Bồi dưỡng học sinh giỏi toán đại số giải tích 12 tập 2

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a

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1234 Bai tap ttj luan dien hinh Dai so giai tich

1234 Bai tap tu luan

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Cuon sdch nay nam trong bo sdch 6 cuon gom:

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do nha gido uu tu, Thac si Le Hoanh Phd to'chiec bien soqn Noi dung sdch duoc bien soqn theo chuong trinh phdn ban: co bdn vd ndng cao mdi cua bo GD & DT, trong dd mot so"van deduac md rong vdi cdc dang bdi tap hay vd khd dephuc vu cho cdc em yeu thich mud'n ndng cao todn hoc, cd dieu kien phdt trien tot nhat khd ndng ciia minh Cud'n sdch la sir ke thira nhirng hieu biet chuyen mdn vd kinh nghiem gidng day ciia chinh tdc gid trong qua trinh true tiep dirng ldp bdi dudng cho hoc sinh gidi cdc ldp chuyen todn

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Cluing tdi hy vong bd sdch nay se la mdt tai lieu thie't thuc, bd' ich cho ngudi day va hoc, dqc biet cdc em hoc sinh yeu thich mdn todn vd hoc sinh chuan bi cho cdc ky thi quoc gia (tot nghiep THPT, tuye'n sinh DH - CD) do bd GD & DT to chirc sap tdi

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3

M U C L U C

Chuong I I : Ham so luy thira ham so mu va ham so logarit

§2 Phucmg trinh, he phuomg trinh, bat phuong trinh mu va logarit 5

Dang 1: Phuong trinh mu va logarit 5

Dang 2: Bat phuong trinh mu, logarit 21

Dang 3: He phuong trinh mu, logarit 31

Chuong H I : Nguyen ham, tich phan va ung dung

§ 1 Nguyen ham 46 Dang 1: Dinh nghia va tinh chat 47

Dang 2: Phuong phap bien doi bien so 55

Dang 3: Nguyen ham tirng phan 62

§2 Tich phan 70 Dang 1: Dinh nghia va tinh chat 71

Dang 2: Tich phan da thuc, phan thuc 81

Dang 3: Tich phan luong giac 89

Dang 4: Tich phan can thiic 100

Dang 5: Tich phan mu - logarit 108

§3 Ung dung cua tich phan 124

Dang 1: Tinh dien tich hinh phang 124

Dang 2: Tinh the tich vat thS 130

Chirong I V So phuc

§1 S6 phuc 139 Dang 1: Phep toan ve so phuc 140

Dang 2: Bieu dien va tap hop diem 143

§2 Can bac hai va phucmg trinh 151

Dang 1: Can bac hai ciia so phuc 151

Dang 2: Phuong trinh nghiem phuc 156

§3 Dang luong giac 165

Dang 1: Viet dang luong giac 165

Dang 2: Toan ung dung 171

Phuong phap chung:

- Dua ve cung mot co so

- Dat an phu

- Logarit hoa, mu hoa

- Su dung tinh chat cua ham so

B P H A N D A N G T O A N

DANG 1: PHUONG TRlNH M U VA LOGARIT

, a ^ = ag ( x ) ( a > 0 )

logaf(x) = logag(x), (a > 0, a * 1) o

- Phuong trinh mu co ban: ax = b (a > 0, a * 1)

Neu b < 0, phuong trinh vo nghiem

Neu b > 0, phuong trinh co nghiem duy nhat x = logab

a = l

a * 1 , f(x) = g(x)

- Phuong trinh logarit co ban: logax = b ( a > 0 , a ^ l )

Phuong trinh logarit co ban luon co nghiem duy nhat x = ab

f ( x ) > 0 h a y g ( x ) > 0 f(x) = g(x)

Chii y: Ngoai 4 phuong phap chinh de giai phuong trinh mu, lograrit, ta co the dung dinh nghTa, bien doi thanh phuong trinh tich so, dung do thi, bit dang thuc,

V i du 1: Giai cac phuong trinh sau:

a) PT o 2*2-3 x + 2 = 22o x2- 3 x + 2 = 2 » x2 - 3 x = 0 o x = 0 hoac x = 3 b) P T < ^ ( 2 + V 3 )2 x = ( 2 + S ) ' 1 <=>2x = - l o x = - i

