chuyen de pt, bpt, hpt mu va logarit - chuyên Đề Ôn thi 5 2 2 2 2 .2 .2 2 .2 2 .2 2 23 2 2 2 25...

LUYN THI I HC MN TON Thy Hng Chuyn HM S M V LOGARITH

Hc trc tuyn ti: www.moon.vn Mobile: 0985.074.831 1

1) Khi nim v Ly tha

Ly tha vi s m t nhin: . . ... ,=na a a a a vi n l s t nhin.

Ly tha vi s nguyn m: 1

, =nn

aa

vi n l s t nhin.

Ly tha vi s m hu t: ( )= =m m

n m nna a a vi m, n l s t nhin.

t bit, khi m = 1 ta c 1

.= nna a

2) Cc tnh cht c bn ca Ly tha

Tnh cht 1: 0

1

1,

,

=

=

a a

a a a

Tnh cht 2 (tnh ng bin, nghch bin): 1:

0 1:

> > >

< < > b > 0 th 0

0

> >

<

m mn m n nn n

n n n

n

nn

a a a a a a a a a

ab a b a b

a aa b

b b

V d 1: Rt gn cc biu thc sau :

a) 2 1

2 1.aa

b) 2 44. :a a a

c) ( ) 33a d) 32. 1,3 3 2. :a a a Hng dn gii:

a) ( )2 1

2 12 2 1 2 1 21.a a a a a aa

= = =

.

b)

112

2 4 4 2

. :a

a a a a a aa

= = =

c) ( ) 33 3. 3 3a a a= = d)

2. 1,332. 1,3 3 2 1,3

2

.. :

a aa a a a

a= =

V d 2: n gin cc biu thc :

Ti liu bi ging:

01. M U V LY THA Thy ng Vit Hng



a) ( )

2 2 2 3

22 3

1a b

a b

+

b) ( )( )2 3 2 3 3 3 3

4 3 3

1a a a a

a a

+ +

c) 5 7

2 5 3 7 2 7

3 3 3 3

a b

a a b b

+ + d) ( )

12 4a b ab

+

Hng dn gii:

a) ( )

( )( )( ) ( )

2 3 2 32 2 2 3 2 3 2 3 2

2 2 2 32 32 3 2 3

21 1

a b a ba b a b a b a

a ba ba b a b

+ + + + = + = =

b) ( )( ) ( )( ) ( )

( )( ) ( )2 3 2 3 3 3 3 3 3 3 3 2 3

3

4 3 3 3 3 3 2 3

1 1 1 11

1 1

a a a a a a a a aa

a a a a a a

+ + + + += = +

+ +

c)

5 7 2 5 3 7 2 7

3 3 3 3 3 3

5 75 73 3

2 5 3 7 2 7 2 5 3 7 2 73 3 3 3 3 3 3 3

a b a a b ba b

a b

a a b b a a b b

+ + = =

+ + + +

d) ( ) ( )1

2 2 2 2 4 2 4a b ab a b a b a b a b a b

+ = + + = =

V d 3: Vit di dng ly tha vi s m hu t cc biu thc sau :

a) 5 32 2 2A = b) ( )11

16: 0B a a a a a a= >

c) 24 3C x x= d) ( )5 3 0b aD aba b

= >

Hng dn gii:

a)

1 11 1 15 5

3 1 31 3 13 3 55 3 2 5 102 2 22 2 2 2 .2 .2 2 .2 2 .2 2 2A

= = = = = =

b)

11 12

1 151 12 211 11 11 7 113 3 12 162 2 1116 16 6 8 162 4 4

11

16

: . : . : :a

B a a a a a a a a a a a a a a a

a

++

= = = = = =

V d 4: Rt gn biu thc sau :

a)

1 111 12 2

4 43 1 1 1 14 2 4 4 4

:a b a b

A a b

a a b a b

= + +

b)

3 3 3 3

4 4 4 4

1 1

2 2

a b a b

B ab

a b

+

=

Hng dn gii:

a)

1 1 1 1 1 11 1 1 12 2 2 2 2 24 4 4 4

3 1 1 1 1 1 11 1 1 1 1 1 1 14 2 4 4 4 4 42 4 4 2 4 4 4 4

1: : .

a b a b a b a b a b a a bA a b a b

a a b a b a ba a b a a b a b

+ = = = = + + + + +

1 1 1

2 2 2

1 1 12 2 2

b a b b

aa a b

= =



b)

( )3 3 3 3 3 3 1 1 1 1 1 1

4 4 4 4 2 2 2 2 2 2 2 2

1 1 1 1 1 12 2 2 2 2 2

a b a b a b a b a b a b a b

B ab a b

a b a b a b

+

= = = =

V d 5: n gin cc biu thc sau (vi gi thit chng c ngha)

a)

32

1 13 24 4

3 3:

a b aA a b

b a a b

= + +

b) 2

22

4

44

2

aB

aa

a

+= +

Hng dn gii:

a)

3 3 121 1 1 1 2 23 2 2 2 34 4 4 4

3 1 1 12 3 1 13 332 2 4 4 4 4

11

: :

aa b a a b a a bb abA a b a b

a bb a a b b a a b ab a b

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

b) ( )

2 2

2 22 2

2

2 024 4

2 04 44

2 4

aaa aB

aaa aa aa a

+ += = = = +a a c)

0,221 .

a a e) ( ) ( )

32

42 2 . > a a f)

1 1

2 21 1.

> a a

Bi 3: Tnh gi tr cc biu thc sau:

a) ( ) ( )11 1

2 23 2 3 2 3 2 3 2

= + + +

A

b) 4 10 2 5 4 10 2 5 .= + + + +B



Bi 4: Cho hm s 4

( ) .4 2

=+

x

xf x

a) Chng minh rng nu a + b = 1 th f(a) + f(b) = 1.

b) Tnh tng 1 2 2010

... .2011 2011 2011

= + + +

S f f f

Bi 5.1: So snh cc cp s sau

a)

52

2

v

103

2

b) 2

2

v 3

5

c)

1043

5

v

524

7

d) 3

7

6

v

2

8

7

e)

5

6

v 2

5

Bi 5.2: So snh cc cp s sau a) 3 30 v 5 20 b) 4 5 v 3 7

c) 17 v 3 28 d) 4 13 v 5 23 Bi 6: Tm x tha mn cc phng trnh sau?

1) 54 1024=x 2) 15 2 8

2 5 125

+ =

x

3) 1 31

832

=x

4) ( )2

2 13 3

9

=

xx

5) 2 8 27

.9 27 64

=

x x

6)

2 5 631

2

+ =

x x

7) 2 81 0,25

.320,125 8

=

xx 8) 0,2 0,008=x 9)

3 7 7 39 7

49 3

=

x x

10) ( ) ( ) 112 . 36

=x x

11) 1 11

7 .428

=x x



1) Khi nim v Logarith

Logarith c s a ca mt s x > 0 c k hiu l y v vit dng log= = yay x x a

V d 1: Tnh gi tr cc biu thc logarith sau ( )2 3 2 2log 4; log 81; log 32; log 8 2 Hng dn gii:

2 2log 4 2 4 2 log 4 2= = = =yy y

y 43 3log 81 y 3 81 3 y 4 log 81 4= = = = =

( ) ( )y 1052 2log 32 y 2 32 2 2 y 10 log 32 10= = = = = = ( ) ( ) ( ) ( )732 2log 8 2 2 8 2 2 . 2 2 7 log 8 2 7= = = = = =

yy y

V d 2: Tnh gi tr ca a)

2 2log 32 = ..........................................................................................................................................................

b) 32

log 128 2 = .....................................................................................................................................................

c) 3

log 81 3 = ........................................................................................................................................................

d) 3 3log 243 3 = ......................................................................................................................................................

Ch : Khi a = 10 th ta gi l logarith c s thp phn, k hiu l lgx hoc logx Khi a = e, (vi e 2,712818) c gi l logarith c s t nhin, hay logarith Nepe, k hiu l lnx, (c l len-x)

2) Cc tnh cht c bn ca Logarith

Biu thc logarith tn ti khi c s a > 0 v a 1, biu thc di du logarith l x > 0. log 1 0 ;log 1,= = a a a a

Tnh ng bin, nghch bin ca hm logarith: 1

log log0 1

> >> < <



a) 3 5logaA a a a= b) 23 5logaB a a a a= c)

5 33 2

1 4log

a

a a a

a a

Hng dn gii:

a) 1 1

33 5 2 5 1 1 37log log 32 5 10a a

A a a a a+ +

= = = + + =

b)

1

31

1 11 2 3

23 2 55

3

27 3log log 1 1

10 10a aB a a a a a

+ + +

= = = + = +

c)

3 215 33 2 5 3

1 1 142 4

34 3 91log log

15 4 60aa

a a a a

a aa

+ +

+

= = =

V d 4: Tnh gi tr cc biu thc sau:

a) 15

log 125 .....................................................= b) 2

log 64 ....................................................................=

c) 16log 0,125 ..................................................= d) 0,125log 2 2 ..........................................................=

e) 3 33log 3 3 ................................................= f) 78 7

7log 7 343 ............................................................=


a) ( )3 5log ..................................................................................................................................aP a a a= =

b) ( )23 54log ............................................................................................................................= =aQ a a a a Cng thc 2: log , 0= >a xa x x , (2) Chng minh: t ( )log , 2= = =t t ta x t x a a a

V d 1: ( ) ( ) ( ) ( )3

3 352

log 4 11 1log 4 log 4log 6log 3 22 22 3, 5 6, 3 3 3 4 2... = = = = = =


1) 8log 152 .....................................................= 2) 2 2log 64

2 ....................................................................=

3) 81log 51

.....................................................3

= 4) ( ) 3

log 43 9 ....................................................................=

Cng thc 3: ( )log . log log= +a a ax y x y , (3) Chng minh:

p dng cng thc (2) ta c log

log log log log

log. . +

= = ==

a

a a a a

a

xx y x y

y

x ax y a a a

y a

p dng cng thc (1) ta c : ( ) log loglog . log log log+= = + a ax ya a a ax y a x y dpcm V d 1: Tnh gi tr cc biu thc sau: a) ( ) 32 2 2 2 2 2 2log 24 log 8.3 log 8 log 3 log 2 log 3 3 log 3= = + = + = + b) ( ) 33 3 3 3 3 3log 81 log 27.3 log 27 log 3 log 3 log 3 3 1 4= = + = + = + = V d 2: Tnh gi tr cc biu thc sau:

a) 4

23 3 32 2 2 2 2

4 10log 4 16 log 4 log 16 log 2 log 2 2 .

3 3= + = + = + =

b)

131

333 3 3

1 1 1 1 1 1 1

3 3 3 3 3 3 3

1 1 1 10log 27 3 log 27 log 3 log 3 log 3 log log 3 .

3 3 3 3

= + = + = + = =

c) ( ) ( )6 2

35 52 2 2 2 2 2 2

log 8 32 log 8 log 32 log 2 log 2 log 2 log 2 6 2 8.= + = + = + = + =



V d 3: Cho bit log 2;log 2a ab c= = Tnh gi tr ca loga x vi

a) 3 2x a b c= .................................................................................................................................................................

........................................................................................................................................................................................

b) 3 3x ab a bc= ......................................................................................................................................................

........................................................................................................................................................................................

