Tìm hiểu các phương pháp giải các chủ đề căn bản Giải tích 12: Phần 2

Chia sẻ: Nhân Y | Ngày: | Loại File: PDF | Số trang:201

Thêm vào BST

Báo xấu

19
lượt xem 2
download

Download Vui lòng tải xuống để xem tài liệu đầy đủ

Cuốn sách đưa một số bài tập cùng phương pháp giải nhằm rèn luyện kỹ năng thực hành của các em. Cuốn sách này sẽ là tài liệu tham khảo thiết thực và bổ ích giúp các em ôn luyện chuẩn bị cho những kỳ thi tại trường, lớp, các kỳ thi học sinh giỏi,... Mời các bạn cùng tham khảo.

Chủ đề:

Bình luận(0) Đăng nhập để gửi bình luận!

Lưu

Nội dung Text: Tìm hiểu các phương pháp giải các chủ đề căn bản Giải tích 12: Phần 2

X X (15^ b) — > 3" + 4 " + 5 " vai moi x bJ + [5, HDDS a) Xet ham s6 f(x) = e" - x - 1, x > 0 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) = 3 tren doan [-1; 1 ]. b) f(x) = x - Inx + 3 tren khoang (0; + 0 0 ) HD-DS a) Kk qua min f ( x ) = f ( 0 ) = 1 ; max f ( x ) = f ( ± l ) = 3 xe[-l,l] xe[-l,l] b) Kel qua m i n f ( x ) = f (1) = 4 , 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. 2 0 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 duy nhat x = logah. "a = l - Phuang trinh mii i/'"" (f'''^ (a > 0) 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 4 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 )^''= 2 - Vs . Giai 5x 5x a) PT: 0,125.4'"-' = (4 V2 )" « T\' = 2'^" « 2'''-' = 2= 0 2 « 4x - 9 = — « 8x - 18 = 5x » X = 6. 2 b) PT:(2+ V3)''' = 2 - (2 + 73)'" = ( 2 + Vs y ' 2 x =-1 o x = Bai toan 2: Giai cac phuong trinh sau: ylogx _ ^Iogx+1 _ 2 ^logX-1 _ J2 ylOgx-1 a) 9 ^ - 2 2 = 2 2 -3'"-' G/di I ! I x+- x+- a)PT: 9^+-.9^ = 2 ^+2.2 ^ -.9''=3.2 ^ 3 3 ^9Y'^V2 X -1 = log, — 0 thi PT: 218
- 31 - 4 + y =0« - 3t^ - 4t + 12 = 0 (t - 2)(t + 2)(t - 3) = 0. Chon nghiem t = 2 hoac t = 3 nen x = ln2 hoac x = ln3. b) Chia 2 vk cho 8' > 0 thi PT: 27 ^12Y V 8 y + 8 - 2 = 0. Dat t = ,t>0. .2) PT: t^ +1 - 2 = 0 « (t - l)(t^ +1 + 2) = 0 « t = 1 « X = 0. Bai toan 5: Giai cac phuomg trinh: a) 2.25"+ 5.4'= 7.10' b) 4 " + 6 " = 9 " Giai 2\ (2^ f2^ a) PT: 5 -7 2 = 0. Dat t = ,t>0. .5j .5, PT: 5t^ - 7t + 2 = 0 t = 1 hoac t = - (thoa man) , Suy ra nghiem x = 0 hoac x = 1. b) Dieu kien x 0. dat y = — va chia hai vi cho 4^, ta c6: 2v (3^ '3^ y '3^ • i+Vs , 1+V5 — — - 1 =0 .2, .2, -1 1 , 1 + V5 1 , 1 + V5 o - -X = log,2 — 2 - — « -X = log, « X = log^^_, - . 