phuong phap giai phuong trinh

8/7/2019 phuong phap giai phuong trinh

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CC PHNG PHP GIIPHNG TRNH- BT PHNG TRNH- H M- LGARIT

CHNG I:PHNG PHP GII PHNG TRNH- BT PHNG TRNH- H MCH I:PHNG TRNH M

BI TON 1: S DNG PHNG PHP BIN I TNG NGI. Phng php:

Ta s dng php bin i tng ng sau:

( ) ( )

( ) ( )

1

0 1f x g x

a

aa a

f x g x

= < = =

hoc ( ) ( ) ( )0

1 0

a

a f x g x

> =

II. VD minh ho:

VD1: Gii phng trnh: ( ) ( )sin 2 3 cos2 22 2 xx x x x + = + Gii: Phng trnh c bin i v dng:

( ) ( )

2

2

2

1 2(*)2 0

1 0(1)

2 1 sin 2 3 cos 0 sin 3 cos 2(2)

xx x

x x

x x x x x x

< =

+ + = + =

Gii (1) ta c1,2

1 5

2x

= tho mn iu kin (*)

Gii (2):1 3

sin cos 1 sin 1 2 2 ,2 2 3 3 2 6

x x x x x k x k k Z

+ = + = + = + = +

nghim tho mn iu kin (*) ta phi c:1 1

1 2 2 1 2 0,6 2 6 2 6

k k k k Z

< + < < < =

khi ta nhn c 3 6x

=

Vy phng trnh c 3 nghim phn bit 1,2 31 5 ;2 6x x = = .

VD2: Gii phng trnh: ( ) ( )22 43 5 2 23 6 9

x xx xx x x

+ + = +

Gii: Phng trnh c bin i v dng: ( ) ( ) ( )2

2 243 5 2 2 2( 4)3 3 3

x xx x x x

x x x+ + + = =

2 2 2

3 1 44

0 3 1 3 45

3 5 2 2 2 8 7 10 0

x xx

x xx

x x x x x x

= = = < < = + = + + =

Vy phng trnh c 2 nghim phn bit x=4, x=5.

BI TON 2: S DNG PHNG PHP LGARIT HO V A V CNG C SI. Phng php: chuyn n s khi s m lu tha ngi ta c th logarit theo cng 1 c s c 2 v ca phngtrnh, ta c cc dng:Dng 1: Phng trnh:

( )

( )0 1, 0

logf x

a

a ba b

f x b

< >= =Dng 2: Phng trnh :

1


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( ) ( ) ( ) ( )log log ( ) ( ).logf x g x f x f xa a aa b a b f x g x b= = =

hoc ( ) ( )log log ( ).log ( ).f x g xb b ba b f x a g x= =II. VD minh ho:VD1: Gii phng trnh:

22 2 3

2x x =

Gii: Ly logarit cs 2 hai v phng trnh ta c:

2 2 2 22 2 2 2

3log 2 log 2 log 3 1 2 1 log 3 0

2x x x x x x = = + =

Ta c , 2 21 1 log 3 log 3 0 = + = > suy ra phng trnh c nghim

x = 1 2log 3. VD2: Gii phng trnh:

1

5 .8 500.x

x x

=Gii: Vit li phng trnh di dng:

1 1 33 3 2 385 .8 500 5 .2 5 .2 5 .2 1

x x xx x xx x

= = =

Ly logarit c s 2 v, ta c:

( ) ( )3 3

3 32 2 2 2 2

3log 5 .2 0 log 5 log 2 0 3 .log 5 log 2 0

x x

x xx xx

xx

= + = + =

( ) 22

31

3 log 5 0 1

log 5

x

xxx

= + = =

Vy phng trnh c 2 nghim phn bit:2

13;

log 5x x= =

Ch : i vi 1 phng trnh cn thit rt gn trc khi logarit ho.BI TON 3: S DNG PHNG PHP T N PH- DNG 1I. Phng php:Phng php dng n ph dng 1 l vic s dng 1 n ph chuyn phng trnh ban u thnh 1phng trnh vi 1 n ph.Ta lu cc php t n ph thng gp sau:Dng 1: Phng trnh ( 1)1 1 0..... 0

k x x

k ka a + + =

Khi t xt a= iu kin t>0, ta c: 11 1 0...... 0k k

k kt t t + + =

M rng: Nu t ( ) ,f xt a= iu kin hp t>0. Khi : 2 ( ) 2 3 ( ) 3 ( ), ,.....,f x f x kf x k a t a t a t = = =

V( ) 1f x

a t

=Dng 2: Phng trnh 1 2 3 0

x xa a + + = vi a.b=1

Khi t ,xt a= iu kin t0, suy ra ( )1f x

bt

=

2


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Dng 3: Phng trnh ( )2 21 2 3 0xx xa ab b + + = khi chia 2 v ca phng trnh cho 2xb >0

( hoc ( )2 , . xxa a b ), ta c:2

1 2 3 0x x

a a

b b

+ + =

t ,x

at

b

=

iu kin t (hoc ( )2 , . ffa a b )

- tf

at

b

=

iu kin hp t>0

Dng 4: Lng gic ho.Ch : Ta s dng ngn t iu kin hp t>0 cho trng hp t ( )f xt a= v:

- Nu t xt a= th t>0 l iu kin ng.- Nu t

2 12xt += th t>0 ch l iu kin hp, bi thc cht iu kin cho t phi l 2t . iukin ny c bit quan trng cho lp cc bi ton c cha tham s.

II. VD minh ho:VD1: Gii phng trnh: 2 2

1

cot sin4 2 3 0g x x+ = (1)

Gii: iu kin sin 0 ,x x k k Z (*)

V 221

1 cotsin

g xx

= + nn phng trnh (1) c bit di dng:

22 cotcot4 2.2 3 0

g xg x + = (2)

t2cot2 g xt= iu kin 1t v

22 cot 0cot 0 2 2 1g xg x =Khi phng trnh (2) c dng:

22 cot 21

2 3 0 2 1 cot 03

cot 0 ,2

g xt

t t g xt

gx x k k Z

=+ = = = =

= = + tho mn (*)

Vy phng trnh c 1 h nghim ,2

x k k Z

= +

VD2: Gii phng trnh: ( ) ( )7 4 3 3 2 3 2 0x x

+ + =

Gii: Nhn xt rng: ( ) ( ) ( )27 4 3 2 3 ; 2 3 2 3 1+ = + + =Do nu t ( )2 3

x

t= + iu kin t>0, th: ( ) 12 3x

t = v ( ) 27 4 3

x

t+ =

Khi phng trnh tng ng vi:

( ) ( )2 3 2 213

2 0 2 3 0 1 3 03 0 ( )

tt t t t t t

t t t vn

= + = + = + + = + + =

( )2 3 1 0x x + = =Vy phng trnh c nghim x=0Nhn xt: Nh vy trong v d trn bng vic nh gi:

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( )

( ) ( )

2

7 4 3 2 3

2 3 2 3 1

+ = +

+ =

Ta la chn c n ph ( )2 3 xt= + cho phng trnhV d tip theo ta s miu t vic la chn n ph thng qua nh gi m rng ca a.b=1, l:

. . 1a b

a b c c c= = tc l vi cc phng trnh c dng: . . 0x xA a B b C + + =

Khi ta thc hin php chia c 2 v ca phng trnh cho 0xc , nhn c:

. 0

x xa b

A B Cc c

+ + =

t thit lp n ph , 0x

at t

c

= >

v suy ra1

xb

c t

=

VD3: Gii phng trnh:2 22 1 2 22 9.2 2 0x x x x+ + + + =

Gii: Chia c 2 v phng trnh cho 2 22 0x+ ta c:

2 2 2 22 2 1 2 2 2 21 92 9.2 1 0 .2 .2 1 0

2 4

x x x x x x x x + = + =

2 22 22.2 9.2 4 0x x x x + =

t2

2x xt = iu kin t>0. Khi phng trnh tng ng vi:2

2

2 22

21

42 2 2 1

2 9 4 0 1212 2

2

x x

x x

tx x x

t txt x x

= = = = + = == = = Vy phng trnh c 2 nghim x=-1, x=2.Ch : Trong v d trn, v bi ton khng c tham s nn ta s dng iu kin cho n ph ch l t>0

v chng ta thy vi1

2t= v nghim. Do vy nu bi ton c cha tham s chng ta cn xc nh

iu kin ng cho n ph nh sau:

2

2 1

2 4

4

1 1 1 12 22 4 4 2

x xx x x t =

VD4: Gii phng trnh: ( )3

3 1

1 122 6.2 1

22

x x

xx + =

Gii: Vit li phng trnh c dng:

3

33

2 22 6 2 1

2 2x x

x x

=

(1)

t33

3 33

2 2 2 22 2 2 3.2 2 6

2 2 2 2x x x x x

x x x xt t t

= = + = +

Khi phng trnh (1) c dng: 3 26 6 1 1 2 12

x

xt t t t + = = =

t 2 , 0xu u= > khi phng trnh (2) c dng:

21(1)

1 2 0 2 2 2 122

xuu

u u u u xu

= = = = = = =

Vy phng trnh c nghim x=1Ch : Tip theo chng ta s quan tm n vic s dng phng php lng gic ho.

VD5: Gii phng trnh: ( )2 21 1 2 1 2 1 2 .2x x x+ = +

4


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Gii: iu kin 2 21 2 0 2 1 0x x x

Nh vy 0 2 1x< , t 2 sin , 0;2

x t t =

Khi phng trnh c dng:

( ) ( )2 21 1 sin sin 1 2 1 sin 1 cos 1 2 cos sin3 32 cos sin sin 2 2 cos 2sin cos 2 cos 1 2 sin 0

2 2 2 2 2 2

cos 0(1) 12 12 6

203 2

2 1sin22 2

x

x

t t t t t t

t t t t t t t t

tt

x

xtt

+ = + + = +

= + = = = = = = = = ==

Vy phng trnh c 2 nghim x=-1, x=0.BI TON 4: S DNG PHNG PHP T N PH- DNG 2I. Phng php:Phng php dng n ph dng 2 l vic s dng 1 n ph chuyn phng trnh ban u thnh 1

phng trnh vi 1 n ph nhng cc h s vn cn cha x.Phng php ny thng s dng i vi nhng phng trnh khi la chn n ph cho 1 biu thc thcc biu thc cn li khng biu din c trit qua n ph hoc nu biu din c th cngthc biu din li qu phc tp.Khi thng ta c 1 phng trnh bc 2 theo n ph ( hoc vn theo n x) c bit s l mt schnh phng.II. VD minh ho:

VD1: Gii phng trnh: ( )23 2 9 .3 9.2 0x x x x + + =Gii: t 3xt= , iu kin t>0. Khi phng trnh tng ng vi:

( ) ( ) ( )

2 229

2 9 9.2 0; 2 9 4.9.2 2 92

x x x x x

x

tt t

t

= + + = = + = +

=

Khi :+ Vi 9 3 9 2xt t= = =

+ Vi3

2 3 2 1 02

x

x x xt x = = = =

Vy phng trnh c 2 nghim x=2, x=0.

VD2: Gii phng trnh: ( )2 22 29 3 3 2 2 0x xx x+ + =Gii: t

2

3xt= iu kin 1t v22 00 3 3 1xx =


( )2 2 23 2 2 0t x t x+ + =

( ) ( ) ( )2 22 2 2 22

3 4 2 2 11

tx x x

t x

= = + = + =

Khi :

+ Vi2 2

3 32 3 2 log 2 log 2xt x x= = = =

+ Vi22 21 3 1xt x x= = ta c nhn xt:

2

2

1 1 3 10

1 1 1 1

xVT VT x

VP VP x

= = = = =

5


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Vy phng trnh c 3 nghim 3log 2; 0x x= =BI TON 5: S DNG PHNG PHP T N PH- DNG 3I. Phng php:Phng php dng n ph dng 3 s dng 2 n ph cho 2 biu thc m trong phng trnh v kholo bin i phng trnh thnh phng trnh tch.II. VD minh ho:VD1: Gii phng trnh: 2 2 23 2 6 5 2 3 74 4 4 1x x x x x x + + + + +

+ = +Gii: Vit li phng trnh di dng: 2 2 2 23 2 2 6 5 3 2 2 6 54 4 4 .4 1x x x x x x x x + + + + + ++ = +

t

2

2

3 2

2 6 5

4, , 0

4

x x

x x

uu v

v

+

+ +

= >=


( ) ( )1 1 1 0u v uv u v+ = + =

2

2

3 2 2

22 6 5

1

1 4 1 3 2 0 2

1 12 6 54 1

5

x x

x x

x

u x x x

v xx x

x

+

+ +

= = = + = = = = + + = =

Vy phng trnh c 4 nghim.