2 c) PT o 2 10x = 200 o 10x = 100 o x * 2

5x 5x d) PT <=> 2"3 24 x"6 = 2T o 24 x"9 = 2Y

5x

<=> 4x - 9 = — <=> 8x - 18 = 5x o x = 6

2

-BDHSG DSGT12/2- c

V i du 2: Giai cac phuong trinh sau:

a) (1,5) 5x-7 / o \

x + 1

b) 7X-' = 2X

1 3 x+— x+—

c) 9X - 2 2= 2 ->2x-l ^ ylogx _ ^logx+1 _ ^ jlogx-1 _ J3 j^ogx-\

x+-3 x+-3

= — O X - 1 = lOgg — O X = 1 - lOgg 2

Z 2 1 2 d) PT » 7l o g x

DSGT12I2-d) Chia 2 ve cho 8X > 0 thi PT: 27V (12\*

- 7t + 2 = 0 o t = 1 hoac t = | (thoa man)

Suy nghiem x = 0 hoac x = 1

b) Dieu kien x * 0, dat y = — va chia hai ve cho 4y, ta co:

V i du 5: Giai cac phuong trinh:

a) x3 + ( x - 2 )6 = 0 b) V2XV4X(0,125)X = 4^2

c) Ul6-x + $fx~+~l=3 d) 3/ x T l -3/ x ^ T = v/x2 - 1

Giai a) B K : x > 0 , x - 2 > 0 < = > x > 2

Voi x > 2 thi V T > 0 nen PT vo nghiem

b) DK: x # 0, PT o 22

_ 1 7

» 22.23"^ = 23 » + — =

-2 3 -2x 3 , 1

» 5x - 14x - 3 = 0 <=> x = — hoac x = 3

5 c) D K : - 1 < x < 16 D a t u = # L 6 - x , v = 7 x + l thi u, v > 0

DSGT12/2-Giai a) Hai ve deu duong, logarit hoa theo co so 10 ta co :

4x

log3 = 3x

log4 <=> log 4

log 3 « x = log4(log34) 3 3x x-2 _ o2 n,2 ^ TX-2

b) PT o 3X 2X + 1 = 32 2 l o 3X~2 2*+ 1 = 1

,x-2 3.2s 1 <=> x - 2 = 0 hoac 3.2X + 1 = 1

» x = 2 hoac 2X + 1 = - <=> x = 2 hoac x = - 1 - log32

3 c) Hai vd deu duong, logarit hoa hai ve theo co so 2 ta co:

log2(3x-1.2x 2) = log2(8.4 x-2)

<=> (x - l)log23 + x2 = log28 + (x - 2)log24

<=>x2-(2-log23)x + 1 -log23 = 0 o x = 1 hoacx= 1 -log23

d) Hai ve deu duong, logarit hoa hai ve theo co so 5 ta co:

<=> 2 0l o g 3 X = 202 <=> log3x = 2 <s> x = 9 (thoa man)

b) DK: x > 0, PT « 4.22 l n x - 6l n x - 18.32 1 n x = 0

Chia ca hai ve cho 32 1 n x, dat t

-9 _2 Chon nghiem t = — <=> x = e

lnx thi duoc PT <=> 4t - t - 18 = 0

c) DK: x > 0, ta co: xl o g 4 = 4I o g^ ' °e 4 = 4l o g 4 l o^x = 4l o g x

nen PT « 2.4l o g x = 32 o 4l o g x = 16 o logx = 2 o x = 100

-BDHSG

DSGT1Z/2-d) DK: x > 0, dat t = log3x thi x = 4

Vi du 8: Giai cac phuong trinh:

a) (4 - Vl5)tanx + (4 + Vl5)tanx =8 b) 81sin2 x + 81c 30

c) (cos72°)x + (cos36°)x = 3.2~x sin(x—)

d) e 4

= tan x Giai

a) Vi (4 - y/l5 )(4 + 7l5) = 1 nen dat (4 - JTE )tanx = t, t > 0 thi phuong

trinh ot+ ^=8«>t2-8t+ 1 = 0«t = 4±Vl5

Do do tanx = - 1 hoac tanx = 1 nen nghiem x = ±— + leir, k e Z

4 b) Datt= 81smZx, 1 <t<81 thiPT

c) Phuong trinh: (2cos72°)x + (2cos36°)x = 3

Vi:2cos72° 2cos36° 2sin36°.cos36°.cos72

0

Dat t = (2cos72°)x, t > 0 thi PT o t + ± = 3

^ t 2 _ 3 t + 1 = 0 « t = N/5 + 1

Ta co: 2cos72° = 2sinl8° = — — - suy ra nghiem x = ±2

d) Dieu kien cosx * 0, v i sinx = 0 khong thoa man phuong trinh nen PT ,_ %/2sinx 72 cosx

