A C ) duang trung true cua S A cat S H tai O la tarn mat cau Gpi K la trung diem S A Khi hai tarn giac vuong S O K va S A H dong dang nen SH SA SK SO Suy ban kinh mat cau: R = SO = ^ ^ " ^ ^ = 2a SH Vay dien tich mat cau la S = 47iR^ = 16Tca^ Cau Dieu kien x > (Vx + Vx - mVx + mVx + I + \j\{x-\) = 1(1) Vx-1 V o + ijxix-l) = Vx - Vx-1 vx-1 0 m0 /(1)^0 m >1 [1-m^O Vay gia tri can t i m la m < hoac m > Cau Fa c6 sinx + > V x Phuong trinh da cho tuong dumig X X X X 3(sin— - cos —)(1 + sin —cos —) = cosx.(2 + sinx) 2 2 X X — (sin cos — )(2 + sinx) = cosx(2 + sinx) 2 V 411 , X 2X 2X « - (sin cos — ) = cos sin — 2 2 (sin— - cos —)[~ + (sin— + cos — )] = ^ 2 '2 2 X X X X sin cos— = (1) hoac - + (sin— + cos — ) = (2) 2 2 Giai(l): V2 s i n ( | - ^ ) = 0 dt = 2cosx(-sinx)dx = -sin2xdx Doi can: x = >t = ; x = ^ = ^ t = l = > I = - lntdt= Intdt Dat u = lnt dv = dt du = — t v=t Taco: I In/t//= / I n / ' - J W / = n - l I Cau Goi I la trung diSm cua AD Ta c6 lA = ID = IC = a =^ CD ± AC Mat khac CD ± SA Suy CD SC nen tam giac SCD vuong tai C Trong tam giac vuong SAB, ta c6: SH _ SAT SA^ 2a^ SB ~ SB' ~ SA' +AB' ~ 2a' + a ' " 412 Goi di va dj Ian luot la khoang each tir B va H din mat phing (SCD) thi: d, SH „ = = -=^d-,=-d, S d, SB ' ' Ta c6: d, = ^ ^ J I ^ C D ^ ^^\CD -"SCD ^BCD S , , „ = l A B B C = ia^ SscD = - sc.CD = -JSA' +BC-.VlC' + I D ' =a'V2 Suy di = ^ a Vay khoang each tir H den mat phang (SCD) la: di = - di = ~ U=X+ X" , ^-1 = (x + - ) " — , u > — Cau Dat v = y + y- = ( y + He da eho tra thanh: 4 v>-4 ?i+v = uv=3m V= • 4-M uv=3m 17 V i v > - = ^ - u ^ - - => u < 4 He da eho c6 nghiem (*) eo nghiem u e [ — ; — ] 4 Xet ham s6 f(u) = -u^ + 4u, u e 4 Ta CO f'(u) = -2u + 4, f (u) = » u = Lap bang bien thien thi yeu cau bai toan thoa man va chi khi: ~ < 3m < —^ ' + + C;;x", Vx e R Lay dao ham hai ve, ta dugc: n(l + x)""' = C;, + 2C^x + 3Cy- + + n C > " - ' , Vx e R Cho X = ta c6: C'„ - 2C^ + 3C^ 3' - + (-l)nC"„3" = 3n(-2)"-' Theo dk 3n(-2)""' = 33792 ^ (-1)"' n2""' = 11264 « n - Vay gia tri can tim la n = 11 Chu de 1: T I N H D O N D I E U 89 Chii de 4: TIEM CAN CUA D O T H I 61 Chu de 3: G I A T R I LCiN N H A T , N H O N H A T 36 Chii de 2: C U C TRI CUA H A M S O 131 Chii de 6: K H A O S A T V A V E D O T H I H A M HUXJ T I 107 Chu de 5: K H A O S A T V A V E D O THI H A M D A THlTC 194 Chii de 8: L U Y T H l / A , M U vA L O G A R I T 151 Chu de 7: B A I T O A N T H U ' N G G A P V E D O T H I 374 Chii de 16: S O PHLTC L U O N G GlAc 358 Chii