VD2: Cho phng trnh:2 25 6 1 6 5.2 2 2.2 (1)x x x xm m + + = +

a) Gii phng trnh vi m=1b) Tm m phng trnh c 4 nghim phn bit.

Gii: Vit li phng trnh di dng:

( )2 22 2 2 2

2 2 2 2

( 5 6) 15 6 1 7 5 5 6 1

5 6 1 5 6 1

.2 2 2 .2 2 2

.2 2 2 .2

x x xx x x x x x x

x x x x x x

m m m m

m m

+ + + +

+ +

+ = + + = +

+ = +

t:

2

2

5 6

1

2, , 02

x x

x

uu v

v

+

= > =

. Khi phng trnh tng ng vi:

( ) ( )2

2

2

5 6

11

31 2 1

1 0 22

2 (*)

x x

x

x

xu

mu v uv m u v m xv m m

m

+

== = + = + = = = = =

Vy vi mi m phng trnh lun c 2 nghim x=3, x=2a) Vi m=1, phng trnh (*) c dng:

21 2 22 1 1 0 1 1x x x x = = = = Vy vi m=1, phng trnh c 4 nghim phn bit: x=3, x=2, x= 1b) (1) c 4 nghim phn bit (*) c 2 nghim phn bit khc 2 v 3.

(*) 2 22 2

0 0

1 log 1 log

m m

x m x m> >

= = . Khi iu kin l:

( )22

2

00 2

1 log 0 1 11 0;2 \ ;1 log 4 8 2568

11 log 9

256

m

m m

mmm

m

mm

>>

6


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Vy vi ( )1 1

0;2 \ ;8 256

m

tho mn iu kin u bi.

BI TON 6: S DNG PHNG PHP T N PH- DNG 4I. Phng php:Phng php dng n ph dng 4 l vic s dng k n ph chuyn phng trnh ban u thnh 1 hphng trnh vi k n ph.Trong h mi th k-1 th phng trnh nhn c t cc mi lin h gia cc i lng tng ng.

Trng hp c bit l vic s dng 1 n ph chuyn phng trnh ban u thnh 1 h phng trnhvi 1 n ph v 1 n x, khi ta thc hin theo cc bc:Bc 1: t iu kin c ngha cho cc biu tng trong phng trnh.

Bc 2: Bin i phng trnh v dng: ( ), 0f x x =

Bc 3: t ( )y x= ta bin i phng trnh thnh h:( )

( ); 0y x

f x y

=

=II. VD minh ho:

VD1: Gii phng trnh:1 1 1

8 2 18

2 1 2 2 2 2 2

x

x x x x + =+ + + +

Gii: Vit li phng trnh di dng: 1 1 1 18 1 18

2 1 2 1 2 2 2x x x x + =

+ + + +

t:1

1

2 1, , 1

2 1

x

x

uu v

v

= + >= +

Nhn xt rng: ( ) ( )1 1 1 1. 2 1 . 2 1 2 2 2x x x xu v u v = + + = + + = +Phng trnh tng ng vi h:

8 1 18 28 18

99;

8

u vu v

u v u vu v uv u v

u v uv

= = + =+ = + + = = = + = + Vi u=v=2, ta c:

1

1

2 1 21

2 1 2

x

xx

+ = =+ =

+ Vi u=9 v9

8v = , ta c:

1

1

2 1 949

2 18

x

xx

+ = =

+ =Vy phng trnh cho c cc nghim x=1 v x=4.

VD2: Gii phng trnh: 22 2 6 6x x + =Gii: t 2xu = , iu kin u>0. Khi phng trnh thnh: 2 6 6u u + =

t 6,v u= + iu kin 26 6v v u = +Khi phng trnh c chuyn thnh h:

( ) ( ) ( )2

2 2

2

6 00

1 06

u v u vu v u v u v u v

u vv u

= + = = + = + + == +

+ Vi u=v ta c:2 36 0 2 3 8

2(1)x

uu u x

u

= = = = =

+ Vi u+v+1=0 ta c:

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2

2

1 21

21 1 21 125 0 2 log

2 21 21(1)

2

x

u

u u x

u

+= + = = =

=

Vy phng trnh c 2 nghim l x=8 v x=2

21 1log .

2

BI 7: S DNG TNH CHT N IU CA HM SI. Phng php:S dng cc tnh cht ca hm s gii phng trnh l dng ton kh quen thuc. Ta c 3 hng pdng:Hng1: Thc hin cc bc sau:

Bc 1: Chuyn phng trnh v dng: f(x)=kBc 2: Xt hm s y=f(x). Dng lp lun khng nh hm s n iu( gi s ng bin)Bc 3: Nhn xt:

+ Vi ( ) ( )0 0x x f x f x k = = = do 0x x= l nghim+ Vi ( ) ( )0x x f x f x k > > = do phng trnh v nghim

+ Vi ( ) ( )0 0x x f x f x k < < = do phng trnh v nghim.Vy 0x x= l nghim duy nht ca phng trnh.Hng 2: Thc hin theo cc bc:

Bc 1: Chuyn phng trnh v dng: f(x)=g(x)Bc 2: Xt hm s y=f(x) v y=g(x). Dng lp lun khng nh hm s y=f(x) l

L ng bin cn hm s y=g(x) l hm hng hoc nghch bin

Xc nh 0x sao cho ( ) ( )0 0f x g x=Bc 3: Vy phng trnh c nghim duy nht 0x x=

Hng 3: Thc hin theo cc bc:Bc 1: Chuyn phng trnh v dng: f(u)=f(v) (3)Bc 2: Xt hm s y=f(x). Dng lp lun khng nh hm s n iu ( gi s

ng bin)Bc 3: Khi : (3) u v = vi , fu v D

II. VD minh ho:VD1: Gii phng trnh: 2log2.3 3xx + = (1)Gii: iu kin x>0. Bin i phng trnh v dng: 2log2.3 3x x= (2)Nhn xt rng:+ V phi ca phng trnh l mt hm nghch bin.+ V tri ca phng trnh l mt hm ng bin.Do vy nu phng trnh c nghim th nghim l duy nht.

Nhn xt rng x=1 l nghim ca phng t rnh (2) v2log2.3 3 1x =

Vy x=1 l nghim duy nht ca phng trnh.

VD2: Gii phng trnh: ( )23 1

23

1log 3 2 2 2

5

x x

x x

+ + + =

(1)

Gii: iu kin:2 13 2 0

2

xx x

x

+

t 2 3 2u x x= + , iu kin 0u suy ra: 2 2 2 23 2 3 1 1x x u x x u + = =

8


9/54

Khi (1) c dng: ( )21

3

1log 2 2

5

u

u

+ + =

Xt hm s: ( ) ( )21

2

3 3

1 1( ) log 2 log 2 .5

5 5

x

f x x x x

= + + = + +

+ Min xc nh [ 0; )D = +

+ o hm: ( )2

1 1 .2 .5 . ln3 0,2 ln 3 5xf x x D

x= + > + . Suy ra hm s tng trn D

Mt khc ( ) ( )31

1 log 1 2 .5 2.7

f = + + =

Do , phng trnh (2) c vit di dng:

( ) ( ) 2 3 51 1 3 2 12

f u f u x x x

= = + = =

Vy phng trnh c hai nghim3 5

2x

=

VD2: Cho phng trnh:22 2 4 22 2 2

5 5 2

x mxx mx

x mx m

+ ++ +

= + +a) Gii phng trnh vi

4

5m =

b) Gii v bin lun phng trnhGii: t 2 2 2t x mx= + + phng trnh c dng: 2 25 5 2 2t t mt t m+ + = + + (1)Xc nh hm s ( ) 5tf t t= ++ Min xc nh D=R+ o hm: 5 . ln 5 1 0,tf x D= + > hm s tng trn D

Vy (1) ( ) ( ) 22 2 2 2 2 0 2 0f t f t m t t m t m x mx m = + = + + = + + = (2)

a) Vi 45

m = ta c: 2 22

8 4 0 5 8 4 0 25 5

5

x

x x x xx

=+ = = =

Vy vi4

5m = phng trnh c 2nghim

22;

5x x= =

b) Xt phng trnh (2) ta c: 2' m m = + Nu 2' 0 0 0 1m m m < < < < . Phng trnh (2) v nghim phng trnh (1) v nghim.+ Nu ' 0 = m=0 hoc m=1.

vi m=0 phng trnh c nghim kp x=0vi m=1 phng trnh c nghim kp x0=-1

+ Nu 1' 0 0mm

> >


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I. Phng php:Vi phng trnh c cha tham s: f(x,m)=g(m). Chng ta thc hin cc bc sau:Bc 1: Lp lun s nghim ca (1) l s giao im ca th hm s (C): y=f(x,m) v ng thng(d): y=g(m).Bc 2: Xt hm s y=f(x,m)+ Tm min xc nh D+ Tnh o hm y ri gii phng trnh y=0

+ Lp bng bin thin ca hm sBc 3: Kt lun:

+ Phng trnh c nghim ( ) ( )min , ( ) max , ( )f x m g m f x m x D + Phng trnh c k nghim phn bit (d) ct (C) ti k im phn bit+ Phng trnh v nghim ( ) ( )d C = III. VD minh ho:

VD1: Cho phng trnh: ( )22 2 2 22 2 23 2 2 2

x xx x x x m + + + + =

a) Gii phng trnh vi m=8b) Gii phng trnh vi m=27c) Tm m phng trnh c nghim

Gii: Vit li phng trnh di dng: 2 22 2 2 2 23 4 2 2x x x x x x m + ++ + + =S nghim ca phng trnh l s giao im ca th hm s:

2 22 2 2 2 23 4 2 2x x x xy x x + += + + + vi ng thng y=m

Xt hm s2 22 2 2 2 23 4 2 2x x x xy x x + += + + + xc nh trn D=R

Gii hn: lim y = +Bng bin thin: v 3>1, 4>1 nn s bin thin ca hm s ph thuc vo s bin thin cca hm s

2 2 2t x x= + ta c:a) Vi m=8 phng trnh c nghim duy nht x=1b) Vi m=27 phng trnh c 2 nghim phn bit x=0 v x=2c) Phng trnh c nghim khi m>8

VD2: Vi gi tr no ca m th phng trnh:2 4 3

4 21 15

x x

m m

+ = +

c 4 nghim phn bit

Gii: V 4 2 1 0m m + > vi mi m do phng trnh tng ng vi:

( )2 4 215

4 3 log 1x x m m + = +

t ( )4 215

log 1m m a + = , khi : 2 4 3x x a + =

Phng trnh ban u c 4 nghim phn bit phng trnh (1) c 4 nghim phn bit

ng thng y=a ct th hm s2

4 3y x x= + ti 4 im phn bit

Xt hm s:2

2

2

4 3 1 34 3

4 3 1 3

x x khix hoacxy x x

x x khi x

+ = + = +

o hm:2 4 1 3

'2 4 1 3

x khix hoacxy

x khi x

< >= + <


11/54

T , ng thng y=a ct th hm s2 4 3y x x= + ti 4 im phn bit

( )4 2 4 215

10 1 0 log 1 1 1 1 0 1

5a m m m m m < < < + < < + < < phng trnh c vit di dng:

2

2

33 1

1

tt m t m

t

++ = + =

+(1)

S nghim ca (1) l s giao im ca th hm s (C): 23

1

ty

t

+=

+vi ng thng (d):y=m

Xt hm s: 23

1

ty

t

+=

+xc nh trn ( )0;D +

+ o hm:

( )2 2

1 3 1' ; ' 0 1 3 0

31 1

ty y t t

t t

= = =

+ ++ Gii hn: ( )lim 1y t= ++ Bng bin thin:

Bin lun:Vi 1m hoc 10m > phng trnh v nghimVi 1 3m< hoc 10m = phng trnh c nghim duy nhtVi3 10m< < phng trnh c 2 nghim phn bit

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CH II:BT PHNG TRNH MBI TON I: S DNG PHNG PHP BIN I TNG NGI. Phng php:Ta s dng cc php bin i tng ng sau:

Dng 1: Vi bt phng trnh:( ) ( ) ( ) ( )

( ) ( )

1

0 1

f x g x

a

f x g xa a

af x g x

>

= <

Ch : Cn c bit lu ti gi tr ca c s a i vi bt phng trnh m.