Cng thc 4: log log log

=

a a a

xx y

y, (4)

Chng minh:

p dng cng thc (2) ta c log log

log logloglog

= = ==

a a

a a

aa

x xx y

yy

x a x aa

y ay a

p dng cng thc (1) ta c : log loglog log log log

= =

a ax ya a a a

xa x y dpcm

y

V d 1: 45

3 322 2 2 2 23

32 5 4 7log log 32 log 16 log 2 log 2 .

2 3 616= = = =

V d 2: Cho bit 1

log ;log 33a a

b c= = Tnh gi tr ca loga x vi

a) 2

3 2

ab cx

abc= .................................................................................................................................................................

........................................................................................................................................................................................

b) 5 3

34

a bcx

a abc= .........................................................................................................................................................

.......................................................................................................................................................................................

V d 3: Tm tp xc nh ca cc hm s sau :

a) 12

1log

5

xy

x

=+

b) 2

1 55

1log log

3

xy

x

+= + c) 2

3log

1

xy

x

=+

f) 2

0,3 3

2log log

5

xy

x

+= + d) 21 2

2

1log log 6

1

xy x x

x

= +

e) ( )22

1lg 3 4

6y x x

x x= + + +

g)

1log

2 3

xy

x

=

Hng dn gii:

a) 12

1log

5

xy

x

=+

. iu kin : 12

1 1log 0 1 211 0 0 11 1

1 111 1; 1 1; 10011

x xx

xx xx x

xx x x x xxx

+ + + + < > < >>> ++

Vy ( )1;D = +

b) 2

1 55

1log log

3

xy

x

+= + . iu kin :

2 2

1 52

3

2 2

5 2

2

1 2log log 0 03 311

1 5 1430 log 1 03 31

0 5 31 30 53

x x xx xx

x x xxx xx

xx xx

+ + + + + + + ++ < > + +< +

( ) ( )3 1; 2 3; 2 2;73; 2 7

x xx

x x

< < > < <



3) Cc cng thc v logarith (tip theo)

Cng thc 5: log .log=ma ab m b , (5) Chng minh:

Theo cng thc (2) ta c ( )log log .log= = =a a amb b m bmb a b a a Khi .loglog log .log= = am bma a ab a m b dpcm

V d 1: ( )

3 22 2 2 5 5 5

14 4

2 2 2

log 27 log 3 3log 3; log 36 log 6 2log 6

1 5log 32 log 32 log 32

4 4

= = = =

= = =

V d 2: 42

231 1 1 1 1 1 1 1 1

3 3 3 3 3 3 3 3 3

1 6 .45 12log 6 log 400 3log 45 log 6 log 400 log 45 log log 81 log 4.

2 20 3

+ = + = = = =

V d 3: 5 5 5 5 5 5 5 51 50 3

log 3 log 12 log 50 log 3 log 12 log 50 log log 25 2.2 2 3

+ = + = = =

V d 4: Cho bit 1 3

log ;log2 4a a

b c= = Tnh gi tr ca loga x vi

a) 3 2

2 34

a b cx

a bc= ...............................................................................................................................................................

........................................................................................................................................................................................

b) 3 3

3

ab a bcx

bc= .....................................................................................................................................................

........................................................................................................................................................................................

Cng thc 6: 1

log log=n aa b bn, (6)

Chng minh:

t ( )log = = =n yn nya b y a b a b Ly logarith c s a c hai v ta c :

1log log log log= = =nya a a aa b ny b y bn

hay 1

log log= n aa b b dpcmn

V d 1 :

1

2

5 1

5

222

222

1log 16 log 16 log 16 2.4 8.

121

log 64 log 64 log 64 5.6 30.15

= = = =

= = = =

H qu: T cc cng thc (5) v (6) ta c : log log=nm

aa

mb b

n

V d 2: ( ) ( ) ( ) ( )3 1 331 11

34 45 2 2 25 2

5

39 11 114log 125 log 5 log 5 ; log 32 2 log 2 log 2 .

1 4 3 33

= = = = = =

Ti liu bi ging:

02. CNG THC LOGARITH P2 Thy ng Vit Hng



V d 3: Tnh gi tr biu thc 13 3 53

4

133

27log 27 log

9.

1 1log log

81 3

+ = +

A

Hng dn gii:

( )23 3 3 3log 27 log 3 3 2= =

12

1335

1 32533 5

27 3 1 13 26log log log 3 2. .

1 5 593

2

= = = =

12

13 3 54 3

3 433

133

27 26log 27 log 291 45log log 3 4.2log 3 8 .81 8 4 51 1

log log81 3

+ = = = = = =

+ +

A

Cng thc 7: (Cng thc i c s) log

loglog

= cac

bb

a, (7)

Chng minh:

Theo cng thc (2) ta c ( )log log loglog log log .log loglog

= = = = a ab b cc c a c ac

bb a b a b a b dpcm

a

Nhn xt : + cho d nh th i khi (7) cn c gi l cng thc chng c s vit theo dng d nhn bit nh sau log log .log=a a cb c b

+ Khi cho b = c th (7) c dng log 1

log .log log

= =bab b

bb

a a

V d 1: Tnh cc biu thc sau theo n s cho: a) Cho 2 2log 14 log 49 ?= = =a A b) Cho 15 25log 3 log 15 ?= = =a B

Hng dn gii: a) Ta c ( )2 2 2 2log 14 log 2.7 1 log 7 log 7 1.= = = + = a a a Khi ( )2 2log 49 2log 7 2 1 .= = = A a

b) Ta c 3

153 3

5

1 1log 5 1

1 1log 3

log 15 1 log 5log 3

1

= == = = + =

a

a aa aa

a

( ) ( )3

253 3

1 1log 15 1 1

log 15 .1log 25 2log 5 2 1 2 12

= = = = = = a aB B

a a aa

V d 2: Cho log 3.a b= Tnh

a) log .=b

a

bA

a b) log .=

ab

bB

a

Hng dn gii:

T gi thit ta c 1

log 3 log .3

= =a bb a

a) 1 1 1 1

log log loglog log log log

log log

= = = = =

b b bb b a aa a a

b a

bA b a

a b a b ab b

a a



1 1 1 1 3 1 3 1.

21 2log log 2 3 2 3 2 3 213

= = = = b a

Aa b

Cch khc: Ta c c 22

2

2

log log 1 3 1log log log

log 2 3 2log

aa

bb baaa a a

bbb b b aA

ba ba aa

= = = = = =

b) 1 1 1 1

log . log loglog log log log log log

= = = = =+ +ab ab ab b b ba a a

bB b a

a ab ab a b a b

1 1 1 1 2 3 1 2 3 1.

1 1 1 11 log 1 3 3 1 3 1log2 2 22 3

= = = =+ + + ++ +ab

Bba

Cch khc: Ta c ( )2

2

2 2 log 2log 1 2 3 1log log log .

log 1 log 1 3

aa

abab aba a

bbb b b aB

a ab ba a

= = = = = = + +

V d 3: Tnh gi tr ca cc biu thc sau :

a) 9

125 7

1 1log 4 log 8 log 24 281 25 .49

+

b)

2 54

1log 3 3 log 51 log 5 216 4

++ +

c) 7 7

3

1log 9 log 6 log 4272 49 5

+

d) 6 9log 5 log 361 lg 236 10 3+

Hng dn gii:

a) ( )3

9 39125 7 5 7

1 1 1 1log 4 2log 24 log 4log 8 log 2 2log 24 2 4 281 25 .49 3 5 7

+ = +

53 7

12 .3log 21 log 4 log 43 33 5 7 4 4 19

4 = + = + =

b) ( )2 5

4 2 54

1log 3 3log 5 2 1 log 5 log 3 6log 51 log 5 6216 4 4 2 16.25 3.2 592

+ + ++ + = + = + =

c) ( )7 7 5 7 7 51

log 9 log 6 log 4 log 9 2log 6 2log 429 1

72 49 5 72 7 5 72 1836 16

+ = + = + = +

4,5=22,5

d) 6 9 6log 5 log 36 log 251 lg2 log536 10 3 6 10 25 5 30+ = + = + = V d 4: Tnh gi tr ca cc biu thc sau :

a) 9 9 9log 15 log 18 log 10A = + b) 3

1 1 1

3 3 3

12log 6 log 400 3log 45

2B = +

c) 36 16

1log 2 log 3

2C = d) ( )1 3 2

4

log log 4.log 3D =

Hng dn gii:

a) 3 39 9 9 9 9 315.18 1 3

log 15 log 18 log 10 log log 3 log 310 2 2

A = + = = = =

b) 2 431 1 1 1 1 33 3 3 3 3

1 36.452log 6 log 400 3log 45 log log 9 log 3 4

2 20B

= + = = = =

c) 36 1 6 6 66

1 1 1 1 1log 2 log 3 log 2 log 3 log 2.3

2 2 2 2 2C = = + = =

d) ( ) ( ) ( )1 3 2 4 2 3 4 2 24

1 1log log 4.log 3 log log 3.log 4 log log 4 log 2

2 2D = = = = =

V d 5: Hy tnh :

a. ( )2 3 4 2011

1 1 1 1.......... 2011!

log log log logA x

x x x x= + + + + =

b. Chng minh :

+ ( )axlog log

log1 log

a a

a

b xbx

x

+=

+



+ ( )

2

11 1 1.........

log log log 2logka aa a

k k

x x x x

++ + + =

Hng dn gii:

a)2 3 4 2011

1 1 1 1.......... log 2 log 3 ... log 2011 log 1.2.3...2011 log 2011!

log log log log x x x x xA

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

Nu x = 2011! Th A= ( )2011!log 2011! 1=

b) Chng minh : ( )axlog log

log1 log

a a

a

b xbx

x

+=

+

Ta c axlog log log

loglog ax 1 log

a a a

a a

bx b xbx

x

+= =

+ pcm.

