tom luoc huong giai pt vo ti

8/7/2019 Tom Luoc Huong Giai Pt Vo Ti

1/8

Phng trnh v t Trang 1

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TM LC CC HNG XLPHNG TRNH V T

Phng trnh v t l loi phng trnh trong c cha mt hoc vi biu thc nm trong du cn thc.Phng php chung gii loi phng trnh ny l: bin i tng ng a v dng phng trnh v t cbn, hoc i bin i sa v phng trnh a thc theo n s mi (hoc a v phng trnh a thc theo

n s mi m trong h s ca phng trnh mi cn cha n s c), hoc chuyn thnh mt h phng trnhgm 2 n s mi ...Xin gii thiu vi cc em hc sinh 5 dng phng trnh v t cbn sau:

Ta c th p dng cc kt qu sau y :

( ) ( )( )

( ) ( )

0g xf x g x

f x g x

=

=. (Khng cn iu kin ( ) 0f x )

( ) ( )( )

( ) ( )2

0g xf x g x

f x g x

=

= . (Khng cn iu kin ( ) 0f x )

( ) ( ) ( )

( )

( )

( ) ( ) ( ) ( ) ( )

0

0

2 .

f x

f x g x h x g x

f x g x f x g x h x

+ =

+ + =

(Khng cn iu kin ( ) 0h x )

Lu :Hin nhin trong cc cng thc trn ta phi thm vo iu kin ( ) ( ) ( ), ,f x g x h x c ngha.

V d 1: Gii phng trnh :

2

3 4 1 .x x x+ + =

Gii :

Phng trnh cho tng ng vi: 2 3 4 1x x x+ =

2 2

1 0

3 4 2 1

x

x x x x

+ = +

11

5 5

xx

x

=

=

Vy x = 1 l nghim ca phng trnh cho.

V d 2: Gii phng trnh : 4 1 1 2x x x+ =

Gii :Phng trnh cho tng ng vi : 4 1 2 1x x x+ = +

1 2 0

1 0

x

x

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

( )( )

+=

12121

12

1

xxx

x

x

DNG 1 : SDNG PHP BIN I TNG NG


2/8


3/8


4/8



Mt s trng hp khi ta chuyn phng trnh v t v phng trnh a thc theo mt n ph s cho raphng trnh a thc c bc kh cao. Khi , ta c th nghn dng 3 ny, tc l ta ch cn dng n ph khi cc du cn thc, cn h s ca phng trnh theo n ph ny vn c th cn cha x. Ta hy xt qua 2 v dsau y:

V d 1: Gii phng trnh : ( ) 2 24 1 4 1 8 2 1.x x x x + = + + Gii :

t : ( )2 2 2 2 24 1, 1 4 1 8 2 1 2 2 1t x t x t x x t x= + = + + = +

Khi phng trnh cho c dng :

( ) 24 1 2 2 1x t t x = +

( )22 1 4 2 1 0t x t x + + =

4 1 4 32 1

44 1 4 3 1

4 2

x xt x

x xt

+ = =

+

= =

* Vi 2 1t x= , ta c : 24 1 2 1x x+ =

2 2

12 1 0

24 1 4 4 1

0

x xx

x x xx

+ = + =

.

Vy phng trnh cho v nghim.

V d 2: Gii phng trnh : ( ) 2 22 1 2 1 2 1x x x x x + =

Gii :

t : 2 2 1t x x= + 2 2 2 22 1 2 1 4x x t x x t x + = =

Khi , phng trnh cho c dng :

( ) 22 1 4x t t x =

( )2 2 1 4 0.t x t x + =

1 1 2

1 1 2

t x x

t x x x

= + + =

= =

Vi 2t= , ta c :2

2 1 2x x+ = 2 1 62 5 0

1 6

xx x

x

= + + =

=

Vi 2t x= , ta c :2 2 1 2x x x+ =

2 2 2

2 0 0.

2 1 4 3 2 1 0

x xx

x x x x x

+ = + =

Vy nghim ca phng trnh cho l : 1 6x = .

Lu :Hin nhin nu bit s ta tnh c khng phi l mt schnh phng, th vic gii phngtrnh tm t theo x l kh vt v. Do vy, nu bc tnh m ra mt skhng phi schnh phng thta nn chn mt hng gii khc.

DNG 3 :I BIN S KHNG HON TON

(loi)


5/8



Ta xt 2 v d sau:

V d 1: Gii phng trnh : ( )( )3 6 3 6 3.x x x x+ + + =

Gii :

iu kin :3 0

3 66 0

xx

x

+

t : 2 23 0

9.6 0

u xu v

v x

= + + =

=

Khi phng trnh cho tng ng vi h phng trnh sau :

( )22 2 9 2 9

. 3 3

u v u v uv

u v u v u v uv

+ = + =

+ = + = +

( )2

9 6 2 9

3

uv uv uv

u v uv

+ + =

+ = +

( )( )

2 0 44 03 13

uv uvuv uvvn

u v u vu v uv

= = =

+ = + = + = +

0 3

0 6

u x

v x

= =

= =

3 6x x = =

Vy phng trnh cho c nghim l : 3 6x x= = V d 2: Gii phng trnh : 3 2 1 1.x x =

Gii :iu kin : 1 0 1x x

t :3

3 22

1.1

u xu v

v x

= + =

=

Khi phng trnh cho tng ng vi h :3 2 1

1

u v

u v

+ =

=

( )23 1 1 0.