72(stnx-cosx) smx e * e A

<=> = cosx sinx cosx

DSGT12/2-Dat u = sinx, v = cosx, u, v e ( - 1 ; 1), u.v > 0 nen ta co phuong trinh 72u Vgv

Vi u, v cung dau nen u, v cung thuoc mot khoang (-1; 0) hoac (0; 1) do do PT

<=> f(u) = ftVr'o u = v «tanx = 1 <=>x= — + kn (chon)

nghich bien tren R nen co nghiem duy nhat x = 2

c) Vi 0 < - < 1 nen khi x > -1 thi VT < 3, VP > 3 (loai), khi x < -1 thi VT > 3

Giai a) Ta co x = 3 la nghiem ciia phuong trinh, v i ham so

f(x) = ( \ o + Vl5 )x + (^7 - V l 5 )x la tong cua hai ham so mu voi co so

Ion hon 1 nen f(x) dong bien tren R Vay x = 3 la nghiem duy nhat ciia

phuong trinh

b) Ta co x = 1 la nghiem ciia phuong trinh Bien doi PT

2 - V 3 V (z + S = 1 thi ve trai la ham f(x) nghich bien, vay x = 1

la nghiem duy nhat ciia phuong trinh

V i du 11: Chiing minh rang phuong trinh:

a) 4x(4x2 + 1) = 1 co diing ba nghiem phan biet

b) xx + 1 = (x + l )x co mot nghiem duong duy nhat

Giai a) PT « • 4x(4x2 + 1) - 1 = 0 Xet ham s6 f(x) = 4x(4x2 + 1) - 1, D = R

Ta co f'(x) = 4xln4.(4x2 + 1) + 8x 4X = 4x[ln4.(4x2 + 1) + 8x]

f (x) = 0 o ln4.(4x2 + 1) + 8x = 0 O (41n4)x2 + 8x + ln4 = 0 (*)

PT (*) nay co biet thiic A > 0 nen co diing 2 nghiem phan biet Tir bang bien thien ciia f(x) suy ra phuong trinh f(x)= 0 co khong qua ba nghiem phan biet

=> dpcm

Cach khac: Xet ham f(t) = — , t > 0

Vi du 12: Giai cac phuong trinh:

a ) 2x + l- 4x = x - 1 b) 4b g 3 X + 2I o g 3 A =2x

c) 3 ^ - 2 ^ = 4c~^ d) cot2x = tan2x + 2tan2x + l

Giai a) P T« 2X + I + ( x + l ) = 22 x + 2x

Xet ham sd f(t) = 2l + 1.1 e R thi f '(t) = 2'.ln2 + 1

V i f ' (t) > 0, Vt nen f dong bien tren R

PT f(x + 1) = f(2x) o x + 1 = 2 x o x = 1

V i f(t) = | J J + | ± J ta co f '(t) > 0 va f(0) = f ( l ) = 0 nen chi co 2

nghiem t = 0 hoac t = 1 <=> x = 1 hoac x = 3

c) Dat t = Vcosx , 0 < t < 1 thi PT o 3l

- 2' = t <=> [ -1 + \ = 1

,9y 2t l n - + l - t l n 3 Xet f(t) = (| J + - L _ i < t < 1 thi f'(t) = S

Xet g(t) = 2'.ln- + 1 - tln3 voi 0 < t < 1 thi g'(t) = 2t.ln2.1n- - ln3 < 0

3 3 nen f '(t) nghich bien tren [0; 1] Lap BBT thi f(t) = 1 co toi da 2 nghiem

ma f(0) = f(l) = 1 nen PT tuong duong t = 0 hoac t = 1

<=> cosx = 0 hoac cosx = 1 o x = k27i hoac x = — + kn

2 d) DK: 2X * k- Dat t = tan2x thi PT o - = t + o t4 - 6t2 + 1 = 0

ot2 = 3 +2^2 =(V2 ± l)2ot = ±(^ + 1)