Chng minh : ( )

2

11 1 1.........

log log log 2logka aa a

k k

x x x x

++ + + =

( ) ( )2 1log log ...log 1 2 3 ... log2log

kx x x x

a

k kVT a a a k a VP

x

+= + + = + + + + = =

V d 6: Chng minh rng : a) Nu : 2 2 2 ; 0, 0, 0, 1a b c a b c c b+ = > > > , th log log 2log .logc b c b c b c ba a a a+ + + = b) Nu 0



Do : 4

3 36

3 3

log 2 4log 2log 16

log 6 1 log 2A = = =

+. Thay t (*) vo ta c : A=

( )( )

2 3 .2 12 4

3 3

x x x

x x x

=+ +

c) T : 3 23 3 32

log 5 3log 135 log 5.3 log 5 3 3 3

log 3

a a bC

b b

+= = = + = + = + =

d) Ta c : 27 3 3 8 2 21 1

log 5 log 5 log 5 3 ; log 7 log 7 log 7 33 3

a a b b= = = = = = (*)

Suy ra : ( )2 3 22 2 2

62 2 2

3 1log 3.log 5 log 7log 5.7 log 5 log 7 .3 3log 35

log 2.3 1 log 3 1 log 3 1 1

b ab a bD

b b

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

e) Ta c : 2 2 2log 14 1 log 7 log 7 1a a a= + = =

Vy : ( )5

249 2

2 2

log 2 5 5log 32

log 7 2log 7 2 1a= = =

V d 8: Rt gn cc biu thc a) ( )( )log log 2 log log log 1a b a ab bA b a b b a= + +

b) ( ) ( )2log log 12 2 42 2 21

log 2 log log2

x xB x x x x+= + +

c) ( )log log 2 log log loga p a ap aC p a p p p= + + Hng dn gii:

a) ( )( ) ( )2

log 1log log 2 log log log 1 1 log 1

loga

a b a ab b aba

bA b a b b a a

b

+= + + = =

2 2 2log 1 log log 1 log 1 log1

1 1 1 1 1log log log 1 log log 1 log

a a a a a

a a a a a a

b a b b b

b ab b b b b

+ + + = = + +

log 1 11 log

log loga

ba a

ba

b b

+= = =

b) ( ) ( ) ( )( ) ( )2 2log log 12 2 42 2 2 2 2 2 21 1

log 2 log log 1 2log log log 1 4log2 2

x xB x x x x x x x x+= + + = + + + + =

( ) ( ) ( )2 2 22 2 2 2 21 3log log 8 log 9 log 3log 1x x x x x= + + + = + +

c) ( ) ( )2

2

log 1 loglog log 2 log log log log log

log 1 loga a

a p a ap a a aa a

p pC p a p p p p p

p p

+ = + + = = +

( ) ( )2 3log 1 log

log loglog 1 log

a aa a

a a

p pp p

p p

+ = = +

V d 9: Chng minh rng

a) ( ) ( )1log 3 log 2 log log2

a b a b = + vi : 2 23 0; 9 10a b a b ab> > + =

b) Cho a, b, c i mt khc nhau v khc 1, ta c :

+ 2 2log loga ab c

c b=

+ log .log .log 1a b cb c a =

+ Trong ba s : 2 2 2log ;log ;loga b cb c a

c a b

b c a lun c t nht mt s ln hn 1

Hng dn gii:

a) T gi thit ( )22 2 2 23 0; 9 10 6 9 4 3 4a b a b ab a ab b ab a b ab> > + = + = =

Ta ly log 2 v : ( ) ( ) ( )12log 3 2log 2 log log log 3 log 2 log log2

a b a b a b a b = + + = +

b) Chng minh : 2 2log loga ab c

c b= .

* Tht vy : 1 2

2 2log log log log log loga a a a a ab c c b c c

c b b c b b

= = = =



* log .log .log 1 log .log log 1a b c a b ab c a b a a= = =

* T 2 kt qu trn ta c

2

2 2 2log log log log .log log 1a b c a b cb c a b c a

c a b b c a

b c a c a b

= =

Chng t trong 3 s lun c t nht mt s ln hn 1 V d 10: Tnh gi tr cc biu thc sau: a) 36log 3.log 36 ......................................................................=

b) 43log 8.log 81 ......................................................................=

c) 32 251

log .log 2 .................................................................5

=

V d 11: Cho log 7.a b= Tnh

a) 3

log .=a b

aA

b

b) 3 2log .= ba

B ab

V d 12: Tnh cc biu thc sau theo n s cho:

a) Cho 325 2 549

log 7 ; log 5 log ?8

= = = =a b P

b) Cho log 2 log ?= = =ab abb

a Qa

Cng thc 8: log log=b bc aa c , (8) Chng minh:

Theo cng thc (7): ( )loglog log .log log log loglog log .log= = = = bb b a b a bac a c c c ab b ac a c a a a a c dpcm V d 1: ( ) 27 7 2

1log 27log 2 log 49 log 22 249 2 2 4; 2 27 27 3 3...= = = = = =


a) 3

6 9log 4log 5 log 3636 3 3 ..........................................................................................................A = + =

b) 23

3

log 32 log 2

log 4

3 .4.............................................................................................................................

27B

= =

c) 3 9 9log 5 log 36 4log 781 27 3 .........................................................................................................C= + + =


Hc trc tuyn ti: www.moon.vn Mobile: 0985.074.831

I. PHNG TRNH C BN

Cc v d gii mu:

V d 1. Gii phng trnh 1 2 12 2 2 5 2.5x x x x x+ + + + = + . Hng dn gii:

Ta c 1 2 1 21

2 2 2 5 2.5 2 2 .2 2 .2 5 2.5 .5

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

( ) 52

2 7 51 2 4 .2 1 .5 7.2 .5 5 log 5

5 5 2

xx x x x x

+ + = + = = =

Vy phng trnh cho c 1 nghim l 52

log 5.x =

V d 2. Gii cc phng trnh sau

1) 2 3 2 12 16x x x+ += 2)

2 4 13243

x x + = 3) 10 5

10 1516 0,125.8x x

x x

+ + =

Hng dn gii:

1) 2 23 2 1 3 2 4 4 2 2 22 16 2 2 3 2 4 4 6 0

3x x x x x x xx x x x x

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

Vy phng trnh c hai nghim l x = 2 v x = 3.

2) 2 24 4 5 2 113 3 3 4 5

5243x x x x xx x

x + + = = = + = =

Vy phng trnh c nghim x = 1; x = 5.

3) ( )10 5

10 1516 0,125.8 , 1 .x x

x x

+ + =

iu kin: 10 0 10

15 0 15

x x

x x

Do 4 3 31

16 2 ; 0,125 2 ; 8 28

= = = = nn ta c ( )10 5

4. 3.310 15 10 51 2 2 .2 4. 3 3.10 15

x x

x x x x

x x

+ + + + = = +

( )2 04( 10) 60 5 150 15 1502010 15

xxx x x

xx x

=+ = = =

Vy phng trnh c nghim x = 0; x = 20. V d 3. Gii cc phng trnh sau:

1) 2 9 27

.3 8 64

x x =

2) 1 2 14.9 3 2x x += 3) ( ) ( )1

115 2 5 2

xx

x

++ =

Hng dn gii:

1) 3 3

2 9 27 2 9 3 3 3. . 3.

3 8 64 3 8 4 4 4

x x x x

x = = = =

Vy phng trnh c nghim duy nht x = 3.

2) ( )2x 3 02x 1x 1 3 2x2x 1 2x 1 2x 3 2x 32

2x 12

4.9 3 3 34.9 3 2 1 3 .2 1 3 . 2 1 1 x .

22 23.2

+ + +

= = = = = = =

Vy phng trnh c nghim duy nht 3

.2

x =

Cch khc: 2 3

1 2 1 1 2 1 81 81 18.81 9 9 34.9 3 2 16.81 9.2 16. 9.2.4 .81 4 16 2 2 2

x xxx x x x x x + +

= = = = = =

Ti liu bi ging:

04. PHNG TRNH M P1 Thy ng Vit Hng



3) ( ) ( ) ( )1

115 2 5 2 , 1 .

xx

x

++ =

iu kin: 1 0 1.x x+

Do ( )( ) ( ) 115 2 5 2 1 5 2 5 25 2

+ = = = +

+

( ) ( )1 1 11 1 1 1 0 21 1x x

x xxx x

= = + = = + +

Vy phng trnh c hai nghim l x = 1 v x = 2.

V d 4. Gii cc phng trnh sau:

1) ( )2

1 13 22 2 4

xx x

+

=

2) ( ) ( )2 5 6

3 2 3 2x x

+ = 3) ( )2 2 2 21 1 25 3 2 5 3x x x x+ = Hng dn gii:

1) ( ) ( )2

1 13 22 2 4, 1 .

xx x

+

=

iu kin: 0

1

x

x

>

( )( )( ) ( )

( )

3 1

1 23 1

1 2 2 2 2 5 3 0 3 9.1

x

x x xx x x x

x x

+

+ = = = = =

Vy phng trnh cho c nghim x = 9.

2) ( ) ( ) ( )2 5 6

3 2 3 2 , 2 .x x

+ =

Do ( )( ) ( ) ( ) ( )11

3 2 3 2 1 3 2 3 2 .3 2

+ = = = +

+

( ) ( ) ( )2 5 6

2 22 3 2 3 2 5 6 03

x x xx x

x

= + = + + = =

Vy phng trnh cho c nghim x = 2 v x = 3.

3) ( )2 2 2 2 2 2 2 2 2 2 2 21 1 2 2 2 2 25 3 2 5 3 5 3.3 5 3 5 5 3.3 35 9 5 9x x x x x x x x x x x x+ = = = 2 2

2 23

3 25 5 125 5 55 3 3.

5 9 3 27 3 3

x xx x x

= = = =

Vy phng trnh cho c nghim 3.x =

Cc v d gii mu trong video:

V d 1: Gii phng trnh

a) 1 27 7 7 342x x x+ ++ + = b) 1 15 10.5 18 3.5x x x ++ + =

c) 17.5 2.5 11x x = d) 2 214.7 4.3 19.3 7x x x x+ =


a) 2 2 2 21 1 22 3 3 2x x x x + = b)

2 3 2 12 16x x x+ +=

c) 10 5

10 1516 0,125.8x x

x x

+ + = d) ( ) ( )

11

15 2 5 2x

xx

++ =


a) ( ) ( )3 1

1 310 3 10 3x x

x x

+ ++ = b)

2 1 2 49 3x x+ =

c) 3

824 32 8

xx = d) ( )29 32 22 2 2 2

xx x x x

+ = +



e) ( )1

cos cos2 22 2x

x xxx x+

+ = +

II. PHNG TRNH BC HAI, BC BA THEO MT HM S M

Cc v d mu:

V d 1. Gii phng trnh: 25 30.5 125 0x x + = Hng dn gii:

Phng trnh cho tng ng: ( )25 30.5 125 0x x + = . t 5xt = , iu kin t > 0.

Khi phng trnh tr thnh: 25

30 125 025

tt t

t

= + = =

+ Vi 5 5 5 1xt x= = = . + Vi 225 5 25 5 5 2x xt x= = = = . Vy phng trnh cho c 2 nghim l x = 1 v x = 2.

V d 2. Gii phng trnh: 23 3 10x x+ + = . Hng dn gii:

Ta c ( )0

22

2

3 1 3 013 3 10 9.3 10 9. 3 10.3 1 0 1 23 3 3

9

x

x x x x xx x

x

x+

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

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


1) 15 5 4 0x x + = 2) 23 8.3 15 0x

x + = 3) 2 8 53 4.3 27 0x x+ + + =

Hng dn gii:

1) ( )15 5 4 0, 1 .x x + = iu kin: x 0.

( ) ( )2 5 1 0 051 5 4 0 5 4.5 5 0 15 15 5x

x x x

x x

x x

xx

= = = + = + = = = =

C hai nghim u tha mn iu kin, vy phng trnh c hai nghim x = 0 v x = 1.

2) ( ) ( ) ( )( )

22

33

3 3 23 8.3 15 0 3 8. 3 15 0

log 5 log 253 5

xx

x xx

x

x

x

= = + = + = = ==

Vy phng trnh c hai nghim 32 ; log 25.x x= =

3) 4

2 8 5 2( 4) 4 2( 4) 4

4 2

3 3 33 4.3 27 0 3 4.3 .3 27 0 3 12.3 27 0

3 9 3 2

xx x x x x x

x

x

x

++ + + + + +

+

= = + = + = + =

= = =

Vy phng trnh cho c hai nghim l x = 2 v x = 3.

