Một Số Bt Giải Pt Phổ Biến

7/23/2019 Mt S Bt Gii Pt Ph Bin

1/13

M T S B I TO N GI I PHNG TR NH

PH BIN NHT TRONG THI I HC

CM NANG CHO MA THI

NGUY N H U BI N

https://www.facebook.com/ng.huubienEmail: [email protected]

(N THI THPT QUC GIA)


2/13

MT S DNG PHNG TRNH THNG XUT HIN TRONG THI I HC

Trang 1NGUYN HU BIN -https://www.facebook.com/groups/nguyenhuubien

Bi 1:Gii phng trnh ( )2

1 1x

2 x x8 2

2 2

1 79 .log x x 2 3 .log 2 x 0

2 4

+

+ + =

Hng dn+ Khi gp phng trnh mhoc log, trc ht ta bin i theo csnhnht (y lcs3), bin i phng trnh cho, ta c:

( )

( )

( )

( )

2

2

2

2

2

1 1x

2 x x8 22 2

12x 1

242 2x x

1

42

2 22x 1 x x

1x x 2x 124

2 2

1x x

4

1 79 .log x x 2 3 .log 2 x 02 4

1 73 .log x x 2 .log 2x 1 0

43

3 1 7.log x x 2 .log 2x 1 0

43 3

73 .log x x 2 3 .log 2x 1 0

4

3 .log

+

+

+

+ + =

+ + =

+ + =

+ + =

2x 122 21 7 7x x 3 .log 2x 14 4 4

(1) + + = +

+ Xt hm s t 27

f (t) 3 .log t4

= +

, ta thy

2

2 1 1x x x 0t 04 2

2x 1 0

+ =

t t

2

7 1f '(t) 3 .ln 3.log t 3 . 0, t 0

4 t.ln 2

= + + >

f(t) l hm ng bin

+ T(1) ( )2 21 1

f x x f 2x 1 x x 2x 14 4

+ = + =

(2)

+ Xt 2 TH: 2x 1 0;2x 1 0 < bdu GTT(2), gii PT (2) ta c S:1 5

x ;2 2

=

Bi 2:Gii phng trnh 3 2 3x x 19x 16 3x x 1+ = + Hng dn+ K: x 1 + Nhn thy biu thc trong cn VP l: ( ) ( )3 2x 1 x 1 x x 1+ = + + nn rt c thy l cscho ta phn tch VT ca phng trnh, tht vy:

( ) ( ) ( )

( )( ) ( ) ( )

3 2 3 2

3 2 2 2

x x 19x 16 x 1 x x 1 18 x 1

x x 19x 16 x 1 x x 1 x x 1 18 x 1

+ = + + + +

+ = + + + + +

+ Phng trnh cho trthnh:

( ) ( ) ( ) ( ) ( ) ( )( )2 2 2x 1 x x 1 x x 1 18 x 1 3 x 1 1 x 1 x x 1 + + + + + = + + + (1)

+ t n ph2

a x 1 0

b x x 1 0

= +

= +

thay vo (1) ta c: ( )2 2 2 2 2a b b 18a 3 a 1 ab+ = (2)

+ Phng trnh (2) mun gii c th chcn cch phn tch a thc thnh nhn t, cngvic phn tch ny khng phi ddng, chng ta sdng mo sau phn tch:


3/13



- Trc ht ta bin i a (2) vphng trnh bc 2 i vi bin l b:

( ) ( ) ( )2 2 2 2 2 2 2 3 2a b b 18a 3 a 1 ab a 1 b 3a 3a b 18a 0(3)+ = + =

- Phng trnh (3) tnh c ( ) ( )( ) ( )2 2

3 2 2 2 23a 3a 4 a 1 18a ... 9a a 3 = + + = = +

phng trnh (3) c 2 nghim

( )

( )