Vay tan2x = ±{42 ± 1), tu do suy ra nghiem x

Vi du 13: Giai cac phuong trinh sau:

a) log2[x(x - 1)] = 1 b) log2(9 - 2X) = 10log(3-x)

DSGT12/2-V j du 14: Giai cac phuong trinh:

a) log2x + log 2(x - 1) = 1

c) l o g3( 3x- l ) l o g3( 3x + ,- 3 ) = 12

b) log2x + log3x + log4X = 1 d) logx-,4 = 1 + l o g2( x - 1) Giai

a) DK: x > 1, PT o log2x(x - 1) = 1 o x(x - 1) = 2 <=> x - x - 2 = 0 Chon nghiem x = 2

b) DK: x > 0, PT o (1 + log32 + log42).log2x = 1

<=> (3 + log32)log2x = 1 o log2x =

3 +log, 2 Vay nghiem x = 23 + 2 1 o g*2

c) DK: x > 2, phuong trinh tro thanh:

^log2(x - 2) + ilog2(3x -5)=U log2(x - 2)(3x - 5) = 2

2 cs> (x - 2)(3x-5) = 4 o x = 3 hoac x = - Chon nghiem x = 3

3 d) DK: x > 0, x * i x * —, dat t = log3x thi PT o—= 2(2 + t)

ot2 + 3t-4 = 0ot=l hoac t = -4

1 Suy ra nghiem x = 3 hoac x =

Vf du 16: Giai cac phuong trinh:

a) l0g2(4x) + log2^- = 8

2 8

81 b) log2^ x+31og2 x + log1 x = 2 c) log4xlog2x + log 2log4x = 2 d) logx 216 + l o g2 x 64 = 3

V 2 2 J = ( - 2 - l o g2 x )

2= ( 2 + log2 x )2 Dat t = log2x thi PT <=> (2 + t )2 + 2t - 3 = 8

log22 log2 x + log2 log22 x = 2- o ilog2 log2 x + log2 f^log2 x j = 2

11 3

o - l o g2l o g 2x + l o g2- + log2log2x = 2 <=> - l o g 2l o g 2x = 3

o log2log2x = 2 <=> log2x = 4 <=> x = 16 (chon)

DSGT12/2-d) D K : x > 0, x* l , x * i thi

2

PT 02102,2 + = 3o + —- = 3

l + l o g2x l o g2x l + l o g 2x Dat t = log2x thi PT: - + —- = 3 o 3t2

- 5t - 2 = 0

t 1 + t

o t = 2 hoac t = -— Suy ra nghiem x = 4, x = JJ=

3 %/2

V i du 17: Giai cac phuong trinh

a) log5x log3x = log5x + log3x

b) 21og2x log5x + log2x - 101og5x = 5

c) log2x log3x log5x = log2x log3x + log2x log5x + log3x log5x

Giai a) DK: x > 0, ta co x = 1 la mot nghiem

Neu x * 1 thi PT o - = —-— + — - —

logx51ogx3 logx5 logx3

o logx5 + logx3 = 1 o logx15 = 1 o x = 15 (chon)

b) DK: x > 0, PT o log2x (21og5x + 1) - 5(21og5x + 1) = 0

o (log2x - 5)(21og5x + 1) = 0 o log2x = 5 hoac 21og5x = - 1

o x = 32 hoac x = —T= (chon)

V5 c) DK: x > 0, phuong trinh o (lgx)3 = (lgx)2(lg2 + lg3 + lg5)

o (lgx)2(lgx - lg30) = 0 o lgx = 0 hoac lgx = lg30

O x = 1 hoac x = 30 (chon)

V i du 18: Giai cac phuong trinh

a) log2x = 3 - x b) log3x + log 4(2x - 2) = 2

c) log2(l + V x ) = log3x d) l o g 3( l + Vx + Vx) = - l o g2 Vx

Giai

a) DK: 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 > 1 Ta co f(x) = log3x + log4(2x - 2) la ham dong bien nen

f(x) > f(3) = 2 voi x > 3 va f(x) < f(3) = 2 voi 1 < x < 3

Vay x = 3 la nghiem duy nhat

c) D K : x > 0, dat log3x = y thi x = 3y

P T o l o g2( l + V 3 y) = y « l + V3y = 2y o | ± | +

-BDHSG DSGT12/2- 17

Ta co y = 2 thoa man phuong trinh, v i ve trai la ham nghich bien nen PT

co nghiem duy nhat y = 2 nen x = 2

bien tren R nen y = 1 la nghiem duy nhat, do do PT cho co nghiem x = 2

V i du 19: Giai cac phuong trinh:

<=> x = y o x = log2(3x - 1) cs> 3x - 1 = 2X

o2x-3x+l=0 Xet g(x) = 2X 3x 1, x >

-3

Ta co g'(x) = 2x.ln2 - 3, g"(x) = 2x.ln22 > 0 nen g* dong bien tren D Do

do g(x) — 0 co toi da 2 nghiem, ma g( l ) = g(3) = 0 nen suy ra tan nghiem

S = { 1 ; 3 }

18 -BDHSG

DSGTW2-c) DK: x > 0 Xet x = 2 thi PT thoa man

VM ^^i+u-x x + 2 , x + 1 x + 3 ,

Xet x > 2 thi — > > 1 , > > 1

2 4 3 5 nen VT > VP (loai) Xet x < 2 thi VT < VP (loai)

Vay PT co nghiem duy nhat x = 2

<=> log3(x2 + x + 3) + (x2 +x + 3) = log3(2x2 +4x + 5) + (2x2 +4x + 5) Xet ham so f(t) = log31 +1, (t>0) thi f'(t)= —+ l>0,Vt>0

t In 3

Do do f(t) dong bien, nen phuong trinh f(x2 + x + 3) = f (2x2 + 4x + 5)

<=> x2 + x + 3 = 2x2 + 4x + 5 <=> x2 + 3x + 2 = 0

Vay phuong trinh co 2 nghiem x = -1 va x= -2

Vi du 20: Giai cac phuong trinh sau:

a) log2(cotx + tan3x) - 1 = log2(tan3x)

<=> cotx = tan3x <=> tan3x = tan(— - x)

o3x= x + k7i<=>x= — + — ,keZ

Dat log2(cosx) =2t=> cosx = 4l => cos2x = 16l

Do do 21og3(cotx) = 2t => cotx = 3l => cot2x = 9l

V i du 2 1 : Tim dieu kien de phuong trinh:

}>/3 b) Dat t = ^/logg x + 1 , x 1; 3% o l < t < 2

DSGT12/2-Cach khac: Xet ham so va lap bang bien thien

Bieu k i f n co nghiem duy nhat: a < 0 hay a = 12

DANG 2: BAT PHUONG TPJNH MO, LOGARIT

- Bat phuong trinh mu:

Neu a > 1 : af ( x ) < ag ( x ) » f(x) < g(x)

N i u 0 < a < 1: a m

< ag ( x ) <=> f(x) > g(x)

- Bat phuong trinh logarit:

'f(x) > 0 Neu a > 1: logaf(x) < logag(x) <=> \ g(x) >0 <=> 0 < f(x) < g(x)

f ( x ) < g ( x ) f(x)>0 Neu 0 < a < 1: logaf(x) < logag(x) <=> lg(x)>0 • » f(x) > g(x) > 0

f(x)>g(x) Phuong phap chung: Bua ve cung co so, dat an phu, mu hoa, lograrit hoa

va tinh chat don dieu ciia ham so

Chii y: ax < m o x < logam (vod m > 0 va a > 1)

1 a) D K : x ^ - l , d a t t = 3x, t > 0 thi B P T o— ! — < — ! —

t + 5 3t -1 f3t -1 < t + 5 l

DSGT12/2-Giai a) Chia 2 ve cho 22 x > 0, BPT

- 2 x - 2 < 0 o l - V3 < x < 1 + V 3

c) DK: x * 0, xet x < 0 thi V T < 0 < 4 diing

Xet x > 0 thi 4X > 3X nen BPT: 4X < 4(4X - 3X)

c) 4x2 + 3.3^ + x.37* < 2 x2 37 x + 2x + 6

d) 2 -xz-3x-2+4Vxz+3x < x2 - 2 x + 5

Giai a) DK: x > 0, BPT: 22 x - 3.2V x2x - 4.22 V x < 0

4

Chia hai ve cho 2 ^ 2X > 0 BPT: 2 X ~^ - 4.27 x"x -3 < 0

Dat t = 2x _ 7 x , t > 0 thi BPT o t 3 < 0 o t2- 3 t - 4 < 0 o - l < t < 4

t Chon 0 < t < 4 o x - V x < 2 o x - V x - 2 < 0

» 0 < > / x < 2 o 0 < x < 4 b) Dat t = 2 \ t > 0 thi BPT: V8 + 2 t - t2> 5 - 2t

Vay tap nghiem S = (-00; 3] u (0; 1) u (1; +00)