V d 4. Gii phng trnh 2 222 2 3.x x x x + =

Hng dn gii:

t 2

2 ( 0).x x t t = > . Phng trnh tr thnh 4 14

31( ) 2

t xt

t L xt

= = = = =

V d 5. Gii phng trnh 2 25 1 54 12.2 8 0x x x x + = .

Hng dn gii:



t 2

25

2

32 5 1

2 ( 0) 94 5 2 4

x x

xt x x

t tt xx x

== = = > = = =

Cc v d gii mu trong video:

V d: Gii phng trnh

a) 2 21 19 3 6 0x x+ + = b)

2 21 39 36.3 3 0x x + =

c) 2 22 1 24 5.2 6 0x x x x+ + = d) 3 2cos 1 cos4 7.4 2 0x x+ + =

BI TP LUYN TP:

Bi I: Gii cc phng trnh sau:

1) ( )2

6 100,2 5x x x = 2)

2 5 2 33 2

2 3

x x x + =

3) ( ) ( )4 1 2 33 2 2 3 2 2x x ++ =

4) ( )2

19. 3 81x x

x

= 5) 25 4 110 1x x = 6)

2

23

1 1x

xee

=

7) ( )1

31 16. 48

xx

=

8) 2

5 74 1 19

3

xx x

=

9)

1 4 2

1 2127 .819

x x

x x

+ +=

10) 1 1

3 .3 27

x xx =

11) ( ) ( )3 25 3 2 1

10 3 19 6 10x x x

= +

Bi II: Gii cc phng trnh sau

1) ( ) 31 1 xx + = 2) 2 56

22 16 2x x +

=

3) ( )2 12 1 1

xx x

+ = 4) ( ) 22 1xx x =

5) ( )242 2 2 1

xx x

+ = 6) ( ) ( )

2 5 102 2

xx xx x

+ + = + /s: x = -1; x = 5

7) ( )2 42 5 4 1

xx x

+ = /s:

5 13

22

x

x

=

=

8) ( )2 2

3 3x x

x x = /s:

1

2

4

x

x

x

= = =

9) ( ) 31 1xx + = /s: x = 3 Bi III: Gii cc phng trnh sau

1) 1 22 .3 .5 12 x x x = 2) 4 6 3 45 25x x = /s : 75

x =

3) 2 2 19.2 8. 3x x+= 4) 5 17

7 332 0.25.128x x

x x

+ = /s : x = 13

5) ( ) ( )4

410 3 10 3x x

x x

++ = 6) ( ) ( )

33

15 2 5 2x

xx

++ =

7) 1 1 2 1 1 13.4 3 .9 6.4 2 .9x x x x+ + + ++ = /s: 12

x =

8) 3 1

2 12 29 2 2 3x xx x+ + = /s: 9

2

9log

2 2x

=

9) 1 1

2 22 25 9 3 5x xx x+ = /s: 3

2x =

10) 3 2 2 37 9.5 5 9.7x x x x+ = + /s: x = 0



III. PHNG PHP T N PH GII PHNG TRNH M

Dng 1: Phng trnh chia ri t n ph

V d 1. Gii phng trnh: 3.9 7.6 6.4 0x x x+ = .

Hng dn gii:

Phng trnh cho tng ng:2

3 21

2 33 33. 7. 6 0

2 2 33 0

2

x

x x

x

x = = + = =



Dng 2: Phng trnh c tch c s bng 1

Cch gii:

Do ( ) ( ) ( ) ( )1

1 1f x f x

f xab ab b

a= = =

T ta t ( ) ( )1

, ( 0)f x f xa t t bt

= > =

Ch :

Mt s cp a, b lin hp thng gp: ( )( ) ( )( )( )( ) ( )( )

2 1 2 1 1 2 3 2 3 1

5 2 5 2 1 7 4 3 7 4 3 1

;

; ...

+ = + =

+ = + =

Mt s dng hng ng thc thng gp: ( )( )

2

2

3 2 2 2 1

7 4 3 2 3 ...

=

=

V d mu. Gii cc phng trnh sau:

a) ( ) ( )2 3 2 3 4x x+ + = b) ( ) ( )3 33 8 3 8 6x x+ + = c) ( ) ( ) 35 21 7 5 21 2x x x+ + + = d) ( ) ( )

2 2( 1) 2 1 42 3 2 3

2 3

x x x + + =

Hng dn gii:

a) ( ) ( ) ( )2 3 2 3 4, 1 .x x+ + =

Do ( )( ) ( ) ( ) ( )( )

12 3 2 3 1 2 3 . 2 3 1 2 3

2 3

x x x

x+ = + = =

+

t ( ) ( ) 12 3 , ( 0) 2 3 .+ = > =x xt tt

Khi ( ) 21 2 31 4 0 4 1 02 3

tt t t

t t

= + + = + = =

Vi ( ) ( )22 3 2 3 2 3 2 3 2.xt x= + + = + = + = Vi ( ) ( ) ( ) 212 3 2 3 2 3 2 3 2 3 2.xt x= + = = + = + = Vy phng trnh c hai nghim x = 2.

b) ( ) ( ) ( )3 33 8 3 8 6, 2 .x x+ + =

Do ( )( ) ( )( ) ( ) ( ) ( )( )

3 3 3 3 33

3

13 8 3 8 3 8 3 8 1 3 8 . 3 8 1 3 8

3 8

x x x

x+ = + + = + = =

+

t ( ) ( )3 3 13 8 ,( 0) 3 8x xt tt

+ = > = .

Khi ( ) 21 3 82 6 0 6 1 03 8

tt t t

t t

= + + = + = =

Vi ( ) ( )3 33 8 3 8 3 8 3 8 3 8 3.xx

t x= + + = + + = + =



Vi ( ) ( ) ( ) ( )1 13 33 8 3 8 3 8 3 8 3 8 3 8 3.xx

t x

= + = = + = =

Vy phng trnh c hai nghim x = 3.

c) ( ) ( ) ( )3 5 21 5 215 21 7 5 21 2 7. 8, 3 .2 2

x xx x x+ + + + = + =

Ta c 5 21 5 21 5 21 5 21 5 21 1

. 12 2 2 2 2 5 21

2

x x x x

x

+ = = = +

t 5 21 5 21 1

,( 0)2 2

x x

t tt

+ = > =

.

Khi ( ) 211

3 7 8 0 7 8 1 0 1

7

tt t t

tt

= + = + = =

Vi 5 21

1 1 0.2

x

t x += = =

Vi 5 21

2

1 5 21 1 1log .

7 2 7 7

x

t x + + = = =

Vy phng trnh c hai nghim 5 21

2

01

log7

x

x +

= =

d) ( ) ( ) ( )( ) ( )( )2 2 2 2( 1) 2 1 2 1 2 14

2 3 2 3 2 3 2 3 2 3 2 3 42 3

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

( )( )( ) ( ) ( ) ( ) ( )2 2 2 22 2 2 2

2 3 2 3 2 3 2 3 4 2 3 2 3 4, 4 .x x x x x x x x

+ + + = + + =

t ( ) ( )2 22 2 1

2 3 , ( 0) 2 3 .x x x x

t tt

= + > =

Khi ( )( )( )

2

2

22

2

22

2 3 2 32 3 2 114 4 0 4 1 0

2 12 3 2 3 2 3

x x

x x

t x xt t t

t x xt

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

Vi phng trnh 2 22 1 2 1 0 2 2x x x x x = = = Vi phng trnh 2 22 1 2 1 0 1.x x x x x = + = =

Vy phng trnh c hai nghim 1

2 2

x

x

= =

Dng 3: Phng trnh t n ph trc tip bng php quan st

V d 1: Gii phng trnh: 1 1 1

8 2 18

2 1 2 2 2 2 2

x

x x x x + =+ + + +

Hng dn gii:

Vit li phng trnh di dng: 1 1 1 1

8 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

= + >= +



Ta c ( ) ( )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 2

8 189

9;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 9

9;8

u v= = , ta c:

1

1

2 1 949

2 18

x

xx

+ = =

+ =

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

V d 2: Gii phng trnh: 22 2 6 6x x + = Hng dn gii:

t 2 ; 0.xu u= > 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: 23

6 0 2 3 82( )

xuu u xu L

= = = = =

+ Vi u + v + 1 = 0 ta c 2 2

1 2121 1 21 125 0 2 log

2 21 21(1)

2

xu

u u x

u

+= + = = = =

Vy phng trnh c 2 nghim l x = 8 v 221 1

log .2

x=

Cc v d gii mu trong video: V d 1: Gii phng trnh

a) 13250125 +=+ xxx

b) 1 1 1

4 6 9x x x

+ =

c) (H khi A 2006): 3.8 4.12 18 2.27 0x x x x+ =


a) ( ) ( )3 5 3 5 7.2 0x x x+ + = b) lg10 lg 2lg1004 6 3x x x =


a) 07.022)12()12( =++ Bxx

b) ( ) ( )2 2 1

10 3 10 3 10 4x x

+ + = +

c) ( ) ( ) ( )2 22 1 2 1 101

2 3 2 310 2 3

x x x x + + + =


a) ( ) ( )sin sin7 4 3 7 4 3 4x x+ + =



b) ( ) ( )( )7 5 2 2 5 3 2 2 3(1 2) 1 2 0x x x+ + + + + + = V d 5: Gii phng trnh a) 3 1 5 35.2 3.2 7 0x x + =

b) 3 14.3 3 1 9x x x+ =

BI TP LUYN TP:

Bi 1: Gii cc phng trnh sau:

a) ( ) ( )5 24 5 24 10x x+ + = b) 7 3 5 7 3 57 82 2

x x + + =

c) ( ) ( ) 25 21 5 21 5.2xx x+ + = d) ( ) ( )4 15 4 15 8 + + =x x e) ( ) ( )( ) ( )3243234732 +=+++ xx Bi 2: Gii cc phng trnh sau:

a) 04.66.139.6111

=+ xxx b) 1 1 1

2.4 6 9x x x+ =

c) 2 26.3 13.6 6.2 0x x x + = d) + =3.16 2.81 5.36x x x

e) + =64.9 84.12 27.16 0x x x Bi 3: Gii cc phng trnh sau:

a) ( ) ( ) 13 5 1 5 1 2x x x++ = b) ( ) ( ) ( )26 15 3 2 7 4 3 2 2 3 1 0x x x+ + + + = Bi 4: Gii cc phng trnh sau:

a) 2 1 1 15.3 7.3 1 6.3 9 0 x x x x + + + = b) 4 4 2 2 10x x x x + + + = c) 1 13 3 9 9 6x x x x + + + = d) 1 3 38 8.(0,5) 3.2 125 24.(0,5)x x x x+ ++ + =



IV. PHNG PHP LOGARITH HA GII PHNG TRNH M

Khi nim:

L phng trnh c dng ( )( ) ( ). , 1f x g xa b c= trong a, b nguyn t cng nhau, f(x) v g(x) thng l hm bc nht hoc bc hai. Cch gii: Ly logarith c s a hoc c s b c hai v ca (1) ta c

( ) ( ) ( )( ) ( ) ( ) ( )1 log . log log log log ( ) ( ) log log , 2 .f x g x f x g xa a a a a a aa b c a b c f x g x b c = + = + = (2) thu c l phng trnh bc nht ca x, hoc phng trnh bc hai c th gii n gin. Ch : Nhng dng phng trnh kiu ny chng ta c gng s dng tnh cht ca hm m bin i sao cho c = 1. Khi vic logarith ha hai v vi c = 1 s cho phng trnh thu c n gin hn rt nhiu.


a) 13 .2 72x x+ = b) 2

5 .3 1x x = c) 3 2 2 37 9.5 5 9.7x x x x+ = + Hng dn gii:

a) 1

1 2 2 23 .23 .2 72 1 3 .2 1 6 1 2.9.8

x xx x x x x x

++ = = = = =

Vy phng trnh c nghim x = 1.

b) ( )2 2 2 23 3 3 3 35 .3 1 log 5 .3 log 1 log 5 log 3 0 log 5 0x x x x x x x x= = + = + = ( )3

3

0log 5 0

log 5

xx x

x

= + = =

Vy phng trnh cho c hai nghim x = 0 v x = log35.

c) ( ) ( )3 2 2 3 3 2 3 2 3 27 9.5 5 9.7 8.7 8.5 7 5 lg 7 lg 5 3 .lg7 2 .lg5 0x x x x x x x x x x x x+ = + = = = = ( )3lg7 2lg5 0 0.x x = =

Vy phng trnh cho c nghim x = 0.


a) 1

5 .8 500x

x x

+

= b) 2 1

15 .2 50x

x x

+ = c)

23 5 62 5x x x += d) 2lg 10xx x= Hng dn gii:

a) ( )1

5 .8 500, 1 .x

x x

+

= iu kin: x 0.