II. VD minh ho:VD1: Gii cc bt phng trnh:

a) 21

2

12

2

x

x x

b) ( ) ( )3 11 310 3 10 3

x x

x x

+ ++ < +

Gii:a) Bin i tng ng bt phng trnh v dng:

( )

2 22 12

22

1 0

2 01 12 1 21 02 2

2 1

x x x

x

x x

x x x xx

x x x

>

Vy nghim ca bt phng trnh l 2x Ch : trnh sai st khng ng c khi bin i bt phng trnh m vi c s nh hn 1 cc emhc sinh nn la chn cch bin i:

2

2

1 2 1 2 2

2

12 2 2 2 1 2 1 2

2

x x x x

x xx x x x x x x

b) Nhn xt rng: ( ) ( ) ( ) 110 3 10 3 1 10 3 10 3

+ = = +

Khi bt phng trnh c vit di dng:

( ) ( ) ( )

( ) ( )

3 1 3 11 3 1 3

2

10 3 10 3 10 3 1

3 53 1 50 0

1 3 1 3 1 5

x x x x

x x x x

xx x x

x x x x x

+ ++ + ++ + + 0)( )

( )

1

log

0 1log

a

a

a

f x b

af x b

> > > < > hoc c ths dng logarit theo c s a hay b.II. VD minh ho:VD: Gii bt phng trnh:

2

49.2 16.7x x>Gii: Bin i tng ng phng trnh v dng: 4 22 7x x >Ly logarit c s 2 hai v phng trnh ta c:

( )2 4 2 2 2

2 2 2 2 2log 2 log 7 4 2 log 7 ( ) log 7 2 log 7 4 0x x x x f x x x > > = + >

Ta c: ( ) ( ) 222 2 2 2log 7 8log 7 16 log 7 4 4 log 7 = + = = . Suy ra f(x)=0 c nghim:

( ) 12 2

1,22 2 1

2log 7 4 log 7

log 7 22

xx

x x

= = = 2 hoc 2log 7 2x <

BI TON 3: S DNG PHNG PHP T N PH- DNG 1I. Phng php:Mc ch chnh ca phng php ny l chuyn cc bi ton cho v bt phng trnh i s quenbit c bit l cc bt phng trnh bc 2 hoc cc h bt phng trnh.II. VD minh ho:

VD1: Gii bt phng trnh : ( ) ( ) ( )22

2 2 2 2 1 2 1x x x < + Gii: iu kin 2 1 0 0x x .t 2 1xt= , iu kin 0t , khi : 22 1x t= + . Bt phng trnh c dng:

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

2 22 22 2 2 2

2 2 2 2 22 2 2

1 2 1 2 1 1 3 1

1 3 1 0 1 1 3 0

t t t t t t

t t t t t t

+ < + + < +

+ < + +


14/54

( ) ( ) ( )2 31 2 2 0 1 1

2 1 1 2 2 1x x

t t t t

x

< (*)

t 5xu = , iu kin u>2, khi bt phng trnh c dng: 22

3 54

uu

u+ >

(1)

Bnh phng 2 v phng trnh (1) ta c:

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15/54

2 2 2 22

2 22 2

4 445 4. 45

4 44 4

u u u uu

u uu u+ + > + >

(2)

t2

2, 0

4

ut t

u= >

. Khi bt phng trnh (2) c dng:

22 4 2

2

2 5

22

4 45 0 5 5 25 100 0

4log 2020 20 5 20(*)

15 log 55 5 5

2

x

x

ut t t u u

ux

u u

u xu

+ > > > + >

> > > > < <

Vy nghim ca bt phng trnh l ( )5 51log 2; log 20;2x +

BI TON 4: S DNG PHNG PHP T N PH- DNG 2I. Phng php:Phng php ny ging nh phng trnh m.II. VD minh ho:

VD1: Gii bt phng trnh: 214 2 4 0x x x+ + Gii: t 2xt= iu kin t>0Khi bt phng trnh c dng:

22 2 4 0xt t + . Ta c:2

' 1 4 0x =

Do :

22' 004 11 4 0

(2) 001 2 1

2

xx

x

xxb

xt ta

= == = = == = = Vy bt phng trnh c nghim duy nht x=0.

VD2: Gii bt phng trnh : ( ) ( )9 2 5 .3 9 2 1 0x xx x + + + Gii: t 3xt= iu kin t>0. khi bt phng trnh tng ng vi:

( ) ( ) ( )2 2 5 9 2 1 0f t t x t x= + + + . Ta c ( ) ( ) ( )2 2' 5 9 2 1 4x x x = + + = .Do f(t)=0 c 2 nghim t=9 hoc t=2x+1

Do bt phng trnh c dng: ( ) ( )9 2 1 0t t x

3 99 0 2

2 1 0 3 2 1 0 1 2

0 19 0 23 92 1 0 0 13 2 1

x

x

x

x

t x

t x x Bemouli x x x

xt x

t x xx

+ +

Vy bt phng trnh c nghim 2x hoc 0 1x

BI TON 5: S DNG PHNG PHP T N PH- DNG 3I. Phng php:S dng 2 n ph cho 2 biu thc m trong bt phng trnh v kho lo bin i bt phng trnhthnh phng trnh tch, khi lu :

15


16/54

0

0. 0

0

0

A

BA B

A

B

> >> 0 v 0v . Khi bt phng trnh c bin i v dng:

( ) ( ) ( )2 22 2 2 2

2 2 2 2 02 2 1x

u v u v u v u v u v

u v x+ < + + < + >

+

Ta xt phng trnh: 20

2 02 2 1 2 2 1 1

2 12

x x

xx

x xx x

== = + = + = =

Vy bt phng trnh c nghim1 1

; / 0;2 2

x +

VD3:Bt phng trnh : 52 log 2 15 1 5 3 5 2.5 16xx x x+ + + + c nghim la) 1x

b) x>1Gii: Vit li bt phng trnh di dng:

( ) ( )

2 1

2

5 1 5 3 2.5 10.5 16

5 1 5 3 2 5 3 2 5 1

x x x x

x x x x

+ + +

+ +

iu kin: 5 1 0 0x x . t5 1 0

5 3

x

x

u

v

=

= . Bt phng trnh c bin i v dng:

16


17/54

( ) ( )2 2

2 22 2

2

0 02 2 5 1 5 3

2 2 0

5 3 0 5 31

5 7.5 10 05 1 5 3

x x

x x

x xx x

u v u vu v u v u v

u v u v u v

x

+ + + + = = + +

= + = =

Vy bt phng trnh c nghim x=1.

CC BT PHNG TRNH M C GII BNG NHIU CCHI. T VN :Nh vy thng qua cc bi ton trn, chng ta bit c cc phng php c bn gii btphng trnh m v thng qua cc v d minh ho chng ta cng c th thy ngay mt iu rng, mtbt phng trnh c th c thc hin bng nhiu phng php khc nhau. Trong mc ny s minhho nhng v d c gii bng nhiu phng php khc nhau vi mc ch c bn l:+ Gip cc em hc sinh tip nhn y kin thc ton THPT tr nn linh hot trong vic lachn phng php gii.

+ Gip cc em hc sinh lp 10 v 11 la chn c phng php ph hp vi kin thc ca mnh.II. VD minh ho:VD: Tm m dng bt phng trnh sau c nghim:

( ) ( )2 2 2 22 1 2 1

2 3 2 3 8 4 3x x m m m x x m m m+ + + + + + +

+ + +

Gii: Nhn xt rng: ( ) ( )2 3 . 2 3 1+ =Nn nu t ( )

2 222 3

x x m m m

u+ + +

= + iu kin u>1

Th ( )2 22 1

2 3x x m m m

u

+ + + = . Khi bt phng trnh c dng:

Ta c th la chn 1 trong 2 cch gii sau:Cch 1: S dng phng php t n ph.

t t=x-m, bt phng trnh c dng: ( )2 22 2 1 0t t mt m m+ + + + (2)+ Vi 0t th (2) ( ) ( )2 22 1 2 1 0f t t m t m m = + + + + (3)Vy (2) c nghim (3) c t nht 1 nghim 0t

f(t)=0 c t nht 1 nghim 0t 1 2(0 t t hoc 1 20 )t t

( ) 2 22

2

1 2

11 2 1 0' 022 1 0(0) 0 1

111 0 2

0 12

2 1 0 1(0) 0 12

m

m m mm

m mafmm

msm

m maf m

+ + + +

17

( ) ( )( )

2 2

2

22 2

2 32 3 4 2 3 4 1 0

2 3 2 3 2 3 2 3 2 1(1)x x m m m

u u uu

u x x m m m+ + +

++ + + +

+ + + + + +


18/54

+ Vi 0t th (2) ( )2 2( ) 2 1 2 1 0g t t m t m m = + + + (3)Vy (2) c nghim (3) c t nht 1 nghim 0t

phng trnh g(t)=0 c t nht (1) nghim 1 21 2

00

0

t tt

t t

( ) 2 22

2

1 21 2 1 0' 0

12 1 0(0) 0 12 11 0 1 2

02 12 1 0 1(0) 0 2

mm m m

mm magm

ms m

m m mag

+ + +

Vy bt phng trnh c nghim khi1

02

m<

Cch 2: S dng phng php t n ph

t t x m= , iu kin 0t . Bt phng trnh c dng: 2( ) 2 2 1 0h t t t mx m= + + + (4)Vy bt phng trnh c nghim min ( ) 0( 0)h t t (5)Nhn xt rng h(t) l 1 Parabol c nh t=-1


19/54

CH 3: H PHNG TRNH MBI TON 1: S DNG PHNG PHP T N PHI. Phng php:Phng php c s dng nhiu nht gii cc h m l vic s dng cc n ph. Tu theo dngca h m la chn php t n ph thch hp.Ta thc hin theo cc bc sau:Bc 1: t iu kin cho cc biu thc trong h c ngha

Bc 2: La chn n ph bin i h ban u v cc h i s bit cch gii ( h bc nht 2 n,h i xng loi I, h i xng loi II v h ng cp bc 2)Bc 3: Gii h nhn cBc 4: Kt lun v nghim cho h ban u.II. VD minh ho:

VD1: Gii h phng trnh:2 2 2 2

1

3 2 17

2.3 3.2 8

x y

x y

+ +

+

+ =

+ =(I)

Gii: t3

2

x

y

u

v

=

=iu kin u, v>0. Khi h (I) c bin i v dng:

2

2 21

19 6 1 0 3 19 4 17 338 6 16 3 8 2 2 23

x

y

u u xuu vu

yu v vv

+ = = = =+ = =+ = = = = Vy h c cp nghim (-1;1)

VD2: Cho h phng trnh:1

1

3 2 2

3 2 1

x y

x y

m m

m m

+

+

+ =

+ = +a) Tm m h c nghim duy nht.b) Tm m nguyn nghim duy nht ca h l nghim nguyn.