2\ Bai toan 6: Giai cac phuong trinh: a) (V2-V3r+(V2+V3 =4 = 6. / a) Ta CO V2-V3.V2 + V3 = 1, dat t = fV2 + V3l , t > 0. 1 PT: t + - ^ = 4 c ^ t - - 4 t + l = 0 t o t = 2 + V3 hoac t = 2 - V3 « X = 2 hoac x = -2. 5 b) D a t t = 2^^^'^' - , t > 0 thi PT: t^ - - t = 6 « 2t^ - 5t - 12 = 0. 2
Chon nghiem t = 4 nen x + Vx^ - 2 = 2 Vx'^ - 2 = 2 - x 3 2-x>0vax^-2 = 4-4x + x^ 0, dat t = log3X thi x = 4' 220
PT: V3 . 3' + 4= V3 • 3' = 2' 4.3' = V3 . 2' V3 V3 '"^^4 t=log, — . V a y x = 4 ' . 2) 4 ; 4 • b) V i (4 - Vis )(4 + Vl5 ) = 1 nen dat (4 - VlS f = t, t > 0 thi phuong trinh: t + | - = 8 < : : > t ^ - 8 t + l = 0 < = > t = 4 ±4\5 . Do do tanx = -1 hoac tanx = 1 nen nghiem x = ±— + krc, k e Z. 4 Bai toan 10: Giai cac phuong trinh: b) 4 ^ - 3 ^ = 1 . a) (sin ^ ) " + (cos ^ ) " = 1 Giai a) V i 0 < sin— < 1 va 0 < cos— < 1 do do: 5 5 N^u X > 2 thi ta c6 (sin ^ f < (sin ^ f va (cos )" < (sin 5~ ) ^ VT < 1 (loai). Neu X < 2 thi ta CO (sin — )^ > (sin ~ )^ va (cos — > (sin ~ f 5 5 5 5 V T > 1 (loai). Neu X = 2 thi PT nghiem dung, do la nghiem duy nhat b) P T: ( — )^ + ( — ) " 1 va ta CO X = 2 thoa man PT. V i ve trai la ham so nghich 4 4 bien tren R nen c6 nghiem duy nhat x = 2. Bai toan 11: Giai cac phuong trinh: a)x.2' = x ( 3 - x ) + 2(2''-1) b) 2"^'- 4" = x - 1. Giai a) PT: X.2'' - x(3 - x) - 2.2" + 2 = 0 2"(x - 2) + x^ - 3x + 2 = 0 o 2"(x - 2) + (x - l ) ( x - 2) = 0 « (x - 2X2" + x - 1) = 0 o X - 2 = 0 hoac 2" + x = l x = 2 hoac x = 1. (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 + 1 Vi f'(t) > 0, V t nen f d6ng b i l n tren R. PT f(x + 1) = f(2x) < ^ x + l = 2 x o x = l . 221
