Tìm hiểu các phương pháp giải các chủ đề căn bản giải tích 12 phần 2

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 PT: log, (x + 2)(x + 3) x-2 x +3 - l o g l < : ^ x ^ - = 16 x^ = 20 « X = ± Vs (chon) b) D K : X > X ^ PT: ^ log2(x + 12) —L_= i log,X « log2(x + 12) = log2X^ « X + 12 = x^ x^ - X - 12 = Chon nghiem x = Bai toan 6: Giai cac phucrng trinh: a)i|„g,(x-2)-i=,|„g,V?x3^ 3 b)-!5fcf ^ J ^ f o ^ log phuomg trinh tra thanh: ^ log2(x - 2) + I log2(3x - 5) = ^ « D O log2(x - 2)(3x - 5) = J o (x - 2)(3x - 5) = x = hoac " ~ • ^hon nghiem x = b) D K : x > 0, X ^ -J x ^ -—, dat t - logsx thi PT: 27 t ^ ( + t) t'^ + t - = < = > t = l hoac t = -4 + t 3(3+ t ) Suy nghiem x = hoac x = — " Bai toan 7: Giai cac phuong trinh: a) l o g U x ) + log, — = 8 b) lo^,,x+31og,x + log, x = 2 Giai a) D K : x > 0, ta c6 log, — = log, x" - log, = 21og, x - logU4x) = log, + l o g , X = ( - - l o g x ) ' = ( + log2 x ) ' V 2 J Dat t = log2X thi PT: (2 +1)^ + 2t - = t ' + 6t - = o t = hoac t = -7 Suy nghiem x = 2'^ hoac x = 227 b) D K : X > 0, dat t = log x thi PT: r + - t - - t =2 « t ^ +t- 2=0 2 t = hoac t = -2 Giai nghiem x = ^ , x = V2 Bai toan 8: Giai cac piiuomg trinh: a) log4log2X + log2log4X = b) log^, 16 + log^^ 64 = Giai a) D K : x > 1, phuong trinh tra (\ log,, log, x + log, log,, x = « - log, log2 X + log, - log, x = \1 « ^ l0g2l0g2X + l o g ^ + l0g2l0g2X = < ^ | log2log2X = log2log2X = logix = X = 16 (chon) b) D K : X > 0, X ^ 1, X ^ ^ thi P t : 21og,2 + + log, X = 3-~-^ + ^ =3 log, X + log, X Dat t = log2X thi PT: ^ + - ^ = « t ' - t - = t 1+t t = hoac t = - - Suy nghiem x = 4, x = ^ • Ml Bai toan 9: Giai cac phuang trinh a) log.sX log3X = logsX + log3X b) 21og2X.log5X + log2X - lOlogsx = Giai a) D K : x > 0, ta c6 x = la mot nghiem N S u x ^ l t h i P T : log^51og,^3 i = -log^5 — + l o^g , logx5 + logx3 = logxlS = x = 15 (chon) b) D K : X > 0, PT: log2X (21og5X + 1) - 5(21og5X + 1) = 0, v i ham so ve trai dong bien, ham so ve phai nghich bien va x = la nghiem nen la nghiem nhat b) D K : X > dat logsx = y thi x = 3^ y PT: log,(l + V F ) = y » + = 2^ — + =1 Ta CO y = thoa man phuong trinh, v i ve trai la ham nghjch bien nen PT c6 nghiem nhat y = nen x = Bai toan 11: Giai cac phuong trinh: a)21og2X = x b) log, ^^"''•^"'"^ = x " + x + • 2x- + x + Giai M-.i^ ^ A n-T- I X a) D K : x > PT: log2X = — w - i /-/ X Xet ham so f(x) = Iri X In In X = X « - X , x > thi f(x) = - In X — X f'(x) = X" = e, lap B B T thi f(x) = In CO toi da nghiem ma f(2) = f(4) = - — nen S = {2; } « X b) Phuong trinh: log, — r = (2x" + 4x + 5) - (x^ + x + 3) " 2x- + x + log,(x" + x + 3) + (x" + X + 3) = log3(2x' + x + 5) + (2x" + x + 5) Xet ham s6 A O = l o g , r + ^ / > t h i / ' ( O = — * — + > 0, V / > ^ln3 Do f(t) dong bien, nen phuong trinh / ( x - + x + 3) = / ( x - + x + 5) « X - + x + = x ^ + x + x ^ + x + = Vay phuong trinh c6 nghiem x = -1 va x = -2 Bai