Gii: t13

2

x

y

u

v

+ =

=iu kin u 3 v v>0. Khi h (I) c bin i v dng:

2

1

mu v m

u mv m

+ = + = +

(II). Ta c:

1

mD = 2

11m

m= ;

2

1um

Dm

=+

21 2 1;

1vm

m m Dm

= = 22

1

mm m

m=

+a) H c nghim duy nht khi:

20 1 01

2 13 3 2 1 2 1

1 1 00

1

u

v

D mm

D mu m m

D m m mD m

vD m

+ = < + < = > +

Vy h c nghim khi 2 1m < .a) Vi m nguyn ta c m=-2 khi h c nghim l:

13 03 3 1 1

2 112 2

x

y

u xx

v yy

+ = == + = = === Vy vi m=-2 h c nghim nguyn (0;1)

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20/54

VD3: Cho h phng trnh:2 cot sin

sin cot

9 3

9 81 2

gx y

y gx m

+ =

=a) Gii h phng trnh vim=1

b) Tm m h c cp nghim (x;y) tho mn 02

y

Gii: Bin i h v dng:

2

. 3

u v m

u v

+ =

= Khi u, v l nghim ca phng trnh 2( ) 2 3 0f t t mt = = (1)a) Vi m=1 ta c:

sin0; 02

2cot

1 3 9 32 3 0

3 1 9 1

y

u v

gx

t ut t

t v

> < = = = = = = =

261 ; 2

sin 5 2 6 ; ,2256

cot 0 ; 22 6

2

y k

x l y y k y

k l Zy k

gx x l y y k x l

= + = + = = + = = + = = + = = +

= +Vy vi m=1 h c 2 h cp nghim.

VD4: Gii h phng trnh:

2 2

2

2 2 2

2 2 2

4 2 4 1

2 3.2 16

x x y y

y x y

+

+ +

+ =

=

Gii: Vit li h phng trnh di dng:( )2 2

2

2 1 1 2

2 1

4 4.4 .2 2 1

2 3.4 .2 4

x x y y

y x y

+ = =

(I)

t

2 1

42

x

yuv

= =iu kin 14

u v v>0.

Khi h (I) c bin i v dng:2 2

2

4 1(1)

4 4(2)

u uv v

v uv

+ =

=(II)

gii h (II) ta c th s dng 1 trong 2 cch sau:Cch 1: Kh s hng t do t h ta c: 2 24 13 3 0u uv v + = (3)

t u=tv, khi : ( )2 23

(3) 4 13 3 0 1

4

t

v t tt

= + = =

+ Vi t=3 ta c u=3v do : 2(2) 8 4v = v nghim.

+ Vi1

4t= ta c

14

4u v v u= = do : 2(2) 4 4 1u u = =

2 211 11 04 1

4 222 4

x

y

u xx

v yy

= = == = === Vy h phng trnh c 2 cp nghim (1;2) v (-1;2)Cch 2: Nhn xt rng nu (u;v) l nghim ca h th 0u

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21/54

T (2) ta c2 4

3

vu

v

= (4). Thay (4) vo (1) ta c: 4 22 31 16 0v v = (5)

t 2 , 0t v t= > ta c: 2 216

1(5) 2 31 16 0 16 41

4(1)2

tu

t t v vvt

= = = = = == 2 21 11 04 1

222 4

x

y

xx

yy

= == === Vy h phng trnh c 2 cp nghim (1;2) v (-1;2)

VD5: Gii h phng trnh:2 1 2

22

2 3.2 2

2 3 2 2

x x

x

y

y y

+ = =

= Gii: t 2 xu = iu kin 1u . H c dng:

( ) ( ) ( )

( ) ( )

2 22 2 2 2

2 2

2 3 22 3

2 3 2

3 1 0 1

u u yu y u y u y

y y u

u y

u y u y y u

= = =

= + = =

+ Vi u=y, h phng trnh tng ng vi:

2 2 2

2 1 0

1 11

22 3 2 3 2 0 12 222

x

x

x

y yu y u y u y

u yu u u u u x

yy

= = = == = = = = = = + = = = ==

+ Vi y=1-u, h phng trnh tng vi:

( ) 2 221 1

3 1 02 3 1 2

y u y u

u uu u u

= =

+ = = v nghimVy h c 3 cp nghim l (0;1), (1;2) v (-1;2).

VD6: Gii phng trnh:

( ) ( )

( ) ( )

22log 3log

2 2

9 3 2 (1)

1 1 1(2)

xyxy

x y

=

+ + + =Gii: iu kin xy>0

+ Gii (1): t ( )2log 2tt xy xy= = . Khi phng trnh (1) c dng:

( ) 2log 3 2 29 3 2 2 3 3 2.3 3 2.3 3 0t t t t t t = = = (3)t 3 , 0tu u= > , khi phng trnh (3) c dng:

2 1(1)2 3 0 3 3 1 2

3t

uu u t xy

u

= = = = = =

+ Gii (2): ( ) ( )22 2 2 2 1 0 2 2 1 0x y x y x y x y xy + + + + = + + + + =

( ) ( )2 2 3 0x y x y + + + = (4)t v=x+y, khi phng trnh (4) c dng:

21 1

2 3 03 3

v x yv v

v x y

= + = + = = + =

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22/54

Vi x+y=1 ta c:1

2

x y

xy

+ = =

Khi x, y l nghim ca phng trnh: 2 2 0X X + = v nghim

Vi x+y=-3, ta c:3

2

x y

xy

+ = =

Khi x, y l nghim ca phng trnh :2 1 1

3 2 0 2 2

X x

X X X y

= = + = = = v

2

1

x

y

= =

Vy h c 2 cp nghim (1;2) v (2;1)

VD7: Gii h phng trnh:

3 1 2 3

2

2 2 3.2 (1)

3 1 1(2)

x y y x

x xy x

+ + + =

+ + = +Gii:

Phng trnh (2) ( )2

1 011 0

0 13 1 03 1 1

3 1 0 1 3

x xxx

x xx x yx xy x

x y y x

= + = + =+ + = + + = =

+ Vi x=0 thay vo (1) ta c: 2 28 82 2 3.2 8 2 12.2 2 log11 11y y y y y y+ = + = = =

+ Vi1

1 3

x

y x

=

thay y=1-3x vo (1) ta c: 3 1 3 12 2 3.2x x+ + = (3)

t 3 12 xt += v 1t nn1

4t

( ) ( )

2 3 1

2 2

3 8(1)1(3) 6 6 1 0 2 3 8

3 8

1

log 3 8 1 2 log 3 83

xt

t t tt t

x y

+ =

+ = + = = += +

= + = +

Vy h phng trnh c 2 nghim:2

0

8log

11

x

y

=

=v

( )( )

2

2

1log 3 8 1

3

2 log 3 8

x

y

= + = +

BI TON 2: S DNG PHNG PHP HM SI. Phng php:Ta thc hin theo cc bc sau:Bc 1: t iu kin cho cc biu thc trong h c ngha.

Bc 2: T h ban u chng ta xc nh c 1 phng trnh h qu theo 1 n hoc c 2 n, giiphng trnh ny bng phng php hm s bitBc 3: Gii h mi nhn cII. VD minh ho:

VD1: Gii h phng trnh:2 2

3 3 (1)

12(2)

x y y x

x xy y

=

+ + =Gii: Xt phng trnh (1) di dng: 3 3x yx y+ = + (3)Xt hm s ( ) 3tf t t= + ng bin trn R.

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23/54

Vy phng trnh (3) c vit di dng: ( ) ( )f x f y x y= = . Khi h c dng:

2 2 2

2

2 212 3 12

x y x y x y x y

x x yx xy y x

= = = = = = = = + + = =

Vy h phng trnh c 2 cp nghim (2;2) v (-2;-2)

VD2: Gii h phng trnh:2 2 3

2 2 3

x

y

x y

y x

+ = +

+ = +Gii: Bin i tng ng h v dng:

2 2 32 3 3 2 3 3

3 2 2

x

x y

y

x yx y

x y

+ = + + + = + ++ = +

(1)

Xt hm s ( ) 2 3 3tf t t= + + l hm ng bin trn R.Vy phng trnh (1) c vit di dng: ( ) ( )f x f y x y= = .

Khi h thnh:2 2 3 2 3 (2)x xx y x y

x y x

= =

+ = + = (II)

+ Gii (2): Ta on c x=1 v 12 3 1= . V tri l mt hm ng bin cn v tri l hm s nghchbin do vy x=1 l nghim duy nht ca phng trnh ny. Khi h (II) tr thnh:

11

x yx y

x

= = = =

Vy h cho c nghim x=y=1.

VD3: Gii h phng trnh:( ) ( )

2 2

2 2 2 (1)

2(2)

x y y x xy

x y

= +

+ =Gii: Thay (2) vo (1) ta c:

( ) ( )2 2 3 33 3

2 2 2 2

2 2 (3)

x y x y

x y

y x x y xy y x

x y

= + + =

=

Xt hm s ( ) 32tf t t= + ng bin trn R.

Vy phng trnh (3) c vit di dng: ( ) ( )f x f y x y= = . Khi h c dng:

2 2 2

1

1 12 2 2

x y x y x y x y

x x yx y x

= = = = = = = = + = =

Vy h c 2 cp nghim (1;1) v (-1;-1)

BI TON 3: S DNG PHNG PHP NH GII. Phng php:Nhiu bi ton bng cch nh gi tinh t da trn cc:

+ Tam thc bc hai+Tnh cht hm s m+Bt ng thc+..Ta c th nhanh chng ch ra c nghim ca h hoc bin i h v dng n gin hn.II. VD minh ho:

VD: Gii h phng trnh:

2 2

2

1 1

1

2 3 2 2 3

2 .3 1

x y x y

x y

+ = + =

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24/54

Gii: t 2 12x

y

u

v

=

=iu kin u>0 v

1

3v . H c dng:

2(1)

1(2)

u v u v

uv

+ + =

=(I)

Bin i (1) v dng:

( ) ( ) ( ) ( )2 2 2 2 2 2 2 2 2 24 2 2 2 2 4 4u v u v u v u v u v u v uv = + + + = + + + =Khi h tng ng vi:

2

2 2

2 21

2 0 2 1 0 01

1 0 13 11

x

y

u v x xu v u v

y yuv

= = = = = = = = = = =

Vy h c 2 cp nghim (0;1) v (0;-1)

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25/54

CH 4: H BT PHNG TRNH MBI TON 1: S DNG PHNG PHP BIN I TNG NGI. Phng php:Da vo cc php ton bin i tng ng cho cc bt ng thc trong h bt phng trnh, ta c

th tm c nghim ca h. Php ton thng c s dng l:A B

A C B DC D

+> + > + >Vic la chn phng php bin i tng ng gii h bt phng trnh m thng c thchin theo cc bc sau:Bc 1: t iu kin cc biu thc ca h c nghaBc 2: Thc hin cc php bin i tng chuyn h v 1 bt phng trnh i s bit cch gii.Bc 3: Kim tra tnh hp l cho nghim tm c, t a ra li kt lun cho h.Vi h bt phng trnh m cha tham s thng c thc hin theo cc bc sau:Bc 1: t iu kin cc biu thc ca h c nghaBc 2: Thc hin cc php bin i tng ng ( phng php th c s dng kh nhiu trongphp bin i tng ng ) nhn c t h 1 bt phng trnh 1 n cha tham s.Bc 3: Gii v bin lun theo tham s bt phng trnh nhn c.Bc 4: Kim tra tnh hp l cho nghim tm c, t a ra kt lun cho h.Ch : i vi h bt phng trnh m 1 n thng c gii tng bt phng trnh ca h, ri kt

hp cc tp nghim tm c a ra kt lun v nghim cho h bt phng trnh.II. VD minh ho:

VD1: Gii h bt phng trnh:

2 22 1 2 2

2

2 9.2 2 (1)

2 5 4 3(2)

x x x x

x x x

+ + + +

< + Gii:Gii (1):

2 2 2 22 22.2 9.2 4.2 0 2.2 9 4.2 0x x x x x x x x+ + = + =

t2

2x xt = iu kin 41

2t . Khi phng trnh c dng:

22

2 2

4

42 9 0 2 9 4 0 2 41(1)

2

12 2 0 (3)

2

x x

t

t t tt t

xx x x x

x

=+ = + = = ==

= = =

Gii (2):

( )

2

22

2

5 512 5 0 2 2

4 3 0 1 3 1451

2 5 0 55 2

1424 3 2 5 255 24 28 0

x xx

x x xxx

xx

x x x xx x


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Bc 3: Gii h nhn c t suy ra nghim x; yBc 4: Kim tra tnh hp l cho nghim tm c, t a ra li kt lun cho h.II. VD minh ho:

VD: Gii h bt phng trnh:( )

2

22 2

3

2 2 2 1

log 2 2 0

x y

x y

=

(I)

Gii: t 22

x

yuv

= =; u, v >

t:2

2

x

y

u

v

=

=, iu kin u, v>0. H c bin i v dng:

( )

( )

222 2

2 2 22

1 (1)2 1

2 1 1 (2)

u v mu v v m

u v u m v u m

+ + + + +

+ + + + + (I)

iu kn cn: Gi s h c nghim (u0;v0) suy ra (v0;u0) cng l nghim ca h. Vy h c nghimduy nht th iu kin cn l u0=v0.