3 2

2

23 2

2 2

3a 3a 3a a 3b 3a

b 3a 0a 1

a b b 12a 03a 3a 3a a 3 12ab

a 1 a 1

+ + = =

=+

+ + = + = = + +

+ Vy (2) : ( ) ( )( )2 2 2 2 2 2a b b 18a 3 a 1 ab b 3a a b b 12a 0+ = + + =

+ Do 2a b b 12a 0 a b 0+ + = = = (do a, b 0 ) 2x 0

x 1 x x 1x 2

= + = +

=

+ Thli vo PT cho thy x = 0; x = 2 khng l nghim (loi)

+ Vy chcn khnng b 3a= 2x x 1 3 x 1 x 5 33 + = + =

Bi 3:Gii phng trnh

( )( )4 3 2 24x 13x 9x 16 2x 3x x 3 x 1 8 + + + + + =

Hng dn

+ K4 3 24x 12x 9x 16 0

x 1

+ +

+ Ta bin i PT ( ) ( ) ( )2

2 22x 3x 16 2x 3x x 1 4 x 1 8 + + + =

+ t2a 2x 3x

b x 1

=

= thay vo phng trnh ta c:

( ) ( )

( )

( )

2 2

2

2

2

2

22

2 2

2 2

a 16 a b 4 b 88

b 4 ba 16 a

8 a 16 ab 4 b

16

a 16 ab 4 b

2

2 b 4 2b a 16 a

2b 16 2b a 16 a (1)

+ + + =

+ + =

+

+ +

+ + =

+ + + + =

+ + = + +

+ + = + +

+ Xt hm s 22

tf (t) t 16 t, t R f '(t) 1 0, t R f (t)

t 16(*)= + + = + >

+

l hm

ng bin.

Vy t(1) f (2b) f (a) 2b a = =( ) ( )

2

22

2

2x 3x 02 x 1 2x 3x (I)

4 x 1 2x 3x

=

=

+ Cc bn tgii h(I) x 2 =


4/13



(Ch : Ta gii thch cho f '(t) 0> bng phng php phn chng nhsau:

Gis 2 22 2

t 0f '(t) 0 t t 16 0 t 16 t

t 16 t

< + + < + <

+


5/13



( )

( )2

3 3

4 27 9 8(6 2 )( 3)(4 27) 0

4 10 216 4 9 37 9 37

x xx x

xx x

+ + + + + =

+ +

- TH1 3 0 3x x+ = = (TMPT)

- TH 2. 3x

( )2

3 3

36 164 27 0

4 10 216 4 9 37 9 37

x

xx x

+ + =

+ +

( )2

3

36 164 27 0

4 10 212 9 37 2x

xx

+ + =+ +

+ Do 5x nn36 16

4.5 27 012 4

VT + + = . ng thc xy ra 5x =

Vy phng trnh c 2 nghim l 3 v 5

Bi 7:Gii phng trnh ( )24 2 1

log 2 2.8 3.2 1 (*)

2.16 2.4 1

+= +

+

x xx x x

x x.

Hng dn:

( )

2 2

2 2

'

2

log (4 2 1) log (2.16 2.4 1) 2.16 3.4 2

log (4 2 1) 4 2 1 log (2.16 2.4 1) 2.16 2.4 1

1Xt ( ) log , 0, ( ) 1 0, 0 ng bin trn 0;

ln 2

+ + = +

+ + + = + + +

= + > = + > > +

x x x x x x x

x x x x x x x x

f t t t t f t t ft

( )2 4 2

4 2 3

2

* (4 2 1) (2.16 2.4 1)

4 2 1 2.16 2.4 1 2 2 2.2 2.2

2.2 3.2 2 0 2.2 3.2 1 0

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

2 2

2 10

1 3 2 1 3

log22

1 32 (loi)

2

+ = +

+ = + =

+ = + =

+ + + =

=

= + = + =

+=

x x x x

x x x x x x x x

x x x x x

x x x

x

x

x

f f

x

x

Bi 8:Gii phng trnh: .16212244 2 +=++ xxxx

Hng dn: iu kin xc nh: .4x Vi iu kin , phng trnh cho tng

ng vi:

( )2

2x 4 x 4 (x 4) (x 4) 12 2 x 16 x 4 x 4 x 4 x 4 12+ + = + + + + + = + +


6/13



+ t t = 44 ++ xx , t > 0 ta c

=

==

4

)(3012

2

t

loaittt

+ Vi t = 4 , ta c

+=

==++

22

2

166416

84816444

xxx

xxxxx

.55

84

=

=

xx

x

Vy nghim ca phng trnh l x = 5.