toan 12: Giai cac phuong trinh: a) log2(4x^ - 7x^ + 1) - log2X = log4(2x^ - ) ^ + b) l o & ( l V ^ + > / ^ ) + x = 2+lo&_(4^+4) Gidi 4x'-7x'+l>0 a) Dieu kien: \ < X^ —f= V2 229 Phuong trinh da cho tuang duong vai log2(4x'' - 7x^ + 1) = log22x I 2x - « x - * - x ^ + - x 2x-1 «4x^+-iT-7 = 22x Dat t = X - t > va phuong trinh tro thanh: t - 2t - = X Chon nghiem t = Vai x - - = « x ' - x - l = « j c = 3±Vi7 X Vai v - - = - « > v ' + x - l = c» x = X ,x = - + Vi7 ru • ' phuong u Chon ngniem cua trmh la: x = + ^ 4 b) Di6u kien -1 < x < Khi phuong trinh tuang duong vai log, (l TVXTT + 34V3^7)= log, 4(4^ + 4) - log^ 2" « log, (l 7y[^ + ^ " ^ ) = logj 4(2^ + 2'-") ( l V ^ + 34V3^)=4(2"+2-'^) Xet ham so f(x) - 17v'x+ T + 34V3 - x fix) = 17 24x + r(x) = « 34 2V3-X x f(3) = 34 Dat 2' = t, ta CO -1 < X < nen ^ < t < 8, ( 4^ Do do: 4(2' + 2-"') = t + Xet ham so g(t) = t + g'(t) = , vai - < t ^ : ,g'(t)=0 « t=2 230 Lap B B T thi g(t) < g( ^ ) = g(8) = 34 •,2-Xx b) l0g2(x - Vx' -1 ) + log3(x + Vx- -1 ) = l0g6(x +Vx' -1 ) Gidi a) Dieu l -1 PFiuong trinh da cho tuang duong vai: log2(x + 2) - log3(x + 1) = Xet ham s6 f(x) = log2(x + 2) - log2(x + 1), x > -1 f(x)- 1 (In3-ln2)x-(ln4-ln3) (x+2)ln2 (x+l)ln3 (A:+l)(x+2)ln21n3 r(x)^0«x^^;4:i;^e(0;2) In3-ln2 Lap B B T thi phuang trinh f(x) = c6 nhieu nhat nghiem Ma X = X = thoa man phuang trinh Vay phuang trinh cho c6 nghicm la x = va x = b) Dieu kien la x > Dat t = x - %/x" -1 x +v'x" -1 = - FT: log2t + log3 Y = Iog6 ~ log2t - log3t + log6t = o log2t( - log32 + log62) = l0g2t = t = Do do: X - Vx" -1 = X - = Vx" -1 x^ - 2x + = x" - x ^ 1: chon Vay nghiem x = Bai toan 14: Lim dieu kien de phuang trinh: a) log^ X + -y/log^ X +1 - 2m -1 = c6 nghiem thuoc doan ];3^'^ b) log -^(x + 3) = log, (ax) CO nghiem nhat Gidi a) Dat t = -Jlog^ X + , x 1; 3^'-^ > 3 + 5x + Xet X = 0: Loai Xet x ;t thi c6: a = Dat f(x) = X' +6x + X- ,x>-3,x;^0,f'(x)= ,x>-3 -9 ^ , f ' ( x ) = 0thix = X" BBT: X f -3 0 - +00 - + +00 f +00 -00 12 Dieu kien c6 ngliiem nhat: a < hay a = 12 DANG TOAN pH|/DfWG TRINH MU VA LOGARIT Viec gidi he phuang trinh mu vd logarit ve ca bdn cung giong nhu gidi cdc he phuang trinh dgi so vai cdc bien doi ve bieu thuc mil vd logarit Phuangphdp chung gidi he: Rut the, cong dgi so, dat an phu Chiiy: 1) Diing dinh nghia, bien doi phuang trinh tich so, dung bat dang thuc, dgo hdm, f ax + by = c 2) He phuang trinh bdc nhdt hai dn: \ [a'x + b'y = c' Phuang phdp the, cong dgi so, dung dinh thuc, dung may tinh, He CO mot phuang trinh bdc nhdt: ta chon rut mot dn theo dn Igi, the vao phuang trinh roi gidi phuang trinh mot dn 3) He doi ximg logi I 'F,(x,y) = ' F , ( x , y ) = 0' Dgt X ^ >' = S vdxy = P vai dieu kien He doi xung logi II > 4P 'F,(x,y) = F2(y,x) = 232 Tru hai phuang trinh dua ve tich so (x - y).A(x, y) = \ax'+bxy + cy' = d 4) He dang cap (thudn nhdt): < \a'x' +b'xy + c' y' = d' Xet x ^ 0, xet x ^ 0, dgt y kx, dua