Khi : ( ) 22 20 0 0 01 2 2 1 0u u m u u m+ + + + (1)

Ta cn (1) phi c nghim duy nht1

02

m = =

Vy iu kin cn h c nghim duy nht l m=1/2

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iu kin : Vi1

2m = h c dng:

( )

( )

22

22

11

21

12

u v

v u

+ + + +

(II)

( ) ( )2 22 2 2 22 2

1 1 1 2 2 2 2 1 0

2 2 12 2 02 2 2

u v u v u u v v

u v u v

+ + + + + + + + +

+ + + = =

Nhn xt rng1

2u v= = tho mn h (II) suy ra x=y=-1

Vy h c nghim duy nht khi m=1/2.

BI TON 4: S DNG PHNG PHP NH GII. Phng php:Nhiu bt phng trnh nh gi tinh t da trn:+ Tam thc bc 2+ Cc bt ng thc c bn nh: Csi, Bunhiacpxki+ Tnh cht tr tuyt iTa c th nhanh chng ch ra c nghim ca n.II. VD minh ho:

VD1: Gii h bt phng trnh:2 1 2 2 (1)

2 2 2 2 1(2)

x y y y

x y y y

+

+

+

+ = (I)

Gii: iu kin: ( )2 11 2 0 2 1 0

22 2 1 02 2 0 2 1

yy y

y xx y y x

y

x+

(*)

Gii (1):(*) 2 11 22 1 0

2 1 2 0

xyx

y yx y = + = =

=(3)

Thay (3) vo (2) thy tho mn. Vy h c nghim duy nht x=y=0.

VD2: Gii h phng trnh:( )

( )

232 3 log 5 4

2

3 5 (1)

4 1 3 8(2)

x x y

y y y

+ =

+ + Gii:

Gii (1) ta c: ( ) ( )2

3 32 3 log 54 log 5 15 3 3 5 4 1 3

x xyy y

+ = = + (3)

Gii (2) vi 3y ta c: ( ) ( ) 2 24 1 3 8 3 0 3 0y y y y y y + + + + (4)T (3) v (4) suy ra y=-3, khi h thnh:

21

1; 32 3 03

3; 333

xx yx x

xx yy

y

= = = = = = = = =

Vy h phng trnh c 2 cp nghim (-1;-3) v (3;-3).

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CHNG II:PHNG PHP GII PHNG TRNH-BT PHNG TRNH- H LGA RIT.

CH 1: PHNG TRNH LGARITBI TON 1: S DNG PHNG PHP LGARIT HO V A V CNG C SI. Phng php: chuyn n s khi lgarit ngi ta c th lgarit ho theo cng 1 c s c 2 v ca phng trnh,

bt phng trnh. Chng ta lu cc php bin i c bn sau:Dng 1: Phng trnh:

( )0 1

log ( )a b

af x b

f x a

< = =

Dng 2: Phng trnh: ( ) ( ) ( ) ( )0 1

log log0a a

af x g x

f x g x

< = = >Ch : Vic la chn iu kin f(x)>0 hoc g(x)>0 tu thuc vo phc tp ca f(x) v g(x).II. VD minh ho:

VD1: Gii phng trnh: ( ) ( )29 3 32 log log .log 2 1 1x x x= +

Gii: iu kin:

0

2 1 0 0

2 1 1 0

x

x x

x

> + > + >

. Phng trnh c vit di dng:

( ) ( )

( ) ( )

( )

( ) ( )

2

2

3 3 3 3 3 3

2

3 3 3 3 3 3

3

3 3

0

2

1 12 log log .log 2 1 1 log log .log 2 1 1

2 2

log 2 log .log 2 1 1 log 2 log 2 1 1 log 0

log 0 1

log 2 log 2 1 1 0 2 1 2 2 1 1

11

4 2 1 22 2 1 2

x

x x x x x x

x x x x x x

x x

x x x x x

xx

x xx x

>

= + = +

= + + = = =

+ = = + + + ==

+ = ++ = +

0

2

1 1

44 0

xx x

xx x

>

= = = = Vy phng trnh c nghim x=1 v x=4.VD2: Gii phng trnh: 3 4 5log log logx x x+ =Gii: iu kin x>0. Ta bin i v cng c s 3:

4 4 3

5 5 3

log log 3. log

log log 3. log

x x

x x

=

=khi phng trnh c dng:

( )3 4 3 5 3

3 4 5 3

log log 3. log log 3. log

log 1 log 3 log 3 0 log 0 1

x x x

x x x

+ =

+ = = =Vy phng trnh c nghim x=1.

BI TON 2: S DNG PHNG PHP T N PH- DNG 1I. Phng php:Phng php t n ph dng 1 l vic s dng 1 n ph chuyn phng trnh ban u thnh 1phng trnh vi 1 n ph.

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Ta lu cc php t n ph thng gp sau:

Dng 1: Nu t logat x= vi x>0 th:1

log ; logk ka x

x t at

= = vi 0 1x<

Dng 2: Ta bit rng: log logb bc aa c= do nu t logb xt a= th logb at x= . Tuy nhin trong nhiu biton c cha logb xa , ta thng t n ph dn vi logbt x= .VD minh ho:

VD1: Cho phng trnh: ( ) ( )2 4log 5 1 .log 2.5 2x x

m = (1)a) Gii phng trnh vi m=1b) Xc nh m phng trnh c nghim 1x

Gii: Bin i phng trnh v dng:

( ) ( ) ( ) ( )2 2 2 21 log 5 1 .log 2 5 1 log 5 1 . 1 log 5 1 22

x x x xm m = + = iu kin: 5 1 0 5 1 0x x x > > >t ( )2log 5 1xt= . Khi phng trnh c dng: ( ) ( ) 21 2 2 0t t m f t t t m+ = = + = (2)

a) Vi m=1 ta c:( )

( )

22

22

log 5 1 11 5 1 22 0

2 5 1 2log 5 1 2

x x

xx

tt t

t

= = = + = = = =

5

5

log 35 3

55log5

44

x

x

x

x

= = ==

Vy vi m=1 phng trnh c 2 nghim 5 55

log 3; log4

x x= =

b)Vi ( )2 21 5 1 5 1 4 log 5 1 log 4 2 2x xx t = =

Vy phng trnh (1) c nghim 1x (2) c nghim 2t 1 2

1 2

2 (*)

2

t t

t t

(loi (*))

( ). 2 0 4 2 2 0 3a f m m + .Vy vi 3m tho mn iu kin u bi.

VD2: Gii phng trnh: ( ) ( )2 2 22 3 6log 1 .log 1 log 1x x x x x x + =

Gii: iu kin:

2

2

2

1 0

1 0 1

1 0

x

x x x

x x

>

+ >

Nhn xt rng:

( ) ( ) ( ) ( )

12 2 2 21 1 1 1 1x x x x x x x x

+ = = + Khi phng trnh c vit di dng:

( ) ( ) ( )( ) ( ) ( )

1 12 2 2

2 3 6

2 2 2

2 3 6

log 1 .log 1 log 1

log 1 .log 1 log 1

x x x x x x

x x x x x x

+ + = +

+ + = +

s dng php bin i c s: ( ) ( )2 22 2 6log 1 log 6.log 1x x x x+ = + v ( ) ( )2 23 3 6log 1 log 6.log 1x x x x+ = +

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Khi phng trnh c vit di dng:

( ) ( ) ( )2 2 22 6 3 6 6log 6.log 1 .log 6.log 1 log 1x x x x x x+ + = + (1)t ( )26log 1t x x= + . Khi (1) c dng: ( )2 3

2 3

0log 6. log 6. 1 0

log 6. log 6. 1 0

tt t

t

= = =

+

Vi t=0 ( )

22 2

6 2

1

log 1 0 1 1 11

x x

x x x x xx x

+

+ = + = = + Vi 2 3log 6. log 6. 1 0t =

( ) ( )( )

( )

6

6

6 6

6

2 22 3 6 2 3

log 22 23 6

log 22log 2 log 2

log 22

log 6.log 6.log 1 0 log 6.log 1 1

log 1 log 2 1 3

1 3 13 3

21 3

x x x x

x x x x

x xx

x x

+ = + =

+ = + =

+ = = + =

Vy phng trnh c nghim x=1 v ( )6 6log 2 log 2

13 32x

= +

BI TON 3: S DNG PHNG PHP T N PH- DNG 2I. Phng php:Phng php dng n ph dng 2 l vic s dng 1 nph chuyn phng trnh ban u thnh phngtrnh vi 1 n ph nhng cc h s vn cn cha x.Phng php ny thng c s dng i vi nhng phng trnh khi la chnn ph cho 1 biuthc th cc biu thc cn li khng biu din c trit qua n ph hoc nu biu din c thcng thc biu din li qu phc tp.Khi thng ta c 1 phng trnh bc hai theo n ph ( hoc vn theo n x ) c bit s l 1 s

chnh phng.II. VD minh ho:VD1: Gii phng trnh: ( )2 2 2lg lg .log 4 2log 0x x x x + =Gii: iu kin x>0.

Bin i phng trnh v dng: ( )2 2 2lg 2 lg lg 2 lg 0x x x x + + =t t=lgx, khi phng trnh tng ng vi: ( )2 2 22 log . 2 log 0t x t x + + =

Ta c: ( ) ( )2 22 2 22 log 8log 2 logx x x = + = suy ra phng trnh c nghim

2

lg 22 lg 2 100

lg

lglog lg 0 1lg 2

xt x x

x

xt x x x

== = = == = =

Vy phng trnh c 2 nghim x=100 v x=1

BI TON 4: S DNG PHNG PHP T N PH- DNG 3I. Phng php:

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Phng php dng n ph dng 3 s dng 2 n ph cho 2 biu thc lgarit trong phng trnh v bini phng trnh thnh phng trnh tch.II. VD minh ho:

Gii phng trnh: ( ) ( )2 22 2 2log 1 log .log 2 0x x x x x + = Gii:

iu kin

( ) 2

2

1 0

0 1

0

x x

x x

x x

>

> > >

. Bin i phng trnh v dng:

( ) ( )( ) ( )

22

22 2 2

2 22 2 2

log log .log 2 0

2 log log .log 2 0

x xx x x

x

x x x x x

+ =

+ =

t( )22

2

log

log

u x x

v x

=

=. Khi phng trnh tng ng vi:

( ) ( )

( )2 222

12 2 0 1 2 02

1( )log 1 2 0

24log 2 4

uu v uv u vv

x Lx x x x

xxx

x

=+ = = == = = = == =

Vy phng trnh c 2 nghim x=2 v x=4.