Bi 9:Gii phng trnh 3 2 2 332 10 17 8 2 5 + + = x x x x x x

Hng dn: Nhn xt: x = 0 khng tha phng trnh cho

+ Chia hai vca phng trnh cho x3, ta c: 32 3 210 17 8 5

2 2 1x x x x

+ + =

+ t ( )1

0t tx

= , phng trnh trthnh: 2 3 232 10 17 8 2 5 1t t t t + + =

( ) ( )

( )

33 2 23 32 1 2 2 1 5 1 2 5 1t t t t + = + ( )

( )

232 1 5 1f t f t =

+ Vi ( ) 3 2 ,= + f t t t t R . Ta c: ( ) 2' 3 2 0,= + > f t t t R nn f ng bin trn R , v vy:

( ) ( )2 23 32 1 5 1 2 1 5 1f t f t t t = =

( )3 2 3 2

0

17 972 1 5 1 8 17 6 0

16

17 97

16

t

t t t t t t

t

=

+ = + = =

=

(loai)

(nhan)

(nhan)

+17 97 17 97

16 12t x

+ = =

+17 97 17 97

16 12t x

+ += =

Vy phng trnh cho c 2 nghim:17 97

12x

=

Bi 10:Gii phng trnh: 3

1 1 3x 2x (x R). (1)2+ + =

Hng dn:iu kin:

1

3x (*).

+ t 1 1 3x2

y = + 1 3x 2y + = (2), (1) trthnh 1 3 2y x+ = (3)


7/13



+ T(2) v (3) ta c h1 3x 2

1 3 2x

y

y

+ =

+ =

2

2

0, 0

1 3x 4 (4)

1 3 4x (5)

x y

y

y

+ =

+ =

.

+ Trvvi v (4) v (5) ta c 2 23( ) 4( )x y x y = ( )(3 4 4 ) 0x y x y y x + + = = (v

0, 0x y ). Thy = x vo (5) ta c 21

4x 3x 1 0 1

4

x

x

= = =

.

+ Kt hp vi 0x , suy ra phng trnh (1) c nghim duy nht x = 1.

Bi 11:Gii phng trnh 41292 22 +=++++ xxxxx

Hng dn:

Ta thy: )4(2)12()92( 22 +=+++ xxxxx + x = -4 khng phi l nghim+ Xt 4x , trc cn thc ta c:

2129241292

8222

22 =++++=+++

+

xxxxxxxxx

x

+ Vy ta c h

=

=

+=++

+=++++

=+++

7

8

0

692241292

212922

22

22

x

x

xxxxxxxx

xxxx

+ Thli vo phng trnh ta thy tha mn.

Vy phng trnh c 2 nghim: x = 0, v x =7

8

Bi 12:Gii phng trnh 0123)13(1 32 =++++ xxxxx

H

ng dn: iu kin: 1

x

=++

=+

=+++

=+++++

=+++++

0131

01

0)131)(1(

0)131()131(1

0311131

2

2

22

32

xx

xx

xxxx

xxxxxx

xxxxxxxx

Kt hp vi iu kin ta thy phng trnh v nghim.