( ) ( ) ( )1 3 3

3 3 2 3 32 2 2

31 5 .2 5 .2 2 5 log 2 log 5 3 log 5

x x xx x xx x x x x

x

+ = = = =

( ) ( )22 25

3log 5 3 log 5 1 3 0 1

log2

xx x

x

= =

=

b) ( )2 1

15 .2 50, 2 .x

x x

+ = iu kin: x 1.

( ) ( )2 1 2 1 2 1

1 12 2 21 1 12 2 2

2 12 5 .2 5 .2 5 .2 1 log 5 .2 log 1 0 1 2 log 5 0

1

x x xx x xx x x x x

x

+ + + = = = = + = +

( )( ) ( ) ( )22 22

22 0 1 log 5 12 2 1 log 5 0

1 1 log 5 0log 5 lg5

xx

x x xx x

= = + + + = + + = = =

Ti liu bi ging:




Vy phng trnh c hai nghim 1

2 ; .lg5

x x= =

c) ( ) ( ) ( )2 23 5 6 3 5 6 22 2 22 5 log 2 log 5 3 5 6 log 5x x x x x x x x x + += = = +

( ) ( ) 2 252 2

2

33 0

3 1 2 log 5 0 log 50log 50log 5 1 2log 5

log 5

xx

x xxx

= = = = == +

Vy phng trnh c hai nghim 53; log 50.x x= = d) ( )2lg 10 , 4 .=xx x iu kin: x > 0.

( ) ( ) ( )2lg 2lg 1 10

4 lg lg 10 2lg lg 1 0 1lg 10

2

x

x xx x x x

x x

= = = = = =

Vy phng trnh c hai nghim 10 ; 10.x x= =

BI TP LUYN TP:

Bi 1. Gii phng trnh

a) 1

5 .8 500

=x

x x

b) 13 .8 36+ =x

x x

c) 4 33 4=x x

Bi 2: Gii cc phng trnh sau :

a) 53 log5 25 =x x b) 9log 29. =xx x

c) 2 2 2log 9 log log 32 .3= xx x x d) ( )3 23 log log 33 100. 10

=

x xx


a) log9 log9 6+ =xx b) 2 2 2log log 3 3log3 6+ =x xx

c) 2

2 2 2log 2 log 6 log 44 2.3 =x xx d) ( ) ( )2lg 100lg 10 lg4 6 2.3 = xx x


a) ( ) ( )2 2

3 32 log 16 log 16 12 2 24 ++ =x x

b) ( )2

2 21 log 2log2 224+ + =x xx

c) 2lg 3lg 4,5 2lg10 =x x xx


a) 2 2 8 24 5x x x+ = b)

9

17 .2 392x x+ = c) 292 .3 8x x =

d) 2 1

15 .2 50x

x x

+ = e)

2 2 32 .32

x x x = f) 21 13 5x x =

HNG DN GII:

Bi 1. Gii phng trnh

a) 1

5 .8 500x

x x

= b) 13 .8 36x

x x+ = c) 4 33 4x x

=

a) ( ) ( )

( )3 1 3 11

23 2 32

2

335 .8 500 5 .2 5 .2 2 5 3 log 5

log 5

== = = = =

x xxx x xx x x

xxx

xx



b) ( )3

13

22 2 31 13

3 3 3

1 2 log 423 .8 36 3 2 .3 3 4 log 4

1 log 4 2 log 41 1 log 4

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

x

xx x

x x xxx

xxx

c) ( )4 3 3 3 4 33

43 4 4 3 .log 4 log 4 log log 4

3 = = = =

x xx

x x x

Bi 2: Gii cc phng trnh sau : a) 53 log5 25 =x x b) 9log 29. =xx x

c) 2 2 2log 9 log log 32 .3= xx x x d) ( )3 23 log log 33 100. 10

=

x xx

GII

a) 5

5

3 log 233 2 2

log

00

5 25 5 555 525

5

> >= = =

==

x

x

xx

x x xxx

b) 9log 29. xx x= Ly loga c s 9 hai v , ta c phng trnh :

( ) ( )2 2 99 9 90 0 0

9 0log 11 log 2log 0 log 1 0

> > > = > =+ = =

x x xx

xx x x

c) 2 2 2log 9 log log 32 .3= xx x x . S dng cng thc : log log=c cb aa b . Phng trnh bin i thnh :

( )2

2 2 2 2 2 2

2

loglog log log log log log2 2 2

log 2

3 09 .3 3 0 3 3 1 0 3 1

3 1 0

> + = + = =

+ =

xx x x x x x

xx x x

x

t : 22log 2 4= = =t tt x x x . Phng trnh :

2log 23 1

3 1 3 4 1 1 04 4

= = = + =

t tx t tx .

Xt hm s3 1 3 3 1 1

( ) 1 '( ) ln ln 04 4 4 4 4 4

= + = +



b) 2

2 2 2 2 2 2 2 2

loglog log 3 3log log log 3log log 3log

3

3 13 6 3 3 6 2.3 6

6 2 + = + = = =

xx x x x x x xx

1

72

1log

2

2 1

72

1log log 2

2 = =x x

c) ( ) ( )2

2 22 2 2 2 2 2 22 1 log 2 2loglog 2 log 6 log 4 log 2log log 2log4 2.3 2 6 2.3 4.2 6 18.3+ + = = =x xx x x x x xx 2

2 2 2 2 2

log

2log log 2log log 2log

2

0 30

4.2 6 18.3 26 34 18.

4 2 18 4 0

> = > = = + =

x

x x x x x

xt

t t

2log 2

2

0

1 3 4 3 10 log 22 2 9 2 44

9

> = x ta c

2

lglg 2lg log 2

2

2

0

3 106 3 3 4 3 104 18. log 22 24 2 2 9 2 4418 4 09

> = > = = >+ = ==

+ = =

xx x x

tt

tt t t

( )2 2 2 23log 16 2 16 3 9 25 5 = = = = =x x x x

b) ( ) ( ) ( )( )222 2 2

2 22 2

log2log1 log log2log log

2

2 02 224 2.2 224 2

2 224 0

+ = >+ = + =

=

xxx xx x tx

t t

( ) ( )2

2

22 2log 4

2224

0 1log 2 2

14 2 2 log 4 4log 2

2 416 2

> = = = = = = = = == =

x

tx x

t xx

xt

c) 2lg 3lg 4,5 2lg10 =x x xx

Ly lg hai v ( ) ( )23 10

lg 3lg 4,5 2 2

3 10

2

1lg 0

3 10lg 2lg lg lg 3lg 4,5 2 0 lg 10

2

3 10 10lg2

+

= =

= + = = =

+ ==

x xx

xx

x x x x x x

xx

V. PHNG PHP HM S GII PHNG TRNH M

C s ca phng php:

Xt phng trnh f(x) = g(x), (1). Nu f(x) ng bin (hoc nghch bin) v f(x) l hm hng th (1) c nghim duy nht x = xo. Nu f(x) ng bin (hoc nghch bin) v f(x) nghch bin (hoc ng bin) th (1) c nghim duy nht x = xo. Cc bc thc hin: Bin i phng trnh cho v dng (1), d on x = xo l mt nghim ca (1).



Chng minh tnh ng bin, nghch bin hay hng s ca (1). Da vo tnh ng bin, nghch bin kt lun trn chng t khi x > xo v x < xo th (1) v nghim. T ta c x = xo l nghim duy nht ca phng trnh. Ch :

Hm f(x) ng bin th > >2 1 2 1x x f ( x ) f ( x ) ; f(x) nghch bin th > 1 th hm s ng bin, ngc li hm nghch bin. Tng hoc tch ca hai hm ng bin (hoc nghch bin) l mt hm ng bin (hoc nghch bin), khng c tnh cht tng t cho hiu hoc thng ca hai hm.

Vi nhng phng trnh c dng ( ) =u( x )f x;a 0, hay n gin l phng trnh c cha x c h s v trn ly tha, ta coi l phng trnh n l hm m v gii nh bnh thng. Bi ton s quy v vic gii phng trnh bng phng php hm s thu c nghim cui cng.

Dng 1: Phng trnh s dng s bin thin ca hm s m


a) 3 5 2x x= b) 22 3 1x

x = + c) ( ) ( )3 2 2 3 2 2 6x x x+ + = Hng dn gii:

a) ( )3 5 2 , 1 .x x= t ( ) 3( ) 5 2 ( ) 2 0

xf x

g x x g x

=

= = 1 th ( ) (1) 3

( ) (1) 3

f x f

g x g

> = < =

(1) v nghim.

Khi x < 1 th ( ) (1) 3

( ) (1) 3

f x f

g x g

< = > =

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

b) ( ) ( )2 3 12 3 1 2 3 1 1, 2 .2 2

x xxx

x x = + = + + =

t 3 1 3 3 1 1

( ) ( ) ln ln 02 2 2 2 2 2

x xx x

f x f x = + = + <

f(x) l hm nghch bin.

Nhn thy x = 2 l mt nghim ca (2). Khi x > 2 th f(x) < f(2) = 1 (2) v nghim. Khi x < 2 th f(x) > f(2) = 1 (2) v nghim. Vy x = 2 l nghim duy nht ca phng trnh cho.

c) ( ) ( ) ( )3 2 2 3 2 23 2 2 3 2 2 6 1, 3 .6 6

x xx x x + + + = + =

t 3 2 2 3 2 2 3 2 2 3 2 2 3 2 2 3 2 2

( ) ( ) ln ln 0.6 6 6 6 6 6

x x x x

f x f x + + + += + = + 1 th f(x) < f(1) = 1 (3) v nghim. Khi x < 1 th f(x) > f(1) = 1 (3) v nghim. Vy x = 1 l nghim duy nht ca phng trnh cho.

V d 2. Gii phng trnh ( )1 13 11 . 3 10 04 2

x x

x x + + + =

.