Bai toan 12: Giai cac phuong trinh: a) 3^'^'^'+2|x| = 3^^' b) 2 ^ ^ V ^ ( x + l) = ^(2^^"^"^'^'+V2^). Giai a) Phuong trinh da cho xac dinh voi mpi x. Xet x < 0. Khi do ta c6 3''"'^' + 2 | x | > 3 > 3"^', nen phuong trinh da cho khong C O nghiem trong khoang (0; + t » ) Xet x ^ 0. Phuong trinh tro thanh \ = 3''"'-(x+l)' Ta CO Vx" + 1 , X + 1 >1 Xet ham s6 f (t) = 3' - 1 ^ voi t e [1; +oo) f (t) = 3' ln3 - 2t, f "(t) = 3' (ln3)^ - 2 Vi 3'(ln3)^ - 2 ^ 3(ln3)' - 2 > 0, Vt > 1, nen f "(t) > 0, Vt ^ 1 Suy ra f '(t) la ham so dong bien tren [1; +oo) Do do f'(t) > f (t) = 31n3 - 2 > 0, Vt ^ 1. nen f(t) la ham so dong bien tren [ 1; +co) Phuong trinh: Vx'+l =x +l fx'+l = x'+2x + l f ( V x ' + l ) = f ( x + 1) « x =0 x>0 x^O Vay phuong trinh c6 nghiem duy nhat x = 0. 2x-l>0 1 b) Dieu kien 3x4-1^0 2 Phuong trinh tro thanh 2>/2x- ) / 3 x + l+l -.1 + A/2X-1 2 2 ,/3x +l +l 2 ,2 '^^/3x + l+l 2 2' 2^^^ 2' 222
Taco V 2 x - 1 + l , V 3 x + l > 1 Xet ham so f(t) = - ^ voi t e [1; +00) f (t) = ln2 - 1 ; f "(t) = 2'^'(ln2)^ - 1 V i t ^ 1 nenf"(t)>(21n2)^ - 1 > 0 Suy ra f '(t) la ham s6 d6ng bi^n tren [ 1 ; +00) Nen f (t) > f ' ( t ) = 41n2 - 1 > 0, Vt > 1. Do do f(t) la ham so d6ng bi^n tren [ 1 ; +00) Phuong trinh f ( V 2 x - 1 +1) = f(V3x + 1 ) V 2 x - 1 + 1 = V3x + 1 o 2x + 2 V 2 x - l =V3x + l X +1 ^ 0 o 2V2JC-I = X + 1 • X = 1 va X = 5 4 ( 2 x - l ) = x^ + 2 x + l Vay phuong trinh da cho c6 2 nghiem x = 1, x = 5. Bai toan 13: Giai cac phuong trinh: a) 5' + 4" + 3" + 2" = ^ ^ + — - 4 x ' + 2 x ' - x + 16 Y y 6' b) 4" - 2^"' + 2(2" - l)sin(2'' + y - 1) + 2 = 0. Giai I I P a) Xet ham s6: fi;x) - 5" + 4" + 3" + 2 ' - -+—+— +4x^-2x^+x-16,xeR 2" 3^ Ta c6: f ' ( x ) = 5'ln5 + 4''ln4 + 3 " ^ + 2''ln2 ln2 ln3 l n 6 -+ -+ + 12x^-4x + 1 > 0 V 2^ 3" 6" y Suy ra ham so dong bien va phucrng trinh f(x) = 0 CO khong qua mot nghiem va f ( l ) = ( Vay phuong trinh da cho c6 nghiem duy nh^t la x = 1. b) Phuong trinh da cho tuong duong v o l (2^' - 2.2' + 1) + 2(2"- l)sin(2" + y - 1) + 1 = 0 o (2" - \f + 2(2" - l)sin(2" + y - 1) + s i n ^ + y - 1) + cos^(2" + y - 1) = 0 o [2" - 1 + sin(2" + y - 1)]^ + cos^(2" + y - 1) = 0 2' + sin(2^ + y - l ) = l cos(2' + y - l ) = 0 Vi cos(2" + y - l ) = 0 = > s i n ( 2 " + y - 1) = ± 1 .