V j du 5: Giai cac bat phuong trinh:

a) l o gM (3 - 2x) > 1 b) log0,2(x2

- 4) > -1 '72

C) log2 l0gQ 5 2X - — |:

16 d) l o g3( 1 6x- 2 1 2x) < 2 x + 1

Giai a) Dieu kien: 3 - 2x > 0 o x = | V i V2 > 1 nen bat phuong trinh tuong

duong voi 3 - 2x > V2 ox< 3~^2" (thoa man)

b) DK: x < - 2 hoac x > 2, vi 0,2 < 1 nen BPT o x2 - 4 < (0,2)"'

<=> x2 - 4 < 5 <=> x2 < 9 » | x | < 3

Ket hop dieu kien thi nghiem -3 < x < - 2 hoac 2 < x < 3

c) V i co s6 2 > 1 nen BPT tuong duong voi

Giai a) DK x > 0, dat t = log0,2x ta co BPT:

c) (2x - 7)ln(x + 1) > 0 d) 2.x— logo x -2 > 22

log2 X

-BDHSG

<=> 1 - (log2x)2 < -4(1 + log2x) o (log2x)2 - 41og2x - 5 > 0

o log2x < -1 hoac log2x >5o0<x< - hoac x > 32

7

x > —

2 -l<x<0

Vay tap nghiem S = ( - 1 ; 0) u ( - ; oo)

2 d) DK: x > 0, logarit hoa theo co s6 2 > 1 ta co:

f 3

w * ' log2 2 2 g 2 X > l o g2 — logo X

DSGT12/2-Giai a) DK: x2- 6 x + 5 > 0 v a 2 - x> 0 o x< l hoac x > 5

-<=>log2(x- 2) + log2(3x- 5 ) > 2 o l o g2( ( x- 2)(3x- 5)) > log24

6t - r > o 6t - V < 5 5 < t < 6

log j ( 6 t - t2) > - 2

S

Tu do giai ra tap nghiem S = (-co; 0] u (log65; 1)

V i du 9: Giai cac bat phuong trinh:

DSGT12/2-Do do logx < - 1 hoac 2 < logx < 3 hoac logx > 5

Vay nghiem x < — hoac 100 < x < 1000 hoac x > 100 000

b) DK: x2 + 2x - 3 > 0, x + 3 > 0, x - 1 > 0 <=> x > 1

BPT <=> log2(x - l)(x + 3) - log2(x + 3) > log2 (x-l)

<=> log2 (x-l) - log2(x -1)<0«0< log2(x - 1) < 1

ol<x-l<2o2<x<3 (chon)

4x + Ft fi c) D K : x > 0 , x # l , — > 0 o 0 < x < - x ^ l

6 - 5 x 5 Neu 1< x < - thi BPT <=> < -

5 6 - 5 x x 4x + 5 1 4x2+5x-6 + 5x

<=> <0c=> < 0

6 - 5 x x ( 6 - 5 x ) x

O 4x2+1°X~6 <0o4x2+10x-6<0o-3<x< - (loai)

(6-5x)x 2 Neu0<x< 1 thi BPT o 4X2+10x-6> Q ^ 1 <x< L

(6-5x)x Vay tap nghiem S = (—; 1)

d) DK: x > 0, x * 1, 4X - 6 > 1 o x > log47

Vi x > log47 > 1 nen BPT o log2(4x - 6) < x

o4x-6<2so4x-2x-6<0o-2<2x<3ox< log23

Ket hop thi tap nghiem S = (log47; log23]

Vi du 10: Giai cac bat phuong trinh:

Neu x > - 1 thi BPT <=> log3(x + 3 ) < 0 o x < - 2 (loai)