BI TON 5: S DNG PHNG PHP T N PH- DNG 4

I. Phng php:Phng php t n ph dng 4 l vic s dng k n ph chuyn phng trnh ban u thnh 1 hphng trnh vi k n ph.Trong h mi th k-1 phng trnh nhn c t cc mi lin h gia cc i lng tng ngII. VD minh ho:

VD1: Gii phng trnh: ( ) ( )2 22 2log 1 3log 1 2x x x x + + =

Gii: iu kin

2

2

2

1 0

1 0 1

1 0

x

x x x

x x

>

+ >

t( )( )

22

22

log 1

log 1

u x x

v x x

= = +

Nhn xt rng: ( ) ( )2 22 2log 1 log 1u v x x x x+ = + + ( ) ( )2 22 2log 1 . 1 log 1 0x x x x= + = =

Khi phng trnh c chuyn thnh:

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32/54

( )( )

22

22

2

2

log 1 10 1

3 2 2 2 1 log 1 1

11 52

4

1 2

x xu v u v u

u v v v x x

x xx

x x

= + = = = + = = = + =

= =

+ =Vy phng trnh c nghim x=5/4.

VD2: Gii phng trnh: ( ) ( )2 22 23 log 4 5 2 5 log 4 5 6x x x x+ + + + = (1)

Gii: iu kin ( )( )

2

2 2 5 22

22

4 5 0

3 log 4 5 0 4 5 2 2 4

5 log 4 5 0

x x

x x x x x

x x

+ > + + +

+ 2 29 2 29(*)x +

t( )( )

22

22

3 log 5

5 log 5

u x x

v x x

= + + = +

iu kin , 0u v . Khi phng trnh c chuyn thnh:

( )

( )( )

( )

( )

22 2 22

22

22

22

22

6 26 22 6 6 2 2

8 5 24 28 06 2 8 14

5

3 log 4 5 2

5 log 4 5 2 log2; 2

14 2 14; 3 log 4 55 5 52

5 log 4 55

u v

u vu v u v v

u v v vv vv

x x

x xv u

v v x x

x x

= = + = = = + = + = + = =

+ + = + == = = = + + =

+ =

( )( )

22

22

2 2

121 1212 225 25

12125

4 5 1

121log 4 5

25

4 5 2 4 3 0

34 5 2 4 5 2 02 2 1

x x

x x

x xx x x x

xx x x x

x

+ = + =

= + = + =

= + = + = =

Vy phng trnh c 4 nghim phn bit.


32


33/54

Phng php t n ph dng 5 l vic s dng 1 n ph chuyn phng trnh ban u thnh 1 hphng trnh vi 1 n ph v 1 n x.Ta thc hin theo cc bc sau:Bc 1: t iu kin c ngha cho cc biu thc trong phng trnh

Bc 2: Bin i phng trnh v dng: ( ),f x x =0

Bc 3: t ( )y x= , ta bin i phng trnh thnh h:( )

( ); 0

y x

f x y

=

=II. VD minh ho:

VD1: Gii phng trnh: 22 2log log 1 1x x+ + = (1)

Gii: t 2logu x= . Khi phng trnh thnh: 2 1 1u u+ + = (2)

iu kin: 21 0

1 11 0

uu

u

+

t 1v u= + iu kin 0 2v 2 1v u = +Khi phng trnh c chuyn thnh h:

( ) ( ) ( )

2

2 22

1 01 0 1 01

u v u vu v u v u v u v u vv u

= + = = + + + = + == + Khi :

+ Vi v=-u ta c:1 5

2 22

1 51 521 0 log 2

21 5(1)

2

u

u u x x

u

= = = =

+=

+ Vi u-v+1=0 ta c:22

2

1log 00

0 11 log 1

2

xxu

u uu x x

=== + = = = = Vy phng trnh c 3 nghim.

BI TON 7: S DNG TNH CHT N IU CA HM SI. Phng php:S dng tnh cht n iu ca hm s gii phng trnh l dng ton kh quen thuc. Ta c 3hng p dng sau:Hng 1: Thc hin theo cc bc:Bc 1: Chuyn phng trnh v dng: f(x)=k (1)Bc 2: Xt hm s y=f(x). Dng lp lun khng nh hm s n iu (gi s ng bin)Bc 3: Nhn xt:

+ Vi ( ) ( )0 0x x f x f x k = = = do 0x x= l nghim+ Vi ( ) ( )0 0x x f x f x k > > = do phng trnh v nghim+ Vi ( ) ( )0 0x x f x f x k < < = do phng trnh v nghim.Vy x=x0 l nghim duy nht ca phng trnhHng 2: Thc hin theo cc bc:Bc 1: Chuyn phng trnh v dng: f(x)=g(x) (2)Bc 2: Xt hm s y=f(x) v y=g(x). Dng lp lun khng nh hm s y=f(x) l ng bin cn hms y=g(x) l hm hng hoc nghch bin.

Xc nh x0 sao cho f(x0)=g(x0)

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34/54

Bc 3: Vy phng trnh c nghim duy nht x=x0Hng 3: Thc hin theo cc bc:Bc 1: Chuyn phng trnh v dng: f(u)=f(v) (3)Bc 2: Xt hm s y=f(x). Dng lp lun khng nh hm s n iu (gi s ng bin)Bc 3: Khi (3) u v = vi , fu v D II. VD minh ho:

VD1: Gii phng trnh:

( )( )22 2log 4 log 8 2x x x + = +

Gii: iu kin

2 4 02

2 0

xx

x

> >

+ >. Vit li phng trnh di dng:

( ) ( ) ( )2

22 2 2 2

4log 4 log 2 3 log 3 log 2 3

2

xx x x x x x

x

+ = = =

+Nhn xt rng:

+ Hm s ( )2log 2y x= l hm ng bin+ Hm s y=3-x l hm nghch bin+ Vy phng trnh nu c nghim th nghim l duy nht+ Nhn xt rng x=3 l nghim ca phng trnh

Vy phng trnh c nghim x=3.VD2: Gii phng trnh: ( ) ( )4 2 225log 2 3 2 log 2 4x x x x =

Gii: iu kin:2

2

2 3 0 1 5

2 4 0 1 5

x x x

x x x

> < > > +

. Vit li phng trnh di dng:

( ) ( )( ) ( )

2 225

2 25 4

log 2 3 log 2 4

log 2 3 log 2 4 (1)

x x x x

x x x x

=

=

t 2 2 4t x x= khi (1) ( )5 4log 1 logt t + = (2)

t 4log 4y

y t t= = phng trnh (2) c chuyn thnh h:4 4 1

4 1 5 15 51 5

y yy

y y

y

t

t

= + = + = + = (3)

Hm s ( ) 4 15 5

y y

f y = +

l hm nghch bin

Ta c:+ Vi y=1, f(1)=1 do y=1 l nghim ca phng trnh (3)+ Vi y>1, f(y)


35/54

Chia c 2 v cho 5 0t ta c:4 3

15 5

t t

+ =

(2)

Nhn xt rng:+ V tri ca phng trnh l mt hm nghch bin+ V phi ca phng trnh l mt hm hng+ Do vy nu phng trnh c nghim th nghim l duy nht

+ Nhn xt rng t=2 l nghim ca phng trnh (2) v

2 2

4 3 15 5

+ = Vi 22 log 2 4t x x= = =Vy x=4 l nghim duy nht ca phng trnh

VD4: Gii phng trnh: ( )23 1

23

1log 3 2 2 2

5

x x

x x

+ + + =

(1)

Gii: iu kin 21

3 2 02

xx x

x

+

t 2 2 2 2 23 2; 0 3 2 3 1 1u x x u x x u x x u= + + = =

Khi (1) c dng: ( )21

3

1log 2 2

5

u

u + + =

(2)

Xt hm s ( ) ( ) ( )2

21

3 3

1 1log 2 log 2 .5

5 5

u

uf u u u

= + + = + +

Min xc nh [ )0;D = +

o hm: ( ) ( )21 1

.2 .5 . ln5 0,2 ln 3 5

uf u u u Du

= + > + .

Suy ra hm s ng bin trn D

Mt khc ( ) ( )3 11 log 1 2 .5 25f = + + =

Khi (2) ( ) ( ) 2 3 51 1 3 2 12

f u f u x x x

= = + = =

Vy phng trnh c 2 nghim3 5

2x

=

BI TON 8: S DNG PHNG PHP NH GII. Phng php:II. VD minh ho:

VD1: Gii phng trnh : ( )3 2log 4 5 1x x + + = (1)Gii:Cch 1: Theo bt ng thc Bunhiacpski ta c:

( ) ( ) ( )3 24 5 1 1 4 5 3 2 log 4 5 1x x x x x x + + + + = + + Vy phng trnh c nghim khi v ch khi:

4 5 1

1 1 2

x xx

+= = l nghim duy nht

Cch 2: Theo bt ng thc Csi ta c:

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36/54

( ) ( ) ( ) ( ) ( )( )

2

3 2

4 5 4 5 2 4 5 9 4 5 18

4 5 3 2 log 4 5 1

x x x x x x x x

x x x x

+ + = + + + + + + + + =

+ + + +

Vy phng trnh c nghim khi v ch khi:1

4 52

x x x = + = l nghim duy nht ca phng trnh

CH 2: BT PHNG TRNH LGARITBI TON 1: S DNG PHNG PHP BIN I TNG NGI. Phng php: chuyn n s khi loga ngi ta c th m ho theo cng 1 c s c 2 v bt phng trnh. Chngta lu cc php bin i c bn sau:

Dng 1: Vi bt phng trnh: ( ) ( )log loga af x g x >< < < >> < > > <


37/54

VD2: Gii bt phng trnh: ( )2log 5 8 3 2x x x + >Gii:Cch 1: Bt phng trnh tng ng vi:

2

2 2

2

2 22

11 34 8 3 0

5 8 3 20 11 30 1 5 8 3 0 2 50 5 8 3

4 8 3 0

xx

x x xx x x

xx xx xx x x

x x

> > + > > + > < < <

> < <

(*)

Bin i tng ng bt phng trnh v dng:

( ) ( ) ( ) ( )2 2

2

lg 5 1 lg 10. 5 5 1 10. 5

9 3 3 5

x x x x

x x x

> > > > <

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38/54

( )

( )

233

333

2

45 1

5 6 035 5

4 5 2 30 5 1

350 35 5

5 6 0

xx

x xx x

x xx

xx x

x x

+ < + > +

> + + >

t 03 1 1xt x t>= > . Khi bt phng trnh (2) c dng:

( ) ( ) ( ) ( )1 03 2 2

3

0

3

2 0 2 0 1 2 0 2 0

2 1 2 1 8 9

0 1 0 0 11 0

t

x

t t t t t t t t t t t

t x x x

t x xx

+ >

>

> > + > >> > > >

< < <


39/54

( ) ( ) ( ) ( )

2 2 23 3 3 3

22 31 1 1 2 2 22 2 2

222 2 2

1 1 2 22 2

log log log log log log 88 8 8 8

log log log log

x x x xx

x x x x

= = = =

= = =


II. VD minh ho:

Gii bt phng trnh: ( )2 33 2 3 2log log 8 .log log 0x x x x + < (1)Gii: iu kin x>0

Bin i phng trnh tng ng v dng: ( )23 2 3 2log 3 log log 3log 0x x x x + + > < < > > < >>


40/54

( ) ( )

3

2

3

2

2 2 0 1 2 0

log 11 0 3

log 22 0 43 4

1 0 3log 1

2 0 4log 2

uv u v u v

xu x

xv xx

u xx

v xx

< > >


41/54

+ Vi -15

Vy bt phng trnh c nghim: ( )1 3

0; 1; 5;2 2

+

CH 3: H PHNG TRNH LGARITBI TON 1: S DNG PHNG PHP BIN I TNGI. Phng php:Ta thc hin theo cc bc sau:Bc 1: t iu kin cho cc biu thc trong h c nghaBc 2: S dng cc php th nhn c t h 1 phng trnh theo n x hoc y (i khi c th ltheo c 2 n x, y)Bc 3: Gii phng trnh nhn c bng cc phng php bit i vi phng trnh cha cnthcBc 4: Kt lun v nghim cho h phng trnh.II. VD minh ho:

VD1: Gii h phng trnh: ( )3

3 41 3 (1)

log 1(2)

y xxx

y x

+ =

+ =

Gii: iu kin:

1 0

4 0 0 4

0

x

x x

x

+ < >

T phng trnh (2) ta c: 33

1 log3 log

3 31 log 3 3

3xy

xy x

x

= = = = (3)

Th (3) vo (1) ta c:

( )

( ) 2 2

3 3 41 1 1 1 4 1 4 1

2 0 24 2 3 03 04 2

xx x x x x

x x

x xx x x y

x xx x

+ = + = + = +

= = = = =

Vy h phng trnh c 1 cp nghim (3;0).