Ta CO hai truomg hop sau: - NSu sinCZ" + y - 1) = 1 thi 2" = 0, v6 nghiem - NSu sin(2' + y - 1) = -1 thi 2" = 2 » X = 1 TC Suy ra sin(y + l ) = -ly = - — - 1 + k2n. Vay phuong trinh da cho c6 nghiem la: x = 1, y = - — - 1 + 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 duy nhdt. Gidi a) Dat t = 3"""', vi 0 < s i n \ 1 nen 1 < t < 9 9 _ t^-9 m. Xet f(t) = t + ^ , 1 < t < 9 ; f ' ( t ) ; f'(t) = Okhit t BBT: X 1 3 9 f 0 + f 10 10 " ^ ^ ^ 6 Vay diSu kien f(t) = m c6 nghiem thoa l < t ^ 9 1 a 6 < m < 1 0 . X b) p r « + 2m =1 2 , V5+1 Vs laco: . - ' = 1 , datt = 2 2 V PT: t + — = 1 » t^ - 1 + 2m = 0 t Xet t = 0 ^ m = 0 thi PT: t^ - 1 = 0 « t = 0 hay t = 1: thoa man Xet t ^ 0, diSu kien c6 nghiem t > 0: ti < 0 < t2 hoac 0 < ti < t2 « P < 0 hoac (A > 0 , P > 0 , S > 0 ) < = > m < 0 hoac m = - 8 Vay: m < 0 hoac m = - 8 Cach khac: Xet ham so va lap bang bien thien. 224
DANG TOAN PHlTOfNG TRINH LOGARIT 2. - Phuang trinh logarit ca ban: logaX = b (a > 0, a ^ 1) Phuang trinh logarit ca ban luon co nghiem duy nhdt x = a*. - Phuang trinh logarit f(x)>Ohayg(x)>0 log/fx) = logagfx), (a>0, a^l) ^ 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 4 phuang phdp chinh de gidi phuang trinh Idgrarit, ta co the dimg dinh nghia, bien doi thanh phuang trinh tich so, diing hat dang thicc,... Bai toan 1: Giai cac phucmg trinh sau: a) log2|x(x - 1)] = 1 b) log2(9 - 2'') = 1 o'"^*^-''*. Gidi a) PT: log2[x(x - 1)| = 1 « x(x - 1) = 2 » x^ - x - 2 = 0 x = -1 hoac x = 2. b) Dieu kien X < 3. PT: log2(9 - 2") = lO'"^''-"* « 9 - 2' = 2^"^ « 2^" - 9.2'* + 8 = 0 2" = 1 hoac 2" = 8. Chon nghiem x = 0. Bai toan 2: Giai cac phucmg trinh sau: 1 a) = 3 b) S^log.C-x) = logj V x ^ . 5-41ogx 1 + logx Giai a) Vai x > 0. dat t = logx thi F T : 1 j — ^ ^ +— = 3 . t ^ ^ ,t ^ - 1 » 2t - 3t + 1 = 0 t = 1 hoac t = - (chon). Suy ra nghiem x = 10 hoac x = VTo . b) DK: X < 0, PT: 5 V l o g , ( - x ) = l o g , ( - x ) « V l o g , ( - x ) . ( 5 - Vlog^C-"^) = 0 o ^log2(-x) = 0 hoac ^log2(-x) = 5 x = -1 hoac x = -2^"\ 225
Bai toan 3: Giai cac pliuong trinh: a) log2X + log2(x - 1) = 1 b) log2X + log3X + log4X Giai a) D K : x > 1, PT < ^ log2x(x - 1) = 1 o x(x - 1) = 2 x^ - X - 2 = 0. Chon nghiem x = 2. b) D K : X > 0, PT: (1 + log32 + log42).log2X = 1 1 (3 + log32)log2X = 1 - » log2X = 3 + log, 2 Vay nghiem x = 2^''^'"^'". Bai toan 4: Giai cac phuang trinh: a) log3(3^-l). log3(3"^'-3) - 12 b) log,.,4 = 