Hng dn gii:

t 1

0.2

x

t t = >

Khi phng trnh cho tr thnh ( )2 3 103 11 3 10 01

t xt x t x

t

= + + + + = =



+ Vi 1

1 1 02

x

t x = = =

.

+ Vi 1

3 10 3 102

x

t x x = + = +

(*).

Ta c x = 2 tha mn phng trnh (*) nn l nghim ca phng trnh (*).

M hm s 1

2

x

y =

lun nghch bin trn R, hm s y = 3x + 10 lun ng bin trn R. Do x = 2 l nghim duy

nht ca phng trnh (*). Vy phng trnh cho c 2 nghim l 0, 2.x x= = BI TP T LUYN: Bi 1: Gii cc phng trnh sau :

a) 6 8 10+ =x x x b) ( ) ( )5 2 6 5 2 6 10+ + =x x x c) ( ) ( )2 3 2 3 2 + + =x x x d) 1 1 13 2 2 6

3 2 6 + = +

x x xx x x

a) 6 8 6 8 6 6 8 8

6 8 10 1 ( ) 1 '( ) ln .ln 010 10 10 10 10 10 10 10 + = + = = + = +

x x

f x

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

d) 1 1 1 1 1 1

3 2 2 6 3 2 2 63 2 6 3 2 6 + = + + + = + + +

x x x x x xx x x xx

( ) 3 2 2 '( ) 3 ln 3 2 ln 2 0 ; (1) 7= = + + = + > =x x x xVT f x f x f

1 1 1( ) 6

3 2 6 = = + + +

x x x

VP g x . L mt hm s nghch bin, mt khc g(1) = 7

Chng t x = 1 l nghim duy nht ca phng trnh . Bi 2: Gii cc phng trnh sau : a) 4 3 1 =x x b) 2 3 5 10+ + =x x x x c) 3 4 12 13+ + =x x x x d) 3 5 6 2+ = +x x x

a) 1 3 1 3

4 3 1 1 3 4 1 ( ) 1 04 4 4 4

= + = + = = + =

x x x xx x x x f x

Ta c 1 1 3 3

'( ) ln ln 0 ( )4 4 4 4

= + <

x x

f x f x l hm nghch bin.

Mt khc f(1) = 0 nn phng trnh c nghim duy nht l x = 1



b) 2 3 5

2 3 5 10 110 10 10 + + = + + =

x x xx x x x

t 2 3 5 2 2 3 3 5 5

( ) 1 '( ) ln ln ln 010 10 10 10 10 10 10 10 10 = + + = + + x x x xf x f x lim ( ) ; lim ( ) 6+

= + = x x

f x f x

Suy ra '( )f x l mt hm s lin tc , ng bin v nhn c gi tr dng ln gi tr m trn R, nn phng trnh '( ) 0=f x c nghim duy nht x0.

Ta lp bng bin thin s suy ra hai nghim ca phng trnh, s khng cn nghim no khc.

Dng 2: Phng trnh s dng php t n ph khng hon ton

V d. Gii cc phng trnh sau a) 25 2(3 ).5 2 7 0x xx x + = b) 2 23.25 (3 10).5 3 0x xx x + + =

c) 2 22 24 ( 7).2 12 4 0x xx x+ + = d) 2 1 24 .3 3 2.3 . 2 6x x xx x x x++ + = + +

Hng dn gii:

a) ( )225 2(3 ).5 2 7 0 5 2(3 ).5 2 7 0, 1 .x x x xx x x x + = + = Ta coi (1) l phng trnh bc hai n 5x.

Ta c ( ) ( ) ( )2 22 23 2 7 6 9 2 7 8 10 4x x x x x x x x = = + + = + =

Khi , ( ) ( )( )

( )5 3 4 5 1 0

1 5 7 2 , *5 3 4 5 7 2

x xx

x x

x xx

x x x

= + = < =

= =

(*) l phng trnh quen thuc v d 1 xt n, ta d dng tm c nghim x = 1 l nghim duy nht ca (*). Vy phng trnh cho c nghim duy nht x = 1.

b) ( ) ( )22 2 2 23.25 (3 10).5 3 0 3. 5 (3 10).5 3 0, 2 .x x x xx x x x + + = + + = Ta c ( ) ( ) ( )2 22 23 10 12 3 9 60 100 36 12 9 48 64 3 8x x x x x x x x = = + + = + =

Khi , ( )( )

( )

12

22

10 3 3 815 5 , (*).6 32

10 3 3 85 3 , (**)5

6

xx

xx

x x

x xx

+ = = = =

Xt phng trnh 2 5 5 51 1 1 25

(*) 5 2 log 2 log log3 3 3 3

x x x = = = + =

Xt phng trnh 2(**) 5 3 .x x = t 2 2( ) 5 ( ) 5 ln5 0

( ) 3 ( ) 1 0

x xf x f x

g x x g x

= = > = = => < =

(**) v nghim.



Khi ( ) (2) 1

2( ) (2) 1

f x fx

g x g

< =< > =

(**) v nghim.

x = 2 l nghim duy nht ca (**), vy phng trnh cho c hai nghim 525

log ; 2.3

x x= =

c) ( ) ( )2 22 2 24 ( 7).2 12 4 0 4 ( 7).2 12 4 0, 0 3x x t tx x t t t x+ + = + + = = Ta c ( ) ( ) ( )2 22 27 4. 12 4 14 49 48 16 2 1 1t t t t t t t t = = + + = + + = +

Khi , ( )( )

( )

7 12 2 4 2.23

7 1 2 3 , (*)2

2

tt

tt

t tt

t t t

+ += = =

+ = =

Vi 2 2.t x= = Vi 2 3 1 1.t t t x= = = Vy phng trnh cho c 4 nghim 1; 2.x x= =

d) ( )2 1 24 .3 3 2.3 . 2 6, 4 .x x xx x x x++ + = + + iu kin: x 0.

( ) ( ) ( ) ( ) ( ) ( )2 1 24 . 4 2.3 3 2 6 3 0 2 2 3 2 3 3 2 3 0x x x x x xx x x x+ + + = + = ( )( ) ( ) ( )

223 32

2 3 02 3 2 3 0 log 2 log 2 .

2 3 0

xx

o

x x x xx x vn

= + = = = + =

BI TP T LUYN:


a) ( )2 1 13 3 3 7 2 0 + + =x x x x

b) ( )5 525 2.5 2 3 2 0 + =x x x x

c) ( )9 2 2 .3 2 5 0+ + =x xx x Bi 2: Gii cc phng trnh sau :

a) ( )2 3 23 3 10 .3 3 0 + + =x xx x

b) ( )3.4 3 10 .2 3 0+ + =x xx x

c) ( ) ( )2 2log log 22 2 . 2 2 1+ + = +x xx x HNG DN GII:

Bi 1: Gii cc phng trnh sau : a) ( )2 1 13 3 3 7 2 0 + + =x x x x b) ( )5 525 2.5 2 3 2 0 + =x x x x c) ( )9 2 2 .3 2 5 0+ + =x xx x

a) ( )2 1 13 3 3 7 2 0 + + =x x x x .

Ta nhn hai v phng trnh vi 3 ta c ( ) ( ) ( ) ( )2

2

3 03 3 3 7 3 2 0

3 7 3 2 0

= >+ + = + + =

xx x

tx x

t x t x

03 1

6 3( ) 3 3 6 0

1

> = =

= + = =

x

x

t

t xf x x

t

0

'( ) 3 ln3 3 0

= = + >

x

x

f x

Suy ra phng trnh f(x) = 0 c nghim duy nht x = 1. Vy phng trnh cho c nghim x = 0, x = 1.



b) ( ) ( )

55 5

2

05 0

25 2.5 2 3 2 0 12 2 3 2 0

2 3

> = > + = =

+ = =

xx x

tt

x x tt x t x

t x

5 5 55 2 3 ( ) 5 2 3 0 '( ) 5 ln 5 2 0 = = + = = = > + + = = =

+ + = =

xx x x

tt

x x xtt x t x

t x

( ) 3 2 5 0 '( ) 3 ln3 2 0 = + = = + >x xf x x f x Chng t f(x) lun ng bin. Mt khc f(1) = 0 nn phng trnh c nghim duy nht l x = 1. Bi 2: Gii cc phng trnh sau : a) ( )2 3 23 3 10 .3 3 0 + + =x xx x b) ( )3.4 3 10 .2 3 0+ + =x xx x c) ( ) ( )2 2log log 22 2 . 2 2 1+ + = +x xx x

a) ( ) ( ) ( ) ( )

22 22 3 2 2

2

3 03 3 10 .3 3 0 3.3 3 10 .3 3 0

3 3 10 3 0

= >+ + = + + = + + =

xxx x x

tx x x x

t x t x

2 12

22

013 31

'( ) 3 ln3 1 0( ) 3 3 03 3 3

3

> = = = + >= = + == =

xx

xx

tx

f xtf x xx

t x

Chng t f(x) lun ng bin. Mt khc f(2) = 0 nn phng trnh cho c hai nghim l x = 1 v x = 2.

b) ( ) ( )1

2

02 0 2 31

3.4 3 10 .2 3 033 3 10 . 3 0 2 33

> = > = + + = = + + = = =

x xx x

x

tt

x x tt x t x x

t x

2log 3 '( ) 2 ln 2 1 0( ) 2 3 0

= = + > = + =

x

x

xf x

f x x

Chng t f(x) lun ng bin. Mt khc f(1) = 0 nn f(x) = 0 c nghim duy nht x = 1. Vy chng t phng trnh cho c hai nghim x = 1 v 2log 3.= x

c) ( ) ( )2 2log log 22 2 . 2 2 1+ + = +x xx x . V ( ) ( ) ( ) ( )

2 2 22

2

log log loglog

log2 2 . 2 2 2 2 2

2 2+ = = =

+

x x xx

x

xx

Khi , phng trnh cho tr thnh :( )

( ) ( )( )( )

22

2

log log

2 2 2 2 2 log22

2 2 0 2 2 10 1

1 0 2 21 0

= + > + => = + + = = + =+ + =

x x

x

t t t

x t x t x t x xt xt

( )2

2 2 2

log 0 11

log 2 2 2log 2 log 0

= = =+ = =

x xx

x x x



V. PHNG PHP HM S GII PHNG TRNH M (tip)

Dng 3: S dng hm c trng gii phng trnh m

Phng php:

+ Bin i phng trnh cho v dng [ ] [ ]( ) ( )=f u x f v x ri xt hm c trng f(t) + Chng minh rng f(t) lun ng bin hoc nghch bin, khi ta thu c u(x) = v(x).