Bi 13:Gii phng trnh 11 33 2 =+ xxx

Hng dn: iu kin: 3 2x + Nhn thy x = 3 l nghim ca phng trnh , nn ta bin i phng trnh

52

)93)(3(

412)1(

31)3(52321

3

2

3 23 22

33 2

+

++=

++

++=+

x

xxx

xx

xxxxx

+ Ta chng minh :52

932

3)11(

31

412)1(

31

3

2

23 23 3 222 +

++


8/13



Bi 14:Gii phng trnh: 31

21

=+

+ x

x

x

x

Hng dn:iu kin: 01 >< xx , t

2

1

10,

1

tx

xt

x

xt =

+>

+=

3 2

2

1t1

2t 3 2t 3t 1 0 2tt 1

= = + =

=

+ Ly3

4

2

11

2

1==

+= x

x

xt

Bi 15:Gii phng trnh 82315 22 ++=+ xxx

Hng dn:

1

0)338

1

415

1)(1(

38

1)1(3

415

1

3833415

222

2

2

2

22

=

=

++

+

++

+

++

+=

++

++=+

x

x

x

x

xx

x

xx

x

x

xxx

Bi 16:Gii phng trnh : 1221)14( 22 ++=+ xxxx

Hng dn:t 1)0(1 222 =>+= txtxt

Phng trnh trthnh 12)1(2)14( 2 ++= xttx 012)14(2 2 =+ xtxt

Ta c: 2)34( = x

3

4

3

4

0

1)12(

)(4

3

12

2

1

22

2

=

=

=

=

=

=

= x

x

x

xx

Lx

xt

t

Bi 17: )(01713264 2 Rxxxxx =++

Hng dn: iu kin: 64 x

0)15132()16(

)16)(16(

)14(

)14)(14(

015132)16()14(1713264

2

22

=++

++

+

+

=+++=+

xxx

xx

x

xx

xxxxxxxx

x 5 5 x(x 5)(2x 3) 0

x 4 1 6 x 1

1 1(x 5)( (2x 3)) 0x 4 1 6 x 1

x 5

1 1(2x 3) 0

x 4 1 6 x 1

+ =

+ +

= + +

=

= + +

+ Ta c: )1(3216

1

14

10)32(

16

1

14

1=

+

+=

+

+x

xxx

xx


9/13



114

1

16

1

14

1

++++=++++=+++ 03

4

3)

2

1(3

4

3

4

1

2

1.23 2222222 yyxyyxyxxyyx

Bi 19: 4384212)47(23)14( 2 ++=+ xxxxxx (vi x )R

Hng dn: iu kin:2

3

2

1 x

t )1(2)0(12

)0(23 22=+

=

= babxb

axa

+ V baxxxxaxbx .23.12384;1247;214 222 ==++==

+ Tphng trnh cho ta c )2(42)12()12( 22 +=+++ abbaab

+ Kt hp (1) v (2) ta c hphng trnh

+=+++

=+

42)12()12(

2

22

22

abbaab

ba

+ y l hphng trnh i xng loi I vi n a, b.+ Gii hphng trnh trn ta c nghim l : a = b =1

+ Vi a = b = 1 khi ta c: 1112

123 =

=

= xx

x (tha mn iu kin)

Vy nghim ca phng trnh cho l x = 1

Bi 20:Gii phng trnh: 83)23(2 32 +=+ xxx

Hng dn: iu kin 2x

+ Bin i phng trnh vdng: )42)(2(3)2(2)422(2 22 ++=++ xxxxxx

+ Do 2=x khng phi l nghim ca phng trnh nn chia chai vcho (x+2) ta c

02

2

423

2

)42(2 22=

+

+

+

+

x

xx

x

xx

+ t )0(2

422=

+

+tt

x

xxta c

=

==

2

2

1

0232 2

t

ttt (loi)

Vi t = 2 133046844242

42 222

==+=+=+

+ xxxxxx

x

xx

Bi 21:Gii phng trnh 3)6)(3(63 =+++ xxxx


10/13



Hng dn: t 2

9)6)(3(63

2

=+++= t

xxxxt

+ Phng trnh trthnh

=

==

3

13

2

92

t

ttt

+ Vi t = -1 163 =++ xx (v nghim)