Vay tap nghiem S = (-2; -1)

c) DK: x > 0 Xet x > 4 thi log2x > 2 con 6 - x < 2 (loai)

Xet 0 < x < 4 thi log2x < 2 < 6 - x nen BPT nghiem diing

Vay tap nghiem S = (0; 4]

d) Dieu kien x>0, x^ l , bat phuong trinh tuong duong

Do do (x - 1) f(x) > O.Tuong tu khi 0 < x < 1 thi f '(x) < 0 nen f(x) nghich bien: x > 1 suy ra f(x) < f ( l ) = 0 Do do (x - 1) f(x) > 0

Vay bat phuong trinh co nghiem voi moi x > 0, x ^ 1

V i d u l l : Tim tap xac dinh cua cac ham so sau:

a)y = log[l - l o g ( x2- 5 x + 16)]

b) y = ^/log0 5( - x2 + x + 6) +

x2 +2x

Giai a) Dieu kien xac dinh ciia ham so la log(x2 - 5x + 16) < 1

o 0 < x2- 5 x + 1 6 < 1 0 o x2- 5 x + 6 < 0 o 2< x < 3

V£y D = (2; 3)

Xet x < -1 thi VT < 3 < VP (loai)

Xet x > - 1 thi V T > 3 > VP (thoa man) Vay D = [ - 1; +00)

d) DK: 3X - x > 0 » 3X > x

Xet f(x) = 3X - x, x e R thi f *(x) = 3x.ln3 - 1

f ' ( x ) = 0 o 3xln3 = 1 0 x = - l o g3 (m3)

Lap BBT thi f(x) > 0, Vx nen D = R

V i du 12 Tim tham so m de bat phuong trinh co nghiem

a) 49x - 5.7X + m < 0 b) x2 - (m + 3)x + 3m < (m - 2)log2x

Giai a) Dat t = 7 \ (t > 0) thi BPT <=> t2 - 5t + m < 0, t > 0

Tu do suy ra dieu kien co nghiem la m * 2

YI du 13: Tim tham so de bat phuong trinh co nghiem voi moi x:

a) 3x

- ( 2 m + 5)(V3 )x

+ m2 b) 1 + log5(x 2 + 1) > log5(mx2 + 4x + m)

5m > 0

- 4x + Giai a) Datt = ( V 3 )x

t > 0 Bai toan tro thanh tim m de: t2 - (2m + 5) + m2 + 5m > 0, Vt > 0

2

<=> m < -5

-BDHSG

Ta tim m de he thoa man voi moi x

Xet m = 5 thi (1): -4x > 0, Vx (loai)

Xet m = 0 thi (2): 4x > 0, Vx (loai)

Xet m * 5, m ^ 0 thi dieu kien

[A,<0, 5-m>0 f4-(5 + m)2 <0,m<5

A2< 0 , m > 0 4 - m2< 0 , m > 0

DANG 3: H E P H U O N G T R I N H M U , LOGARIT

Viec giai he phuong trinh va logarit ve co ban cung giong nhu giai cac

he phuong trinh dai so voi cac bien doi ve bieu thiic mu va logarit

Phuong phap chung: Rut the, cong dai so, dat an phu

Vi du 1: Giai cac he phuong trinh:

Vf du 3: Giai cac he phucmg trinh:

a) x + y = 20

Llog4 x + log4 y = 1 + log4 9 b)

l o g1( y - x ) - l o g4- = l (1)

x2+ y2= 2 5 (2) Giai

x + y = 20 a) D K x > 0, y > 0, he tuong duong

[log4 xy = log 4 36 Tir do giai duoc 2 nghiem (2; 18) va (18; 2)

3 x - y = l j x = l 3x + y = 5 ° ]y = 2

(3x + y)(3x-y) = 5 <

3x + y = 5(3x-y)

b) DK:x + y>0

[log2(x + y) + log2(x-y) = l

[log2 (x + y) - log3 2 log2 (x - y) = 1

DSGT12/2-V j du 5: Giai cac he phucmg trinh:

l o g2( x - y ) = 5-log2(x + y)

a) {logx-log4 , b)

logy-log 3

[21og2x-3 y=15 [3y.log2x = 21og2x + 3y Giai

xy = 16

x = 4

y = 4 b) DK: x, y > 0, x, y * 1 He tuong duong

|6x + 4y = x2 [6x + 4y = x2

|6y + 4x = y2 ° l ( x - y X x + y- 2 ) = 0

fy = x f y = 2- x

x2 -10x = 0 hoac x 2 - 2 x - 8 = 0

Tir do giai ra nghiem (5; 5)