1 + 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 t ' + t - 12 = 0 « t = - 4 hoac t = 3. « log3(3' - 1) = -4 hoac log3(3' - 1) = 3 « 3' - 1 = ~ hoac 3' - 1 = 27 81 8? o 3' = — hoac 3' - 28 X = log382 - 4 hoac x = log328 81 b) D K : x > 1.x ^ 2 , PT: 1 + l o g 2 ( x - 1) log^Cx-l) Dat t = log2(x - 1) thi PT: - = 1+ t « t^ + t - 2 = 0 t = 1 hoac t = -2. Giai ra nghiem x = — hoac x = 3. 3 • Bai toan 5: Giai cac phuong trinh: a) log4L(x + 2)(x + 3)1 + ~ log2 - 2 2 x+3 b) l o g 4 ( x + 1 2 ) . I o g x 2 = l . Giai (x + 2)(x + 3 ) > 0 x0 x>2 x +3 226
x-2 PT: log, (x + 2)(x + 3) - l o g 4 l 6 < : ^ x ^ - 4 = 16. x +3 x^ = 20 « X = ± 2 Vs (chon). b) D K : X > 0. X ^ 1. PT: ^ log2(x + 12) —L_= i . 2 log,X « log2(x + 12) = log2X^ « X + 12 = x^ x^ - X - 12 = 0. Chon nghiem x = 4. Bai toan 6: Giai cac phucrng trinh: a)i|„g,(x-2)-i=,|„g,V?x3^ b)-!5fcf ^ J ^ f o ^ . 6 3 3 log 2. phuomg trinh tra thanh: ^ log2(x - 2) + I log2(3x - 5) = ^ « log2(x - 2)(3x - 5) = 2 D O J 2 o (x - 2)(3x - 5) = 4 x = 3 hoac " ~ • ^hon nghiem x = 3. b) D K : x > 0, X ^ -J . x ^ -—, dat t - logsx thi PT: 3 27 t ^ 2 ( 2 + t) t'^ + 3 t - 4 = 0 < = > t = l hoac t = -4. 1 + t 3(3+ t ) 1 Suy ra nghiem x = 3 hoac x = — . " 8 1 Bai toan 7: Giai cac phuong trinh: a) l o g U 4 x ) + log, — = 8 b) lo^,,x+31og,x + log, x = 2. 1 8 2 Giai a) D K : x > 0, ta c6 log, — = log, x" - log, 8 = 21og, x - 3 8 logU4x) = log, 4 + l o g , X = ( - 2 - l o g 2 x ) ' = ( 2 + log2 x ) ' 2 V 2 2 J Dat t = log2X thi PT: (2 +1)^ + 2t - 3 = 8 t ' + 6t - 7 = 0 o t = 1 hoac t = -7. Suy ra nghiem x = 2'^ hoac x = 2. 227
b) D K : X > 0, dat t = log x thi PT: r + - t - - t =2 « t ^ +t- 2=0 2 2 t = 1 hoac t = -2. Giai ra nghiem x = ^ , x = V2 . Bai toan 8: Giai cac piiuomg trinh: a) log4log2X + log2log4X = 2 b) log^, 16 + log^^ 64 = 3 . Giai a) D K : x > 1, phuong trinh tra thanh 1 (\ log,, log, x + log, log,, x = 2 « - log, log2 X + log, - log, x = 2 2 \1 « ^ l0g2l0g2X + l o g 2 ^ + l0g2l0g2X = 2 < ^ | log2log2X = 3. log2log2X = 2 logix = 4 X = 16 (chon). b) D K : X > 0, X ^ 1, X ^ ^ thi P t : 21og,2 + = 3-~-^ + ^ =3 1 + log, X log, X 1 + log, X Dat t = log2X thi PT: ^ + - ^ = 3 « 3 t ' - 5 t - 2 = 0 t 1+t t = 2 hoac t = - - . Suy ra nghiem x = 4, x = ^ 3 • Ml Bai toan 9: Giai cac phuang trinh a) log.sX. log3X = logsX + log3X b) 21og2X.log5X + log2X - lOlogsx = 5. Giai a) D K : x > 0, ta c6 x = 1 la mot nghiem. i N S u x ^ l t h i P T : log^51og,^3 = -log^5 — + l o^g , 3 logx5 + logx3 = 1 logxlS = 1 x = 15 (chon) b) D K : X > 0, PT: log2X (21og5X + 1) - 5(21og5X + 1) = 0