ve giai theo an k Hoac ngiiac Igi, xet y xely ^0, vd dgt x - Ay Bai toan 1: Giai cac he phuang trinh: X 2^+2.3"" =55 a) 3.2^+3^^^'' b) =84 +y=1 4-2^+4-'^ =0,5 Giai a) Dat u = \ = 3"^^ thi u v > fu + 2v = 55 he: CO b) y =1- X He < 4^^ +4 .4'^ =0,5 " 3u + 3v = 84 rx = o u =l Ia V = 27 y =3 2x = l X [y = l - x Cach khac: Dat u = \ = 4* thi x + y = « =y = ' uv = Bai toan 2: Giai cac he phuong trinh: x + y = 20 a) b) logj x + log^ y = + log4 log,(y-x)-log,- = l y x' + y ' = (1) (2) Giai a) D K X > 0, y > 0, he tuong ducrng: x + y = 20 x +y= log4 xy = log^ 36 20 xy = 36 Tir giai dugc nghiem (2; 18) va (18; 2) b) DK: y > X, y > Ta c6: (1): log I (y - x ) - log,, - = y - l o g , ( y - x) = - l o g , - = y 0, 3x - y > < (3x + y)(3x - y) = « 3x + y = 5(3x-y) Giai 9x'-y'=5 i 3x + y > x - y >0 3x + y = 5(3x-y) ly = 3x + y = x=l 3x-y = l b) I)K: x + y ^ He tuang duong: [log, (x - y) = log,(x + y) - log, 2.log,(x - y) = f log, (x + v) = log, (x + y) + log, (x - y) = X = - (chon) ''2 B a i toan 4: Giai cac he phuang trinh: 'log,(x-y) = 5-log,(x + y) a) logx-log4 logy-log3 21og,x-3^ =15 b) = -1 3Mog2X = 21og,x+3^'' Giai a) DK: x > y > He tuang duang: 12 12 X = - x^-y==32 xv = 12 X = s 144 , „ - v - y =22 y'+22y'-144=0 rCr giai dugc nghiem x = 6, y = b) DK X > dat u = logix va v = 3^ (y > 0) He tuong duang: u =9 hoac u =-2 (loai) v = -10 v = 2u-15 u - v = 15 uv = 2u + 3v i 2u'-23u + 45 = v =3 Tu giai nghiem (512; 1) Giai cac he phuang trinh: B a i toan 5: a) [log^(6y + 4x) = 21og,x + log,y = jlog, (6x + 4y) = log, (x' + y ' ) = Giai a) DK x y > 234 He tuong duong: log,(x-+y') = fx-+y'=32 < logj x + logj y = [xy = 16 (x + y ) ' - x y = 32 f(x + y ) ' = < xy = 16 b) DK: x, y > 0, x, y 7t fx = " y=4 [xy = 16 |6x + 4y = x" f6x + 4y = x ' He tuong duong: T a c ( l ) < » ' " ^ - ' = 2'- « va (2) o x - 3y = ^ -logax + l = log3(9y) « • log3(xy) = -1 - » xy = | Tird6c6S = { ( ; - ) } Bai toan 7: Giai cac he phuong trinh: (1) a) y-=x^ log, ( x- + y - ) = l + l o g , ( x y ) b) (2) >X - XV + V -81 Giai a) DK: x, y > Ta c6 (2) 0, Vx nen f d6ng bi6n tren R Ta CO PT: f(x) = f(y) o X = y Do 2" + 2x = + X 2" + X - = Xet ham g(x) = 2" + x - 3, x e R, tir suy he c6 nghiem (1; 1) b) DK: X , y > nen xy + 2x + > Vi ca so > 1, e > nen vai (1): NSu x > y thi VT > > VP, mu x < y thi v r < < VP, Neu X = y thi thoa man Do (2) x^ + X - = 0, chon x = => y = Vay he c6 nghiem (2;2) Bai toan 9: Giai cac he phuang trinh: a) 'log,x + log,>; = log,(x + 2) (x + Vl+7)(y + V l + ) = l 0) [4x + y + l = -^"'' b) (2) 3-^'"' + = 5.3 ^ Giai a) PT (1) bien doi thanh: X + Vl + X " = ^j\ y' - y va y + y]\ y^ = ^l\ x' - x 236 ... 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 -... logx-log4 logy-log3 21 og,x-3^ =15 b) = -1 3Mog2X = 21 og,x+3^'''' Giai a) DK: x > y > He tuang duang: 12 12 X = - x^-y== 32 xv = 12 X = s 144 , „ - v - y =22 y'' +22 y''-144=0 rCr giai dugc nghiem x = 6,