+ Vi t = 3

=

=

=++ 6

3363

x

xxx

Cch khc:

Bi ton c thgii bng cch a vhphng trnh vi 3+= xu v xv = 6

Bi 22:Gii phng trnh: )1(055)3(66138 223 =++++ xxxxxx

Hng dn:iu kin: 0552 + xx

+ T(1) 055)3(6)25)(3( 22 =++ xxxxxx

=++

=

)2(055625

)(3

22 xxxx

loaix

+ Gii (2): t ,552 txx =+ iu kin 0t

(2)

=

==+

7

10762

t

ttt , t = -7 (loi), t = 1 (tha mn)

+ Vi t =1

=

==+

4

11552

x

xxx (tha mn iu kin)

Vy phng trnh c hai nghim x = 1 v x = 4

Bi 23: 16924422 2 +=++ xxx

Hng dn:

Bnh phng hai v, lc phng trnh :

169)2(16)4(216)42(4 22 +=+++ xxxx

xxxx 8)4(216)4(8 222 +=+

+ t t = )0()4(2 2 tx

+ Phng trnh trthnh : 08164 22 =+ xxtt

+ Gii phng trnh trn vi n t ta tm c

=

=

42

2

2

1

xt

xt

+ Do 2x nn 02


11/13



+ Khi phng trnh cho trthnh )1(03638 32 =+++ xxxx

+ Nhn thy x = 0 khng tha mn nn (1) 03

63

8 =+++x

xx

x

+ t )12(3 4

=+ ttx

x , ta c

=

==+

4

20862

t

ttt (tha mn)

+ Vi t = 2

=

=31

xx

+ Vi t = 4

=

+=

618

618

x

x

Bi 25: 1635223132 2 +++=+++ xxxxx

Hng dn:

Bin i phng trnh vdng:20)1)(32(2132132 ++++++=+++ xxxxxx

+ t )0(132 +++= txxt

+ Khi )1)(32(21322 ++++++= xxxxt + Ta c phng trnh: 5;40202 === tttt + Ly t = 5 tha mn, thay t = 5 c x = 3

Bi 26:Gii phng trnh 0192

29

22 =

+

+

x

x

x

Hng dn: iu kin: 0x

+ Phng trnh cho tng ng 0392

292

22

2

=

+

++

x

x

x

x

+ t 092 2

+

=

x

xt . Phng trnh trthnh

=

=

=+

2

1

1

03212 t

t

tt

+ Vi2

23

924

0922

2

122

2=

+=

+= ,053)(' 2 , suy ra hm strn ng bin trn R

+ M phng trnh (*) c dng: )92()5( 3 = xfxf

=

=

=+

=+=

2

511

4

0)2911)(4(

0116731592)5(

2

233

x

x

xxx

xxxxx

Bi 32:Gii phng trnh 612243 =++ xx

Hng dn:iu kin: 12x

+ t xvxu =+= 12;243 vi 0v

+ Ta c

=+

=+

=+

=+

=+

=+

=+

=+

0)12(

6

012

6

36)6(

6

36

6

2232323uuu

vu

uuu

vu

uu

vu

vu

vu


13/13


=

=

=

=

=

=

=

=

=

3

88

24

3

3

10

4

6

0

x

x

x

v

u

v

u

v

u

(tha mn)

Bi 33:Gii phng trnh 2132727122 233 += xxxx

Hng dn:

)13(2)13(122)12( 33 +=+ xxxx

+ Xt hm 32)( aaaf += ta suy ra 0072727)13(12 233 ==+= xxxxxx

p s: x = 0

Bi 34: Gii phng trnh 3 233 23 1 xxxxx ++=++

Hng dn:

+ Xt x = 0 khng phi l nghim ca phng trnh+ Vi 0x , ta chia hai vcho x:

10)1(11

111 33333 ==

+++=+

+xx

x

xxx

x

x

(lm bi tp kt hp vi hc trong vghi)

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