VD2: Gii h phng trnh:( ) ( )

2 2

2 3

4 2

log 2 log 2 1

x y

x y x y

=

+ =

Gii: iu kin:2 0

2 0

x y

x y

+ > >

(*)

41


42/54

T phng trnh th nht ca h ly lgarit c s 2 hai v ta c:

( ) ( ) ( )( ) ( )

2 22 2 2 2

2 2

log 4 log 2 log 2 log 2 1

log 2 1 log 2

x y x y x y

x y x y

= + + =

+ = Th vo phng trnh th hai ta c:

( ) ( ) ( ) ( )

( )

2 3 2 3 2

2

1 log 2 log 2.log 2 1 1 log 2 log 2 0

log 2 0 2 1

x y x y x y

x y x y

= + =

= =

Vy ta c h mi:2 2

32 24 2 42 1 12 1

2

xx yx y

x yx yy

=+ = = = = =

tho mn iu kin (*)

Vy h phng trnh c 1 nghim.

BI TON 2: S DNG PHNG PHP T N PHI. Phng php:Phng php c s dng nhiu nht gii cc h lgarit l vic s dng cc n ph. Tu theodng ca h m la chn php t n ph thch hp.Ta thc hin theo cc bc sau:Bc 1: t iu kin cho cc biu thc ca h c ngha.Bc 2: La chn n ph bin i h ban u v cc h i s bit cch gii (h i xng loi I,loi II v h ng cp bc hai)Bc 3: Gii h nhn cBc 4: Kt lun v nghim cho h ban u.II. VD minh ho:

Gii h phng trnh:

( ) ( )3 3

4 32

log 1 log

x y

y x

x y x y

+ =

= +

Gii: iu kin:

0

0

; 0

x y

x y

x y

> + >

Bin i h phng trnh v dng:

( )2 2 2 23

2 5 2 5(1)

log 1 3(2)

x y x y

y x y x

x y x y

+ = + = = =

Gii (1): t1x y

t

y x t

= = . Khi (1) c dng:

2

221

2 5 2 5 2 0 12

2

tx y

t t ty xt t

= = + = + = ==

+ Vi x=2y2 2 1 2(2) 4 3

1 2(1)

y xy y

y x

= = = = =

+ Vi y=2x 2 2(2) 4 3x y = v nghimVy h phng trnh c 1 cp nghim (2;1)

42


43/54

BI TON 3: S DNG PHNG PHP HM SI. Phng phpTa thc hin theo cc bc sau:Bc 1: t iu kin cho 2 biu thc ca h c nghaBc 2: T h ban u chng ta xc nh c 1 phng trnh h qu theo 1 n hoc theo c 2 n, giiphng trnh ny bng phng php hm s bit.Bc 3: Gii h mi nhn c.

II. VD minh ho:

Gii h phng trnh:2 3

2 3

log 3 1 log

log 3 1 log

x y

y x

+ = +

+ = +Gii: iu kin x; y>0. Bin i tng ng h v dng:

( ) ( )( ) ( )

( ) ( )( ) ( )

2 3 2 3

2 3 3 2

log 3 2 1 log log 3 2 1 log

log 3 2 1 log 2 1 log log 3

x y x y

y x x y

+ = + + = + + = + + = +

(I)

( ) ( )2 3 2 3log 3 2log log 3 2logx x y y + + = + + (1)Xt hm s: ( ) ( )2 3log 3 2 logf t t t = + +

Min xc nh ( )0;D = + .o hm ( ) ( )

1 20,

3 ln 2 .ln 3f t t D

t t= + >

+ hm s lun ng bin.

Vy phng trnh (1) c vit di dng: ( ) ( )f x f y x y= =

Khi h (I) tr thmh: ( ) ( )2 3log 3 2 1 log (2)x y

x x

= + = +

(II)

+ Gii (2): ( )2 2

3 3 3 22 1 log log log 2.log3 2 3 4.2 3 4.2x x xx x x+ + = + = + =

( ) 3 3 3 3log 2 log 4 1 log 4 log 423 4. 3 4. 3. 4x x x x x x + = + = + = (3)Xt hm s ( )

3 31 log 4 log 43.g x x x = +Min xc nh ( )0;D = +o hm: ( ) ( ) 3 3log 4 1 log 43 3' 1 log 4 . 3log 4. 0g x x x x D = < hm s lun nghch binVy phng trnh (3) nu c nghim th nghim l duy nhtNhn xt rng nu x=1 l nghim ca phng trnh bi khi :

3 31 log 4 1 log 41 3.1 4 4 4 + = = ng

Khi h (II) tr thnh: 11

x yx y

x

= = = =

Vy h cho c nghim duy nht (1;1)

BI TON 4: S DNG PHNG PHP NH GII. Phng php:II. VD minh ho:

VD1: Gii h phng trnh:( ) ( )2 2

2 2

log log 1 (1)

1(2)

x ye e y x xy

x y

= +

+ Gii: iu kin x; y>0*) Gii (1) ta c nhn xt sau:

43


44/54

- Nu 2 2log logx y x y> > , khi :( )

( )

1

1

0

0

VT

VP

> + >

+ > + > < + + T phng trnh th nht ca h vi vic s dng n ph t=x+y>0, ta c: 2log 1t t=

t 2log 2uu t t= = khi phng trnh c dng:

2

2

log 00 12 1

1 log 1 2Bernoulliu

tu x yu

u t x y

== + = = + = = + =

+ Vi x+y=1 h c dng: ( )3

1 1 1 0; 1

log 1 0 1 1 0 1; 0

x y x y x y x y

xy xy xy x y

+ = + = + = = = + = + = = = =

+ Vi x+y=2 h c dng: ( )42 2 2

log 1 1 1 4 3x y x y x y

xy xy xy+ = + = + = + = + = =

Khi x; y l nghim ca phng trnh: 2 2 3 0t t + = v nghimVy h c 2 cp nghim (0;1) v (1;0)

44


45/54

PHNG TRNH M1) 6224 241 +=+ +++ xxx

2) 0273.43 5284 =+ ++ xx

3) 26.52.93.4x

xx =4) xxx 6242.33.8 +=+

5) ( ) 77.0.61007 2

+=x

x

x

6) 13250125 +=+ xxx

7) 623.233.4 212 ++=++ + xxxx xxx

8) 5008.51

=x

x

x

9) 7503333 4321 =++ + xxxx

10) 3421 5353.7 ++++ = xxxx

11) 09.66.134.6 =+ xxx

12) 1284 = xx

13) 1105.35 1212 = + xx

14)xxx

6.59.24.3 =+15) 0273.43 582 =+ ++ xx

16) 3421 5353.7 ++++ = xxxx

17) 04.66.139.61

.611

=++

xxx

18) ( ) ( ) ( )3210101

32321212 22

=++

+ xxxx

19) 02525 21 =++ + xxxx

20) 5332 242 + = xxx

21) 1223

2

1

3229 ++= x

xxx

22) ( ) 329log2 =+xx

23) ( ) ( )( ) ( )3243234732 +=+++ xx

24) xxx 9.21525 =+25) 22 2.10164 =+ xx

26) 022.92 221222

=+ +++ xxxx

27) ( ) 12

12

2

12.62

13

3 =+ xxxx

28) xx

231 2 =+29) 1282

=

x

30) 0624 =+ xx

31) 055.625 31 =+ +xX

32) 073.59 =++ xx

33) 0543.259 = xx

34) 3033 22 =+ + xx

35)( )

093.82312

=++ xx

36) xxxx 3223 7.955.97 +=+37) 033.369 31

22

=+ xx

38) 0639 1122

= ++ xx

39) 123

694 ++

=+ xxx

40) xxxx 3.25.235 22 ++=41) 211

2222

2332 + = xxxx

42) xxx = 21 105

15.2

43) ( ) ( )3

2531653+

=++

xxx

44) xxx 36.281.216.3 =+45) ( ) 2

log

122222

2

xx

xxlo

+=

++

46) 8444242 22 +=+ xxxxx

47) 3loglog29log 222 3. xxx x =

48) 68.3 2 =+xx

x

49) 052.2 82 log3log =+ xx xx50) 5log3log 22 xxx =+

51) ( ) ( ) ( ) 324log 242 2 = xx x

52) xxx 100lglg10lg 3.264 =

53) 626

1

2

12

3

13 +=

+

x

xx

x

x

x

54) 093.613.73.5 1112 =++ + xxxx

55) 20515.33.12 1 =+ +xxx

56)2

222 4log6log2log 3.24xx

x =57) 2653 +=+ xxx

58) ( ) 21 1222

= xxxx

PHNG TRNH LGARIT

1) ( ) ( ) 8log21log3log 444 =+ xx 42) ( xxx 4846 loglog.2 =+

45


46/54

2) ( 2652log 25 =+ xxx3)

( ) ( ) ( )12lg2021lg110lg5lg =++ xxx

4)

+

+=

8

1lg

2

1

2

1lg

2

1lg

2

1lg xxxx

5) 4lglg3lg 22 = xxx

6)02log3log

31

31 =+

xx

7) ( ) 88

log4log2

2

2

2

1 =+x

x

8) ( ) ( ) 222log64log 255 = xx

9)3

22

4

2

log3log2log4 xxx xxx =+

10) 1log2log2

33 = xx

11) 0log14log40log3

1642

2

=+ xxx xxx

12)

2 3 5

2 3 2 5 3 5

log . log . log

log .log log .log log .log

x x x

x x x x x x= + +

13) ( ) xx

xx 2

3

323 log2

1

3loglog.3log +=

14) ( ) ( 02lglglglg 3 =+ xx15) ( ) ( ) 212log1log 53 =+++ xx

16) ( ) ( ) 01106log3log 222 =+ xx17)

(xxx 4

46 loglog.2 =+

18) 1log 22 = xxx19) 12log.4log 22

2 =xxx20) ( ) 05,4lg1log =+xx

21) 33

log3

log 22 =

+

+

x

x

x

x

22) ( ) ( )2lg46lg 2 ++=+ xxxx23) ( ( ) 01106log3log 222 =+ xx24) 633log33log.log 33 =+xx

25)( ) ( )32log22log 232

2

322= ++ xxxx

26) 112log.loglog.2 332

9 += xxx

27) 013loglog.3 33 = xx

28)

xxxx x 24

2

44

2 log2log2log2log =++

29) ( ) 4lg2lg2

110lg 2 =++ xx

43) ( ) ( ) 61log1log 232

22

32=++++ + xxxx

44) ( ) 34log2log 22 =+ xx

45) ( ) 33logloglog4

3

3

3

13=++ xxx

46) 06

74log2log =+ xx

47) 225log.3logloglog 9535 =+ xx

48) ( ) ( )1log22log

113log 2

3

2 ++=++

xx

x

49) ( ) ( )32log44log1

2

12 =++xx x

50) xxxx 7272 log.log2log2log +=+51)

1log1log.1log 2202

52

4 =+ xxxxxx

52) ( ) 43.59log 2 =+ xx

53) [ ] 1323.49log1

+=+

xxx

x

54) ( 169loglog 3 =xx55) ( ) ( ) 1122log42log 22 +=+ xx x56) ( ) 16log1log 12 +=+ xx

57) ( ) ( ) 2loglog 12222 22 xx xx +=++

58) ( ) ( )1log1log2

1

2

2 = xx

59) ( )( ) 1logloglog 232 =x60)

61) ( ) ( ) ( ) ( ) 01lg.1241lg1 22222 =+++ xxxx62) ( ) ( ) ( ) 0621log51log 3

23 =++++ xxxx

63)

( ( ( 1log1log.1log 262322 =+ xxxxxx64) 5logloglog

3

8

16

14 =++ xxx

65) ( ) ( ) 155log.15log 1255 = +xx66) 225log.3logloglog 9535 =+ xx

67) ( ) ( ) 0226log8log 39 =+++ xx

68)4log.27log.