Bai toan 10: Giai cac phirong trinh a) log2X = 3 - X b) l o g 2 ( l + Vx ) = logax. Giai a) D K : x > 0, v i ham so ve trai dong bien, ham so ve phai nghich bien va x = 2 la nghiem nen do la nghiem duy nhat. b) D K : X > 0. dat logsx = y thi x = 3^ y PT: log,(l + V F ) = y » 1 + = 2^ =1 — + 2 Ta CO y = 2 thoa man phuong trinh, v i ve trai la ham nghjch bien nen PT c6 nghiem duy nhat y = 2 nen x = 9. Bai toan 11: Giai cac phuong trinh: a)21og2X = x b) log, ^^"''•^"'"^ = x " + 3 x + 2 . • 2x- + 4 x + 5 Giai M-.i^ ^ A n-T- I X In X In 2 a) D K : x > 0. PT: log2X = — = . 2 X 2 w 1 - i /-/ X Iri X « 1 - X 1 - In X Xet ham so f(x) = , x > 0 thi f(x) = — X X" f'(x) = 0 = e, lap B B T thi f(x) = 0 « X In 2 CO toi da 2 nghiem ma f(2) = f(4) = - — nen S = {2; 4 } . b) Phuong trinh: log, — r = (2x" + 4x + 5) - (x^ + x + 3) " 2x- + 4 x + 5 log,(x" + x + 3) + (x" + X + 3) = log3(2x' + 4 x + 5) + (2x" + 4 x + 5) Xet ham s6 A O = l o g , r + ^ / > 0 t h i / ' ( O = — * — + 1 > 0, V / > 0 ^ln3 Do do f(t) dong bien, nen phuong trinh / ( x - + x + 3) = / ( 2 x - + 4 x + 5) « X - + x + 3 = 2 x ^ + 4 x + 5 x ^ + 3 x + 2 = 0 Vay phuong trinh c6 2 nghiem x = -1 va x = -2 Bai toan 12: Giai cac phuong trinh: a) log2(4x^ - 7x^ + 1) - log2X = log4(2x^ - 1 ) ^ + 1 . b) l o & ( l 7 V ^ + 3 4 > / 3 ^ ) + x = 2+lo&_(4^+4). Gidi 4x'-7x'+l>0 a) Dieu kien: \ 0 < X^ —f= V2 229
Phuong trinh da cho tuang duong vai log2(4x'' - 7x^ + 1) = log22x I 2x - 1 « 4 x - * - 7 x ^ + 1 - 2 x 2x-1 «4x^+-iT-7 = 22x-- Dat t = 2 X - 1 . t > 0 va phuong trinh tro thanh: t - 2t - 3 = 0. X Chon nghiem t = 3. 3±Vi7 Vai 2 x - - = 3 « 2 x ' - 3 x - l = 0 « j c = X 4 Vai 2 . v - - = - 3 « > 2 . v ' + 3 x - l = 0 c» x = X 4 ru Chon • ngniem cua u ' phuong trmh la: x = 3 + ^ ,x = - 3 + Vi7 4 4 b) Di6u kien -1 < x < 3 Khi do phuong trinh tuang duong vai log, (l TVXTT + 34V3^7)= log, 4(4^ + 4) - log^ 2" « log, (l 7y[^ + 3 4 ^ 3 " ^ ) = logj 4(2^ + 2'-") ( l 7 V ^ + 34V3^)=4(2"+2-'^) Xet ham so f(x) - 17v'x+ T + 34V3 - x x f(3) = 34 Dat 2' = t, ta CO -1 < X < 3 nen. ^ < t < 8, ( 4^ Do do: 4(2' + 2-"') = 4 t + - Xet ham so g(t) = 4 t + - , vai - < t ^ 8 2 g'(t) = 4 1 : ,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) = 1 Xet ham s6 f(x) = log2(x + 2) - log2(x + 