V i du 7: Giai cac he phuong trinh:

Giai a) D K : x + y > O T a c o ( 2 ) o x + y = 5 o y = 5- x

The (1) ta co 3"x 25 _ x = 1152 <=> 6"x = 36 o x = - 2

Do do y = 7 Thu lai diing Vay nghiem la (-2; 7)

- Q b) DK: x, y > 0 Ta co (1) o 2x - 3 y = 22 o x - 3 y = -

va (2) <=> -log3x + 1 = log3(9y) o log3(xy) = -1

<=> xy = - Tir do co S = {(2; — ) }

b) log2(x

2 + y2) = l + log2(xy) 3x 2-xy+y2

= g l

a) DK: x, y > 0 Ta co (2) « x = y 2

1 2

(1) o y3 X + = y1 2 Xet y = 1 thi x = 1 (diing)

Xety * 1 thi -(x +y)2 = l2<=>x + y = 6

3

Do do y6 = x3 <=> x = y2 nen y2 + y - 6 = 0

Chon y = 2 => x = 4 Vay S = {(1; 1), (4; 2)}

f x2 + y2 = 2xy b) DK: xy > 0, he tuong duong

Giai a) Trir 2 phuong trinh ve theo ve thi duoc: 2X + 3x = 2y + 3y

Xet f(t) = 21 + 36, t e R thi f(t) = 2l ln2 + 3 > 0, Vx nen f ddng bien tren

R

Ta co PT: f(x) = f(y) <=> x = y

Do do 2X + 2x = 3 + x <=> 2X + x - 3 = 0

Xet ham g(x) = 2X + x - 3, x e R tir do suy ra he co nghiem (1; 1)

b) Trir 2 phuong trinh ve theo ve thi duoc: (2X - 2y) + (3X - 3y) + 3(x - y) = 0 Xet x > y thi V T > 0 (loai), x < y thi V T < 0 (loai)

Xet x = y = t thi duoc: 2l + 3l - 3t - 2 = 0

34 -BDHSG

log2 V l + 3sinx = log3(3cosy)

log2 y l + 3 cos y = log3(3 sin x)

Do do log2(l + 3u) - 21og3(3u) = log2(l + 3v) - 21og3(3v)

Xet f(t) = log2(l + 3t) - 21og3(3t), 0 < t < 1

f'(t)= + —— > 0 nen f dong bien tren (0; 1], do do PT

Giai a) PT (1) bien doi thanh:

Do do (2) <=> 3x + 1 = 22- 3x » 8x(3x + 1) = 4

-BDHSG DSGT12 /2- 35

PT nay co nghiem duy nhat x = - nen S = { ( - ; - - ) }

3 3 3

,t+i b) Datt = 2 x - y t h i ( l) o( l + 4t) 5M= 1 + 2

Bang each xet f(y) = 1 - e~y - y thi phuong trinh f(y) = 0 co nghiem duy nhat y = 0, do do x = y = z = 0

Neu x * 0 thi y * 0, z * 0

y.ey Dat f(t) t.e

1

t * 0 thi he

e =

-ey -1 z.eL

ez - 1 x.e"

f(t) > 1, Vt > 0 nen he tuong duong x = y = z = t, do do e' - t - 1 = 0 (vo nghiem)

Vay he co nghiem duy nhat x = y = z = 0

V i du 13: Giai cac he bat phuong trinh

Gia su (x, y) la mot nghiem thi (-x, y) cung la nghiem, ma he co nghiem duy nhat nen x = 0

Xet y = 3 - x - m t a c 6 phuong trinh: x + (m - 3)x + m(m - 3) = 0

V i A = - 3 ( m - 3)(m + 1) > 0 o - 1 < m < 3 thoa (1) nen suy ra dieu kien

co nghiem la m > -3 va m * - 2

C B A I L U Y E N TAP

Bai 1: Giai phuong trinh:

.25 J b)(42 - 1)X + (V2 + 1 )X- 2 V 2 = 0

H D : a) Chia so hang roi dat an phu

j i b) Bien doi tich so

38 -BDHSG