1

u u

v u

+ =

=

3 2 2 0

1

u u u

v u

+ =

=

0 1 2

1 0 3

u u u

v v v

= = =

= =

3 3 32 0 2 1 2 2

1 1 1 0 1 3

x x x

x x x

= = =

= = =

2 1 10x x x = = = Vy phng trnh cho c 3 nghim l : 2 1 10x x x= = = .

DNG 4 : CHUYN V MT H GM 2 PHNG TRNH 2 N


6/8



Gi s cn gii phng trnh : ( ) ( )f x g x= . Nu nh ta th cch gii ca cc dng trc u

gp kh khn v khng thi n ch, l lc ta hy suy nghn dng ny.

Nu ta ch ra s tn ti ca mt hng s k thoiu kin :( )

( )

f x k

g x k

hoc

( )

( )

f x k

g x k

.

Khi , phng trnh cho tng ng vi h phng trnh :( )

( )

f x k

g x k

=

=.

Lu :Hng sk ni trn thc cht l GTLN , GTNN ca 2 hm s ( )f x v ( )g x .

Do vy tm ra sk ta c th:+ Sdng btng thc (Csi , Bunhiacopxki...).+ Sdng tnh cht tng, gim ca hm s.

Chng ta s tm hiu phng php ny qua cc v d sau:

V d 1: Gii phng trnh : 22 4 6 11.x x x x + = + Gii :

Cch 1:(Dng btng thc)

Xt phng trnh : 22 4 6 11.x x x x + = + (*)

iu kin : 2 4x

Ap dng bt ng thc Buhiacopxki cho 4 s : 1 , 2 , 1 , 4x x , ta c :

2 4 1 2 1 4x x x x + = +

( ) ( )2 21 1 2 4x x+ +

= 2 (1)Mt khc, ta c :

( )22 6 11 3 2 2 ,x x x x + = + (2)

Do , phng trnh (*) tng ng vi h lm cho du = (1) v (2) xy ra:

2 43

3

x xx

x

= =

=(nhn)

Vy phng trnh cho c nghim l : x = 3.Cch 2: (Phng php hm s, chdng cho hc sinh c kin thc vo hm, cc tr hm s...)

t : ( ) 2 4f x x x= + , ( ) 2 6 11g x x x= +

Ta c : ( )' 1 12 2 2 4

f xx x

=

( ) ( )' 0 4 2 2 , 4f x x x x x= = ( )3 3 2x f = =

Bng bin thin ca ( )f x l :

x 2 3 4 +

( )'f x + + 0 - -

( )f x 2

Theo bang bien thien ta thay :

DNG 5 : SDNG BT NG THC CAUCHY,BUNNHIACOPXKI, HOC HM SNH GI TNG V


7/8



[ ]2 , 4x th ( ) ( )3 2f x f = (1)

Xt hm s ( ) 2 6 11g x x x= + , ta c : ( )' 2 6g x x=

( ) ( )' 0 3 3 2g x x g= = =

Bng bin thin ca ( )g x l :

x 2 3 4 +

( )'g x - 0 +

( )g x

2Theo bang bien thien ta thay :

[ ]2 , 4x th ( ) ( )3 2g x g = (2)Do , phng trnh (*) tng ng vi h :

( )( )

2 3.2

f x xg x

= ==

Vy phng trnh cho c nghim l : 3x = .Nhn xt:Cch gii th2 kh di v rm r, nhng n l phng php rt mnh tm GTLN, GTNN camt hm sbt k. Cch 1 srt kh khn nu gp phng trnh sau:

Gii phng trnh : 2 2316 1 3.x x+ + = Gii :

Phng trnh cho tng ng vi : 2 2316 3 1x x+ =

t :( )

( )

2

23

16

3 1

f x x

g x x

= +

=

Xt hm s : ( ) 2 16f x x= +

( )'2 16

xf x

x =

+

( ) ( )' 0 0 0 4.f x x f = = =

Bng bin thin ca ( )f x l :

x 0 +

( )'f x - 0 +

( )f x 4

Da vo bng bin thin ta thy :

x th ( ) ( )0 4f x f = (1)

Xt hm s: ( ) 233 1g x x

( )( )

'

223

2

3 1

xg x

x

=

( ) ( )' 0 0 0 4.g x x g= = =


8/8



Bng bin thin ca ( )g x l :

x -1 0 1 +

( )'g x + + -

-

( )g x 4

Da vao bang bien thien ta thay:

x th ( ) ( )0 4g x g = (2)

T (1) v (2) ta thy phng trnh cho tng ng h :

( )

( ){

40.

4

f xx

g x

= =

=

Vy phng trnh cho c nghim l : 0.x =

Chc cc em n tp hiu qu. Cho cc em!

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