V d 1: Gii phng trnh sau:

a) ( )2 212 2 1 = x x x x \

b) ( )2 21 14 2 1 = x x x V d 2: Gii phng trnh sau:

a) 2 3 1 2 22 2 4 3 0 + + + =x x x x x \

b) 2 2cos sin cos2 =x xe e x

BI TP T LUYN


a) 21 2 1 23 3 4 .3 + =x x xx b)

2 24 2 2 8 4 25 5 4 2+ + + + = + +x x x x x x c) ( )2 2 2 2sin sin os os2 3 2 3 2cos2+ + =x x c x c x x Bi 2: Gii cc phng trnh sau :

a) 2 5 11 1

2 5 1 =

x xe e

x x b)

2

2 2

1 1 2 1 12 2

2

= x x

x x

x c)

2 3 1 2 22 2 3 3 0 + + + =x x x x x x

Bi 3: Gii phng trnh ( ) ( )0 0cos36 cos 72 3.2+ =x x x HNG DN GII:


a) 21 2 1 23 3 4 .3 + =x x xx b)

2 24 2 2 8 4 25 5 4 2+ + + + = + +x x x x x x c) ( )2 2 2 2sin sin os os2 3 2 3 2cos2+ + =x x c x c x x a)

2 2 21 2 1 2 2 1 2 13 3 4 .3 3 3 4 + + + + = =x x x x x x xPT x x Ta c ( ) ( )2 22 1 2 1 4 4+ + + = =x x x x x u v x . Phng trnh cho c dng 3 3 3 3 = + = +v u u vu v u v Xt hm s ( ) 3 '( ) 3 ln3 1 0= + = + >t tf t t f t . Suy ra f(t) ng bin, do ta c 4 0 0= = =u v x x b) ( ) ( )2 24 2 2 8 4 2 2 25 5 4 2 2 8 4 4 2+ + + + = + + = + + + +x x x xPT x x x x x x

( ) ( )2 24 2 2 2 8 4 25 4 2 5 2 8 4 ( ) ( )+ + + + + + + = + + + =x x x xx x x x f u f v Xt hm s ( ) 5 '( ) 5 ln5 1 0= + = + >t tf t t f t .

Suy ra f(t) ng bin, do ta c 22 2

4 2 02 2

= = + + =

= +

xu v x x

x

c) ( ) ( )2 2 2 2 2 2 2 2sin sin os os sin sin 2 os os 22 3 2 3 2cos 2 2 3 2sin 2 3 2cos + + = + + = + +x x c x c x x x c x c xPT x x x . Xt hm s ( ) 2 3 2 , '( ) 2 ln 2 3 ln3 2 0= + + = + + >t t t tf t t t R f t

Ti liu bi ging:




Suy ra ( )2 2 cos sin cos2 0 2 ;2 4 2

= = = = + x x x x k x k k Z


a) 2 5 11 1

2 5 1 =

x xe e

x x b)

2

2 2

1 1 2 1 12 2

2

= x x

x x

x c)

2 3 1 2 22 2 3 3 0 + + + =x x x x x x

a) 2 5 12

1 1 1 1( ) ; 0 '( ) 0

2 5 1 = = > = + >

x x t te e f t e t f t e

x x t t.

Chng t hm s f(t) ng bin. Do 3

2 5 14

= = =

xx x

x

b)

2

2 2

1 1 2 2 2

2 2 2

1 1 1 2 1 2 2 1 12 2 ; 1 2

2 2

= = = =

x x

x xx x x x

x x x x x x.

Cho nn phng trnh cho c dng ( )1 1 12 2 2 22 2 2

= + = +a b a bb a a b

Xt hm c trng 1 1

( ) 2 '( ) 2 .ln 2 02 2

= + = + >t tf t t f t .

Chng t hm f(t) lun ng bin. Suy ra 1 1

0 22

= =

xx

c) 2 23 1 2 2 3 1 2 22 2 3 3 0 2 3 1 2 2 + + + + = + + = + x x x x x xPT x x x x x x

Bng cch xt nh cc bi trn ta c kt qu

2 2 3 33 1 2 3 3 33 6 9 3

+ = = = = + =

x xx x x x x x x

x x x

Bi 3: Gii phng trnh ( ) ( )0 0cos36 cos 72 3.2+ =x x x Do 0 0 0 0 0cos72 sin18 ;cos36 sin54 sin3.18= = = . Cho nn t t= 0sin18 0= >t , v dng cng thc nhn ba ta c :

0 0 2 0 0 3 0 3 2cos36 sin 54 1 2sin 18 3sin18 4sin 18 4 2 3 1 0= = + =t t t

( )( )2 2 00

1 50

5 141 4 2 1 0 4 2 1 0 cos3645 1

sin184

=



HNG DN GII:

a) ( ) ( ) ( ) ( ) ( ) ( )( )2.2 2 2 1 3 .2 2.2 2 3 0 2 2 1 2 0= + + + = + =x x x x xx x x x x x x x x

( )( ) 2 0 2 22 2 1 0(0) 0( ) 2 1 0 '( ) 2 ln 2 1 0

= = = + = == + = = + >

x

x x

x x xx x

ff x x f x

D dng tm dc hai nghim ca phng trnh l x = 0 v x = 2.

b) ( )22 2 2 2 211 2 2 1 2 14 2 2 1 2 2 2 1 ++ = + + = +xx x x x x x x x .

t : 2 2 22 2 ; 1 2 1= = + = +a x x b x a b x x . Khi phng trnh c dng :

( ) ( ) ( )( ) 2 1 02 2 2 1 2 1 2 1 2 0 2 1 1 2 002 1

+ = = + = + + = = ==

aa b a b a b a a b

b

a

b

( )22

2 0 0; 10; 1

1; 11 0

= = = = = = = =

x x x xx x

x xx

c) ( ) ( ) ( ) ( )3 2 3 4 3 4 3 2 3 22 1 2 1 2 1 1 2 1 2 2.2 2 .2 2 .2 2 .2 2 2 4 1 2 4 1 + + + + ++ + + = + = = x x x x xx x x x xx x x x x x

( )( )2

3 22 1

3 2 1

1 1 14 1 02 24 1 2 2 0 2

2 2 0 3 2 1 3 3 3

+ +

= = = = = = + = =

x x

x x

x xx xx

x x x x x

Dng 2: Phng php nh gi hai v


a) xx 2cos32

= b) ( )2 1cos 22 2 += + xx x c) 242 2 16+ = x x x V d 2: Gii phng trnh

a) 2

22 12

+=x x xx

b) 3 22 8 14 = + x x x c) 32.6 4 3.12 2.8 2.3 + =x x x x x

BI TP T LUYN:

Gii cc phng trnh sau

a) = 42 cos ,x x vi x 0 b) + = + 2 6 10 23 6 6x x x x

c) =sin3 cosx x d) = +

322cos 3 3

2x xx x

e) sin cos=

xx f)

2 2sin os3 3 2 2 2+ = + +x c x x x



VII. PHNG TRNH M C THAM S

V d 1. Tm m phng trnh sau c nghim duy nht: .2 2 5 0+ =x xm

/s: 25

; 04

= m m b) m > 2

c) m < 1 d) 39

1 log 24

+m

Bi 4. Tm m phng trnh 1 3 1 34 14.2 8+ + + + + =x x x x m c nghim duy nht.

Hng dn:

t 1 3 2 2 2= + + t x x t

Ti liu bi ging:




Bi 5. Tm m phng trnh 2 21 19 8.3 4+ + + =x x x x m c nghim duy nht.

Hng dn:

t 21 1 2= + t x x t



I. PHNG TRNH C BN

Khi nim: L phng trnh c dng ( )log ( ) log ( ), 1 .=a af x g x trong f(x) v g(x) l cc hm s cha n x cn gii. Cch gii:

- t iu kin cho phng trnh c ngha

0; 1

( ) 0

( ) 0

> > >

a a

f x

g x

- Bin i (1) v cc dng sau: ( ) ( ) ( )11

=

=f x g x

a

Ch :

- Vi dng phng trnh log ( ) ( )= = ba f x b f x a

- y ly tha bc chn: 2log 2 log=na ax n x , nu x > 0 th log log=n

a an x x

- Vi phng trnh sau khi bin i c v dng [ ]2

( ) 0( ) ( )

( ) ( )

= =

g xf x g x

f x g x

- Cc cng thc Logarith thng s dng: ( )

loglog ;

log log log ; log log log

1log log ; log

log

= =

= + =

= =

a

n

xxa

a a a a a a

ma aa

b

a x a x

xxy x y x y

y

mx x b

n a

V d 1. Gii phng trnh

a) log5(x2 11x + 43) = 2 b) log3(2x + 1) + log3(x 3) = 2

c) ( )2log 2 3 4 2 =x x x d) ( )21log 3 1 1+ + =x x x V d 2. Gii phng trnh

a) ( ) ( )4 4 4log 3 log 1 2 log 8+ = x x b) ( )lg 9 2 lg 2 1 2 + =x x

c) 2 21

log log ( 1)( 4) 24

+ + =+

xx x

x d) 28 8

42log (2 ) log ( 2 1)

3+ + =x x x


a) 2 34 82log ( 1) 2 log 4 log (4 )+ + = + +x x x b) 2 2

4 4 4log ( 1) log ( 1) log 2 = x x x

c) ( )29 3 32log log .log 2 1 1= + x x x d) 115

log (6 36 ) 2+ = x x


Ti liu bi ging:

05. PHNG TRNH LOGARITH P1 Thy ng Vit Hng



a) 4 2 2 4log (log ) log (log )=x x b) 2 3 4 20log log log log+ + =x x x x

BI TP T LUYN

Bi 1. Gii cc phng trnh sau:

a) 2log ( 1) 1x x = b) 2 2log log ( 1) 1x x+ =

c) =2 18

log ( 2) 6.log 3 5 2x x d) 2 2log ( 3) log ( 1) 3x x + =


a) lg( 2) lg( 3) 1 lg5x x + = b) 8 822 log ( 2) log ( 3)3

x x =

c) lg 5 4 lg 1 2 lg0,18x x + + = + d) 23 3log ( 6) log ( 2) 1x x = +


a) 2 2 5log ( 3) log ( 1) 1/ log 2x x+ + = b) 4 4log log (10 ) 2x x+ =

c) + =5 15

log ( 1) log ( 2) 0x x d) 2 2 2log ( 1) log ( 3) log 10 1x x + + =


a) 9 3log ( 8) log ( 26) 2 0x x+ + + = b) 3 1/33log log log 6x x x+ + =

c) 2 21 lg( 2 1) lg( 1) 2 lg(1 )x x x x+ + + = d) + + =4 1 816

log log log 5x x x


a) 2 22 lg(4 4 1) lg( 19) 2 lg(1 2 )x x x x+ + + = b) 2 4 8log log log 11x x x+ + =

c) + + = + 1 1 12 2 2

log ( 1) log ( 1) 1 log (7 )x x x d) 116

log (5 25 ) 2x x+ =


a) 2log (2 7 12) 2xx x + = b) 2log (2 3 4) 2

xx x =

c) 22log ( 5 6) 2x x x + = d) 2log ( 2) 1

xx =


a) 23 5log (9 8 2) 2x x x+ + + = b) 2

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

c) 15log 21 2x x

=

d) 2log (3 2 ) 1x

x =

e) 2 3log ( 3) 1x x x+ + = f) 2log (2 5 4) 2

xx x + =



I. PHNG TRNH C BN (tip theo)


a) 13log2)5(log3

182 =+ xx b) 2 2log (4.3 6) log (9 6) 1 =

x x

c) 13

)29(log 2 =

x

x

d) 1lg

2lg

1lg

lg2

+=

xx

x

x


a) 4

21 2log (10 )

log+ =x x x

b)