de 15: C A N B A C H A I vA P H U O N G T R I N H N G H I E M PHLTC 337 Chu del4: S O P H U C 320 Chii de 13: L T N G D U N G C U A T I C H P H A N 294 Chii de 12: T I C H P H A N C A N T H L T C , M U , L O G A R I T 271 Chii de 11: T I C H P H A N D A T H L T C , P H A N T H L T C , L U O N G G I A C 253 Chii de 10: N G U Y E N H A M C U A H A M S O 217 Chu PHU de 9: P H U O N G T R I N H M U V A L O G A R I T LUC: C A C D E O N T H I T O N G HOP 397 416 y- y y y ^ y ^ •w^w^sTv n h i a s a c h i h i o n g a r a c o m v n 20C OT: Email: nhasachhonganOiotmail.com N g u y i n Thj Minh Khai - Q.1 - T P H C M 3824670e - 08083021 - 3910709S * * Fax: 0B083017 * t * r (^a/rt' tt/nv' t£po: HOD hoc T\} \ H Q C GIOI HOCT6T iiliitoi GiAl BfflDUflNG ^ HQGSINHGldl ^ 'S*^ HOCT6T -i- - HOC SINH GI61 HOC T6T 5:.? mm mmt T O A N ID* _ ^ GIAl T O A N Ph&i 1091 va phUdng phdp V glal nhanh bai tap HOAHOCj V A T L I / - N S MINH T A M , 245 T r I n N g u y e n H a n - H P * DT: (0313) 858699 ^ ãN PhSn kôi va ptiuong phap giai nhanh bai t$p \ X ^ ^ ^ ^ : ^ \l - & P h a n B o i C h a u - H^i P h o n g *DT: (0313) 9 - 04 L y Thai T o - TP B a N i n g *BT: 0511.3823421 •'^ - 259 L e Duan - TP V i n h - D T : 0383.554777 Dexacinh sach chinh phlin^ ^ A - 39-41 V T h i S a u - C a n T h d * DT: (0710) 3818891 chiing to! in chim d bia va chff: ^ -^158 TTnh 10 - TT.CCi C h i - T P H C M * B T : ( ) "NS HONG AN" ^ - 51 L y T h U d n g Kiet - T P D o n g Hdi - Q B * DT: (0523) 857868 ^ ' - 1 P h a m Hufu L a u - T P Cao L a n h * D T : (067) 2211794 - 66 L y T h a i T o - T h j x a Q u ^ n g T r j - V T h i S a u , 3/5 T o n Dure T h i n g - L o n g X u y i n - ^ V V 67 N g u y i n K h o a i - H a N^i * DT: (04) 9 - ChUOng D i f d n g D p - H a N o i ^ ISBN: 978-604-939-543-7 N V - ' ^ ^ ^ 935092 756783 Gia: 84.000d • - 8 D U d n g L a n g - H a N p i * D T : (04) 5 >" y y y y _ y ... a)x .2' = x ( - x ) + 2( 2''-1) b) 2" ^'- 4" = x - Giai a) PT: X .2' ' - x(3 - x) - 2. 2" + = 2" (x - 2) + x^ - 3x + = o 2" (x - 2) + (x - l ) ( x - 2) = « (x - 2X2" + x - 1) = o X - = hoac 2" +... (2^ ' - 2. 2' + 1) + 2( 2"- l)sin (2" + y - 1) + = o (2" - f + 2( 2" - l)sin (2" + y - 1) + s i n ^ + y - 1) + cos^ (2" + y - 1) = o [2" - + sin (2" + y - 1)]^ + cos^ (2" + y - 1) = 2' + sin (2^ + y -... 26 5 Bai toan 2: Tinh: a) I = x^cos2xdx b) J = sin Vxdx Gidi a) Dat u = x^ v' = cos2x K h i u' = 2x, v = — sin2x I = x^sin2x - fxsin2xdx = — x^sin2x + — rxd(cos2x) J 2^ = ~ x^sin2x + ~ xcos2x