1), x > -1 1 1 (In3-ln2)x-(ln4-ln3) f(x)- (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) = 1 c6 nhieu nhat 2 nghiem. Ma X = 0. X = 2 thoa man phuang trinh. Vay phuang trinh cho c6 2 nghicm la x = 0 va x = 2. b) Dieu kien la x > 1. Dat t = x - %/x" -1 x +v'x" -1 = - FT: log2t + log3 Y = Iog6 ~ log2t - log3t + log6t = 0 o log2t( 1 - log32 + log62) = 0 l0g2t = 0 t = 1. Do do: X - Vx" -1 = 1 X - 1 = Vx" -1 x^ - 2x + 1 = x" - 1 x ^ 1: chon. Vay nghiem x = 1. Bai toan 14: Lim dieu kien de phuang trinh: a) log^ X + -y/log^ X +1 - 2m -1 = 0 c6 nghiem thuoc doan ];3^'^ b) log -^(x + 3) = log, (ax) CO nghiem duy nhat. Gidi a) Dat t = -Jlog^ X + 1 , x 1; 3^'-^
DiSu kien c6 nghiem: f ( l ) < 2m + 2 < f(2) 2 < 2m + 1 < 6 » 0 < m < 2 rx + 3 > 0 b) PT: 21og3(x + 3) = log3(ax) « log3(x + 3)' =log3(ax) (x + 3)^ = ax, X+ 3 > 0 >--3 3 + 5x + 9 Xet X = 0: Loai. Xet x ;t 0 thi c6: a = ,x>-3 X' +6x + 9 X- -9 Dat f(x) = ,x>-3,x;^0,f'(x)= ^ , f ' ( x ) = 0thix = 3 X" BBT: X +00 -3 0 3 f 0 - - 0 + f +00 +00 -00 12 Dieu kien c6 ngliiem duy nhat: a < 0 hay a = 12. DANG TOAN pH|/DfWG TRINH MU VA LOGARIT 3. 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 thanh 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 con Igi, the vao phuang trinh kia roi gidi phuang trinh mot dn. 'F,(x,y) = 0 3) He doi ximg logi I ' F , ( x , y ) = 0' Dgt X ^ >' = S vdxy = P vai dieu kien > 4P. 'F,(x,y) = 0 He doi xung logi II F2(y,x) = 0 232
Tru hai phuang trinh dua ve tich so (x - y).A(x, y) = 0. \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: 2^+2.3"" =55 X +y=1 a) b) 3.2^+3^^^'' =84 4-2^+4-'^ =0,5 Giai a) Dat u = 2 \ = 3"^^ thi u. v > 0. fu + 2v = 55 u =l rx = o Ia CO he: 3u + 3v = 84 V = 27 y =3 y =1- X 2x = l b) He < X =y = 4^^ +4--.4'^ =0,5 " [y = l - x ' 2 Cach khac: Dat u = 4 \ = 4* thi x + y = 1 « uv = 4. Bai toan 2: Giai cac he phuong trinh: 1 x + y = 20 log,(y-x)-log,- = l (1) a) b) y logj x + log^ y = 1 + log4 9 x' + y ' = 2 5 (2) Giai x + y = 20 x +y= 20 a) D K X > 0, y > 0, he tuong ducrng: log4 xy = log^ 36 xy = 36 Tir do giai dugc 2 nghiem (2; 18) va (18; 2). b) DK: y > X, y > 0. Ta c6: (1): log I (y - x ) - log,, - = - l o g , ( y - x) = - l o g , - = 1 y y y-x 3y