9

2

+=xxx

x

69) ( ) 944log2log 232

3 =++++ xxx

70)( ) ( ) 02144log156log 231221 =++ xxxx xx

71) ( ) ( ) ( ) ( ) 01lg1241lg1 22222 =+++ xxxx72)

( ) ( ) ( )2log22log5log1log25

15

5

1

2

5 +=++ xxx

73) ( ) ( ) ( ) ( ) 0161log141log2 323 =+++++ xxxx

46


47/54

30) ( ) ( ) xxx x 2221log2 log1log233

2 +=+

31) ( ) ( ) ( ) ( ) 162log242log3 323 =+++++ xxxx

32) ( ) ( )2

1213log 23 =++ xxx

33)154

22

2

22

3log81log4log

36log =+ xx

34) ( ) 212log2

1 =++ xx35)36) ( ) 062log1log 2

2

2 =++ xxxx37) 33loglog.4 9 =+ xx

38) ( 13log6log 22 = xx39)

3log3127log23log 22

2

2

2 +=+++++ xxxx

40) ( ( ) 0log211 22 =++ xxxx

41) ( ) ( ) 155log.15log1

255 = +xx

74) ( ) ( ) ( )3

4

1

3

4

1

2

4

1 6log4log32log2

3 ++=+ xxx

75) xx

xx

2

3

323 log2

1

3loglog.

3log +=

76)( ) ( ) 421236log4129log 252273 =+++++ ++ xxxx xx

77) ( ) xxxxxxxx 2325log325log. 226

12

62 +=

78) 3logloglog.log 23

332 += xxxx

79)( ) ( )

( )

2

2 1

2

1 2

3 .log 1 2 log 2

3 .log 2 2log 1

x

x

x x

x x

+

= +

80) ( ) ( ) xxx

xxx 2772

2 log3log22

3loglog

++=++

BT PHNG TRNH M1) 922 7 + xx

2) 123

13

3

11

12

=

+

+

xx

3) 4loglog .3416 aa xx +4) ( ) ( ) xxxxxx ++++ +++ xxxx

6) ( ) 13.43 2242

+ xx x

7) 8log.2164 41 +

x

xx

9) 022

1212

32

12 +

++

x

x

10) xxx111

9.46.54.9


48/54

1) 21

18log

2

2 ++

x

xx

2) ( ) ( ) 224log12log 32 +++ xx

3) ( ) 123log2

2

1 + xx

4) ( ) ( )243log1243log 2329 ++>+++ xxxx

5)3

16

5log 3 x

x

x

6)2

1

1

12log 4 ++ xxxx

15) 11

32log 3

x

x

52)2

1122log 2

12 xx

54)( ) ( )

043

1log1log2

3

3

2

2 >

++xx

xx

55) ( ) 22lglg

23lg 2 >+

+x

xx

56) 216_185log2

3>+ xx

x

57) 316log64log 22 + xx

58) 0loglog2

4

1

2

2

1 ++ 2log1244log22

1

2

2

61) ( ) 2log2log 12 ++ xxx62)

( ) ( )1log.112

1log1log.2

5

15225

xx

x

63) ( ) ( )232log1232log 2224 ++>+++ xxxx

64) xx

xx 2

2

122

32

2

1

4

2 log432

log98

loglog + xxx

66) ( ) ( )73log219log1

2

11

2

1 +>+ xx

a. ( ) + 28log x 4x 3 1b. 2

1 4

3

log log x 5 0

48


49/54

27) ( 12log 2 >+ xxx28)

( ) ( )xxx

xx xlog22421412

1272 22 +

++

29) ( ) 2385log 2 >+ xxx30) =31) ( ) 4316 13log.13log

4

14

x

x

32) ( 015log 3,0 >++ xx

33) ( )3log2

12log65log

3

1

3

12

3 +>++ xxxx

34)( ) ( )

0352

114log114log

2

32

11

22

5 >

++

xx

xxx

35) ( )2

3

2

9

4

1loglog

xx

36)2

1

2

54log 2

x

xx

37) ( ) ( )32

1

2

1 21log1log2

1xx >

38) 1log2

1log

2

32

34 > xx

39)( )

014log

5

2

x

x

40) ( )22log1log2

2

2

+

42)( )

082

1log

2

2

1

44) ( 164loglog 2 xx45) xxxx 5353 log.logloglog 5 xlog 3x 4.log 5 1

m. +

+

2

3 2

x 4x 3log 0

x x 5

n.+ >1 3

2

log x log x 1

o. ( ) + 23x xlog 3 x 1

q.+

+ 223x

x 1

5log x x 1 02

r. + > +

x 6 23

x 1log log 0

x 2

s. + 22 2log x log x 0

t. > x x 216

1log 2.log 2

log x 6

u. + 23 3 3log x 4log x 9 2log x 3

v. ( )+ < 2 41 2 162

log x 4log x 2 4 log x

49


50/54

H PHNG TRNH M LGA

I. H phng trnh m.

1)

=

=

+

13

35

4

yx

yx

xy

xy

2)

( ) ( )

+==

+

yxyx

x

y

y

x

33 l o g1l o g3 24

3)

=+

=

423

99.3

1 21

y

x

x

yx

y

x

y

4)( )

=

=

y

y

y

x

x

y

y

x

12

3

5

2

3.33

2.225)

=+

=

2l gl g

1

22yx

x y

6)

=

=

723

723

2

2

y

x

yx

7)

=

+=2

l o g.2

l o gl o g

43

xx y

yy

yx

xy

8)

=

=++ 132

63.22.3

11 yx

yx

9)

=

=+

33

3

3.55

5

yx

yx

yx

yx

10)

=+

+

=+

y

yy

x

xx

x

22

24

452

1

23

11)( )

( )

=+

=+

068

13.4

4

4

4

yx

xy

yx

yx

12)

( ) ( )

=

=3l g4l g

l gl g

34

43

yx

yx

13)

( )

==

1l o g.

3l o g

4

2

5l o g

xyy

x

y

y

xxy

14)

+=++

=+ ++

113

2.322

2

3213

xx yx

xyyx

15)

( ) ( ) ( )

=++=++

++3

81l o g2l o g

1 42

21 xy

yx yx

yx

16)( ) ( )( ) ( )

=+++

=++

421223

421223

xy

yx

17)+

=

=

x y

3x 2y 3

4 128

5 118)

+

= =

2

x y

(x y) 1

5 125

4 1

19) =

=

2x y

x y

3 2 77

3 2 720)

+ =

+ =

x y2 2 12

x y 5

50


51/54

II.H phng trnh lgarit.

1)

( ) ( )

=+

+=

1 6

2l o gl o g

33

22

yx

x yxyyx

2)

( ) ( )

=

=3l g4l g

l gl g

34

43

yx

yx

3)

( )

=+

+=+

1l o g

3l o g2l o gl o g

7

222

yx

yx

4)

=

=+

8

5l o gl o g2

x y

yxxy

5)

=

=+

1l o gl o g

4

44

l o gl o g 88

yx

yxxy

6)

=

=+

1l o gl o g

4

44

l o gl o g 88

yx

yxxy

7)( )

=+

=+

1l o g

433.11

3

yx

x

xx

y

8)

( ) ( )

( ) ( )

=

=

xx

yx

4224

2442

l o gl o gl o gl o g

l o gl o gl o gl o g

15)( )

( )

=+

+=

0l g.l gl g

l gl gl g

2

222

yxyx

x yyx

16)( ) ( )

( ) ( )

=++

=++++

+

+

14l o g5l o g

612l o g22l o g.2

21

2

21

xy

xxyxx y

yx

yx

17)

( ) ( ) ( )

( ) ( )

=+++

+=++

1l o g4224l o g1l o g

3l o g12l o gl o g

4

2

44

44

22

4

y

xxyyx y

yxxyx

18)

( ) ( )

=+

+=

133

24

22

2l o gl o g 33

yxyx

x yx y

19)

( ) ( )

+==

+

yxyx

y

x

x

y

33 l o g1l o g3 24

20)

( )

=+

=+

yyy

yx

x8 13.1 22

3l o g

2

3

21)

=

=

2l o g

4l o g

2

1

2

y

x

x y

51


52/54

9)

( ) ( ) ( )

=++

=++

++3

81l o g2l o g

1 42

21 xy

yx yx

yx

10) ( )( )

=+=+

223l o g223l o g

xy

yx

y

x

11)

( ) ( )

( ) ( )

=++

=+++

453l o g.53l o g

453l o g53l o g

xyyx

xyyx

yx

yx

12)

=+

=

02

0l o gl o g21

23

32

3

yyx

yx

13)

=

=+

1l o gl o g

22

33

l o gl o g 33

xy

yxxy

14)

=

=

9l o gl o g.5

8l o gl o g.5

4

3

2

242

yx

yx

22)

( )

=++

=++

01422

22

32

222

12

2

xyxxyx

x yyx

x

23.+ =

+ =

2 2

lgx lgy 1

x y 29

24.+ = +

+ =

3 3 3log x log y 1 log 2

x y 5

25.( )

( ) ( )

+ = +

+ =

2 2lg x y 1 3lg2

lg x y lg x y lg3

26. =

+ =

4 2

2 2

log x log y 0

x 5y 4 0

27.( ) ( )

+ = + = +

x yy x

3 3

4 32

log x y 1 log x y28.

= = + y

2x y

2log x

log xy log x

y 4y 3

1) (A07) 3 13

2 log (4 3) log (2 3) 2x x + + ( 3 34

x< )

2) (D305) 22 43

log log ( 4 4 )2

x

x x

x

++ + + >

(x>2 4x < )

3) (D206) 2 4 2 12(log 1) log log 04x x+ + = ( x=2 x= )

4) (B203) 0,5 0,25 2log 2 log ( 1) log 6x x+ +

(x 3)

5) 2 4 12

log 2 log 5 log 8 0x x- + + + = 3 176;3;2

x - -

6)2

2 4 1

2

log ( 2) log ( 5) log 8 0x x+ + - + = 3 176;2

x x = =

7) 84 221 1

log ( 3) log ( 1) log (4 )2 4

x x x+ + = (x = 3 x= 3+ 12 )

52


53/54

8)2

9 1 3

3

log ( 3) log 2 log 2 1x x+ - - - < ( 4; 3) ( 3; 1) (0;2) (2;3)- - - -

9) 33log 1 log 12 .5 400x x+ + < ( -10 < x < 8 )

10) (B104)12 4 16

42

x x

x

+ >

(x 4)

11) (A104)2

2

4

log [log ( 2 )] 0x x x

+ < (x >1 x< - 4)

12) (B204) 3log log 3xx > ( x>3 1/3


54/54

38)(A1-08) 1 23

2 3log (log ) 01

+ +

x

xx < 1

39)(A1-08) sin( )4 tanp-

=x

e xx= /4 + k

40)(A2-08)3

1 63 log (9 )log

+ = -x

x

x xx = 2

41)(B1-08) 122

2log (2 2) log (9 1) 1+ + - =x x x= 1; x = 32

42)(B2-08) 2 1 2 13 2 5.6 0+ +- - x x x 23

1log2

x

43)(D1-08)2 22 4 2 2 12 16.2 2 0- - - -- - x x x x 1 3 1- +x

44)(D1-07)2 2

1 22

1 1log 2 3 1 log ( 1)2 2

- + + - x x x x x

48)(A2-07) 4 22 11 1

log ( 1) log 2log 4 2+- + = + +xx x

x =52

49)(B1-07) 23 3log ( 1) log (2 1) 2- + - =x x x=2

50)(B2-07) 3 93

4(2 log )log 3 11 log

+ - =-

xx

xx= 1

3; x= 81

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