=+x

xx x

11

4

75log

2log

13

232

c) 23

lg( 2 3) lg 01

++ + =

xx x

x d) ( )9 3log log 4 5+ =x x


a) [ ]{ }4 3 2 2log 2log 1 log (1 3log ) 1x+ + = b) 4 82log 4log log 13x x x+ + =

c) 3 9 817log log log2

x x x+ + = d) xx

xx2log

log

log.log125

5

255 =


a) 2 29 331 1

log ( 5 6) log log 32 2

+ = + xx x x

b) 84 221 1

log ( 3) log ( 1) log 42 4

+ + =x x x

c) ( )4 1lg 3 2 2 lg16 lg 44 2

= + x x x

d) 2 2 4 2 4 22 2 2 2log ( 1) log ( 1) log ( 1) log ( 1)+ + + + = + + + +x x x x x x x x

e) 21 1

lg( 5) lg5 lg2 5

+ = +x x xx

II. PHNG TRNH BC HAI, BC BA THEO MT HM LOGARITH

V d 1. Gii phng trnh sau

a) 22 22log 14 log 3 0 + =x x b) 2 32 2log log 4 0+ =x x

c) 3 22 2log (2 ) 2 log 9= x x d) 3 31

log log 3 log log 32

+ = + +x xx x

BI TP T LUYN:


a) ( ) ( )21 13 3

log 3 4 log 2 2x x x+ = + b) ( )1lg lg 12

x x= +

Ti liu bi ging:




c) 2 12

8 1log log

4 2

xx

= d) ( )25log 2 65 2x x x + = Bi 2: Gii cc phng trnh sau: a) ( ) ( )lg 3 2lg 2 lg0,4x x+ =

b) ( ) ( )5 5 51 1

log 5 log 3 log 2 12 2

x x x+ + = +

c) ( )2 12

1log 4 15.2 27 2log 0

4.2 3x x

x

+ + =


a) ( ) ( )222 2log 1 5 log 1x x = + b) ( ) ( )22 14

log 2 8log 2 5x x =

c) 1 13 3

log 3. log 2 0x x + = d) 2

21 2

2

log (4 ) log 88

+ =xx


a) 2 23 3log log 1 5 0x x+ + = b) + + =2

2 122

log 3log log 2x x x

c) 51log log 25x

x = d) 71log log 27x

x =

e) =22 14

log (2 ) 8log (2 ) 5x x f) 25 25log 4 log 5 5 0x x+ =

HNG DN GII:


a) ( ) ( )21 13 3

log 3 4 log 2 2x x x+ = + b) ( )1lg lg 12

x x= +

c) 2 12

8 1log log

4 2

xx

= d) ( )25log 2 65 2x x x + =

a) ( ) ( )2

21 1

3 3 2 2

1

4 13 4 0

log 3 4 log 2 2 2 2 0 1 2.2

33 4 2 2 6 0

x

x xx x

x x x x x xx

xx x x x x

> < >+ >

+ = + + > > == = + = + + =

Vy phng trnh c nghim x = 2.

b) ( )( ) ( ) ( )

2 2

0

0 1 50 01 1 5lg lg 1 1 0 2lg lg 12 21

2lg lg 1 1 5

2

x

x x x xx x x xx x x x

x xx

> > +> > + == + + > = = + = + = + =

Vy phng trnh cho c nghim 1 5

.2

x+=

c) ( )2 12

8 1log log , 3 .

4 2

xx

=

iu kin: 8 0

0 8.0

xx

x

> <



Khi ( ) ( )1

22 2

8 1 8 8 13 log log 8 4

4 2 4 4

x x xx x x x

x

= = = =

( )22 8 16 4 0 4.x x x x + = = = Nghim x = 4 tha mn iu kin, vy phng trnh c nghim x = 4.

d) ( ) ( )25log 2 65 2, 4x x x + =

iu kin:

( )2 2

5 0 55

5 1 44

2 65 0 1 64 0,

x xx

x xx

x x x x R

> < + >

Khi ( ) ( )224 2 65 5 8 40 0 5.x x x x x + = + = = Nghim x = 5 tha mn iu kin, vy phng trnh c nghim x = 5. Bnh lun: Trong cc v d 3 v 4 chng ta cn phi tch ring iu kin ra gii trc ri sau mi gii phng trnh. v d 1 v 2 do cc phng trnh tng i n gin nn ta mi gp iu kin vo vic gii phng trnh ngay.

Bi 2. Gii cc phng trnh sau: a) ( ) ( )lg 3 2lg 2 lg0,4x x+ =

b) ( ) ( )5 5 51 1

log 5 log 3 log 2 12 2

x x x+ + = +

c) ( )2 12

1log 4 15.2 27 2log 0

4.2 3x x

x

+ + =

a) ( ) ( ) ( )lg 3 2lg 2 lg0,4, 1 .x x+ =

iu kin: 3 0 3

2.2 0 2

x xx

x x

+ > > > > >

Khi , ( ) ( ) ( ) ( )( )

( )( )

( ) ( )2 22 23 3 2

1 lg 3 lg 2 lg0,4 lg lg0,4 0,4 2 2 5 3 052 2

x xx x x x

x x

+ + + = = = = + =

27

2 13 7 0 1

2

xx x

x

= = =

i chiu vi iu kin ta c nghim ca phng trnh l x = 7.

b) ( ) ( ) ( )5 5 51 1

log 5 log 3 log 2 1 , 2 .2 2

x x x+ + = +

iu kin:

5 0 5

3 0 3 3.

2 1 0 1

2

x x

x x x

xx

+ > > > > >

+ > >

Khi , ( ) ( ) ( ) ( ) ( )( ) ( )5 5 5 5 51 1 1

2 log 5 log 3 log 2 1 log 5 3 log 2 12 2 2

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

( )( ) 2 25 3 2 1 2 15 2 1 16 4.x x x x x x x x + = + + = + = = i chiu vi iu kin ta c nghim ca phng trnh l x = 4.

c) ( ) ( )2 12

1log 4 15.2 27 2log 0, 3 .

4.2 3x x

x

+ + =



iu kin: 4 15.2 27 0,

4.2 3 0

x x

x

x R + + >

>

Khi ( ) ( ) ( )2

2 2 21 1

3 log 4 15.2 27 2log 0 log 4 15.2 27 04.2 3 4.2 3

x x x xx x

+ + + = + + =

( )2 2

22

2 31 2 15.2 27

4 15.2 27 1 1 15.2 39.2 18 0 24.2 3 16.2 24.2 9 2 0

5

xx x

x x x xx x x x

=+ + + + = = = + = 1.

t ( ) ( ) ( ) ( )2 22 22 2

2 2 2 2log 1 log 1 log 1 2log 1 4t x x x x t = = = =

Khi ( )( )

( )2

2

5 52

4 4

1 3log 1 11 12 21 4 5 0 5 5

log 14 4 1 2 1 2

xt x xt t

t xx x

= = = = = = = = = +

C hai nghim u tha mn iu kin, vy phng trnh cho c hai nghim l 5

43 ; 1 2 .2

x x= = +

b) ( ) ( ) ( )22 14

log 2 8log 2 5, 2 .x x =

iu kin: x < 2.

( ) ( ) ( ) ( ) ( ) ( )( )2 2 22 2 2 2

2

log 2 182 log 2 log 2 5 log 2 4log 2 5 0

log 2 52

xx x x x

x

= = + = =

Vi ( )2log 2 1 2 2 0.x x x = = =

Vi ( )21 63

log 2 5 2 .32 32

x x x = = =

C hai nghim u tha mn iu kin, vy phng trnh cho c hai nghim l 63

0; .32

x x= =

c) ( )1 13 3

log 3. log 2 0, 3 .x x + =

iu kin: 1

3

00 1.log 0

xxx

> <

( )2 1 1

3 31 1

13 3 133

1log 1 log 133 log 3. log 2 0

log 4 1log 281

x x xx x

xx x

= = = + = = = =

C hai nghim u tha mn iu kin, vy phng trnh cho c hai nghim l 1 1

; .3 81

x x= =



d) ( )2

21 2

2

log (4 ) log 8, 4 .8

+ =xx

iu kin: x > 0.

Ta c [ ] ( ) ( )

222 22

1 1 2 2 2 2

2 2

22

2 2 2 2

log (4 ) log (4 ) log (4 ) log 4 log log 2

log log log 8 2log 38

= = = + = +

= =

x x x x x

xx x

Khi ( ) ( ) ( )2 2 22 2 2 2 72

2log 1

4 log 2 2log 3 8 log 6log 7 0 1log 7 2

128

xx

x x x xx x

== + + = + = = = =

Vy phng trnh cho c hai nghim 1

2; .128

x x= =



II. PHNG TRNH BC HAI, BC BA THEO MT HM LOGARITH (tip theo)


a) 2 21 22

log 8 log 4 2+ =x x b) 2

24 2log 16 log 114

+ =xx c) 23log (9 ) log 27 7+ =xx


a) 2820

2log 4 log3

+ =x x b) 2 3 21 3

9

2log (3 ) log 3 log =x

x x c) 2

2

2log 2log 32 10

4+ =x

x


a) 3 2log 10 log 10 6log 10 0 =x x x b) 52log log 125 1 0 =xx

c) ( )22 2log 1 6log 1 2 0+ + + =x x d) 33loglog3 33 = xx d) 2log 5 log 5 2,25 log 5+ =x x xx

BI TP T LUYN:


a) 2

21 422

313log (8 ) 2 log (4 ) log

2 2+ + =xx x (/s: 1

2=x )

b) 2 3

21 4 24

32log log 8 3log

4 16 2+ = x xx (/s: x = 4)

c) 2

2 29 93

log 2log (3 ) log (27 ) 83

+ + =x x x (/s: x = 3)

d) 92327

log (9 ) log log (3 ) 3 0+ + + =xx xx

(/s: 1

3=x )


a) 2411

log (4 ) log (8 )2

+ =xx x (/s: x = 4)

b) 231 21

3log (9 ) log (3 )2 2

+ =xx x (/s: 3=x )

c) 2225log (125 ) 2log (5 ) 5+ =xx x (/s: 5=x )


a) 2 23 3log 1 log 5 0+ + =x x (/s: 33=x )

b) 2 21 24

log log log (4 )+ = xx x x (/s: 1 1

2; ;2 4

= = =x x x )

Ti liu bi ging:




III. PHP PHP T N PH GII PHNG TRNH LOGARITH


a) 2 22log log (4 ) 12+ =xx x b) ( )3 93

42 log log 3 1

1 log =

xx

x

c) 2 25log (5 ).log 5 1=xx d) 3 3

2 2

2log log

3 = x x


a) 2 23 log log (4 ) 0 =x x b) 1 4

35 4lg 1 lg

+ = +x x

c) 22log 64 log 16 3+ =x x d) 2 2log 2 2log 4 log 8+ =x x x


a) 22

327log 3log 0 =xx x x b)

3 32 2log 2 3 log 2+ = x x

c) 2 4log 2log 2 log 2=x x x d) 2

2 2 23log 1 4 log 13log 5+ = + x x x

BI TP T LUYN:


a) 51

2log 2 log5x

x = b) 29log 5 log 5 log 54x x x

x+ = +

c) 2 32 16 4log 14log 40log 0x x xx x x + = d) 3

3 2 3 23 1

log .log log log23

xx x

x

= +

chuyen de pt, bpt, hpt mu va logarit - chuyên Đề Ôn thi 5 2 2 2 2 .2 .2 2 .2 2 .2 2 23 2 2 2 25...

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