Giai 9x'-y'=5 9 x ' - y ' =5 a) He: i 3x + y > 0 . 3 x - y >0 log5(3x + y) = logjSCSx - y) 3x + y = 5(3x-y) 3x + y > 0, 3x - y > 0 3x-y = l x=l < (3x + y)(3x - y) = 5 « 3x + y = 5 ly = 2 3x + y = 5(3x-y) b) I)K: x + y ^ 0. He tuang duong: 3 X = - log, (x + y) + log, (x - y) = 1 f log, (x + v) = 1 2 (chon). log,(x + y) - log, 2.log,(x - y) = 1 [log, (x - y) = 0 1 ''2 B a i toan 4: Giai cac he phuang trinh: 'log,(x-y) = 5-log,(x + y) 21og,x-3^ =15 a) logx-log4 b) = -1 3Mog2X = 21og,x+3^'' logy-log3 Giai a) DK: x > y > 0. He tuang duang: 12 X = - 12 x^-y==32 X = s xv = 12 144 , „ - v - y =22 y'+22y'-144=0 rCr do giai dugc nghiem x = 6, y = 2. b) DK X > 0. dat u = logix va v = 3^ (y > 0). He tuong duang: 5 2 u - v = 15 v = 2u-15 u =9 i hoac u =-2 (loai) uv = 2u + 3v 2u'-23u + 45 = 0 v =3 v = -10 Tu do giai ra nghiem (512; 1). Giai cac he phuang trinh: B a i toan 5: log, (x' + y ' ) = 5 jlog, (6x + 4y) = 2 a) 21og,x + log,y = 4 [log^(6y + 4x) = 2 Giai a) DK x. y > 0. 234
log,(x-+y') = 5 fx-+y'=32 He tuong duong: < logj x + logj y = 4 [xy = 16 (x + y ) ' - 2 x y = 32 f(x + y ) ' = 6 4 fx = 4 < " xy = 16 [xy = 16 y=4 b) DK: x, y > 0, x, y 7t 1. |6x + 4y = x" f6x + 4y = x ' He tuong duong: 0 . T a c 6 ( l ) < » 2 ' " ^ - ' = 2'- « x - 3y = ^ va (2) o -logax + l = log3(9y) « • log3(xy) = -1 - » xy = | . 1 Tird6c6S = { ( 2 ; - ) } . 6 Bai toan 7: Giai cac he phuong trinh: (1) log, ( x- + y - ) = l + l o g , ( x y ) a) b) >X - XV + V y-=x^ (2) -81 Giai a) DK: x, y > 0. Ta c6 (2)
Xet y ^ 1 thi ^ (x +y)^ = 1 2 « x + y = 6 Do do y^' = x'' x = y^ nen y'^ + y - 6 = 0 Chony = 2 = ^ x = 4. Vay S = {(1; 1), (4; 2)} X" +y" = 2xy b) DK: xy > 0. he tirong dirang: X" - x y + y" = 4 X= y x=y X =y=2 < (chon). =4 v = ±2 X = y = -2 Bai toan 8: Giai cac he phuomg trinh: 2'+2x = 3 + y a) T + 2y = 3 + x 3^ - 3^ = (Iny - lnx)(xy + 2x +1) (1) b) X + y' = 6 (2) Giai a) Trir 2 phuomg trinh v l theo v l thi dugc: 2" + 3x = 2- + 3y Xet f(t) = 2' + 36, t e R thi f(t) = 2\2 + 3 > 0, Vx nen f d6ng bi6n tren R. Ta CO PT: f(x) = f(y) o X = y. Do do 2" + 2x = 3 + X 2" + X - 3 = 0 Xet ham g(x) = 2" + x - 3, x e R, tir do suy ra he c6 nghiem (1; 1). b) DK: X , y > 0 nen xy + 2x + 1 > 0. Vi ca so 3 > 1, e > 1 nen vai (1): NSu x > y thi VT > 0 > VP, mu x < y thi v r < 0 < VP, Neu X = y thi thoa man. Do do (2) x^ + X - 6 = 0, chon x = 2 => y = 2. Vay he c6 nghiem (2;2). Bai toan 9: Giai cac he phuang trinh: 'log,x + log,>; = log,(x + 2) (x + Vl+7)(y + V l + 7 ) = l 0) a) b) [4x + y + l = 2---^"'' (2) 3-^'"' + 6 = 5.3--^ Giai a) PT (1) bien doi thanh: X + Vl + X " = ^j\ y' - y va y + y]\ y^ = ^l\ x' - x 236