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8/9/2019 phuongtrinhlogarit

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LUYN TPGII PHNG TRNH

LNGGIC

LUYN TP GIIPHNG TRNHLNG GICxt vn x Phng php av cng c sx

Phng php tn s phx

x Phng php sdng tnh n iux Phng phpnh gi

x

x Phng php th

Cc hot ng

Gii phng trnh lgarit-Diendankienthuc.net slide th 2/ 15
http://diendankienthuc.net/diendan/http://diendankienthuc.net/diendan/


3/61

tvn


Phng php tn s phx


x

x Phng php th

Cc hot ng


Trong bi trc chng ta hc phng trnh lgaritv cch gii mt s phng trnh lgarit thng gp.Hm nay chng ta s luyn tp v cch gii cc phngtrnh lgarit.


4/61

Phngphpavcngcs


Phng php tn s phx


x

x Phng php th

Cc hot ng


Vd1. Gii phng trnhlog3 3x + log9 x + log27 x = 12.


5/61

Phngphpavcngcs


Phng php tn s phx


x

x Phng php th

Cc hot ng


Vd1. Gii phng trnhlog3 3x + log9 x + log27 x = 12.

Hng gii. t iu kin x > 0.Bin i a v dng phng trnh n log3 x.


6/61

Phngphptnsph


Phng php tn s phx


x

x Phng php th

Cc hot ng


Phng php: t iu kin cc biu thc ca phng trnh

c ngha.

La chn n ph cho phng trnh v iu kincho cc n ph.

Gii phng trnh mi nhn c. Kim tra tnh hp l.


7/61


Phng php tn s phx


x

x Phng php th

Cc hot ng


Vd2. Gii phng trnhlog2 (3

x 1). log2 (2.3x 2) = 2.


8/61


9/61


10/61


Phng php tn s phx


x

x Phng php th

Cc hot ng



x 1). log2 (2.3x 2) = 2.

Hng gii. Phng trnh tng ng vilog2 (3

x 1).[1 + log2 (3x 1) = 2.

iu kin 3x 1 > 0 x > 0.


11/61


Phng php tn s phx


x

x Phng php th

Cc hot ng



x 1). log2 (2.3x 2) = 2.


x 1).[1 + log2 (3x 1) = 2.

iu kin 3x 1 > 0 x > 0.t t = log2 (3x 1), phng trnh tr thnh

t(t + 1) = 2


12/61


Phng php tn s phx


x

x Phng php th

Cc hot ng



x 1). log2 (2.3x 2) = 2.


x 1).[1 + log2 (3x 1) = 2.


t(t + 1) = 2

t = 1

t =

2


13/61


Phng php tn s phx


x

x Phng php th

Cc hot ng



x 1). log2 (2.3x 2) = 2.


x 1).[1 + log2 (3x 1) = 2.


t(t + 1) = 2

t = 1

t =

2

x =

1

x = log3

5

4


14/61

Phngphpsdngtnhniu


Phng php tn s phx


x

x Phng php th

Cc hot ng


Vd3. Gii phng trnhlog2 (x

2 4) + x = 3 + log2 (x + 2).


15/61

Phngphpsdngtnhniu


Phng php tn s phx


x

x Phng php th

Cc hot ng



2 4) + x = 3 + log2 (x + 2).

Hng gii. iu kin

x2 4 > 0x + 2 > 0

x > 2.


16/61

Phngphpsdngtnhniu

LUYN TP GIIPHNG TRNHLNG GICxt vn x Phng php av cng c sx Phng php tn s phx


x

x Phng php th

Cc hot ng



2 4) + x = 3 + log2 (x + 2).

Hng gii. iu kin

x2 4 > 0x + 2 > 0

x > 2.

Phng trnh tng ng vi

log2 (x 2) = 3 x.


17/61

Phngphpsdngtnhniu



x

x Phng php th

Cc hot ng



2 4) + x = 3 + log2 (x + 2).

Hng gii. iu kin

x2 4 > 0x + 2 > 0

x > 2.

Phng trnh tng ng vi

log2 (x 2) = 3 x.

Nhn thyx = 3 l nghim.


18/61

Phngphpsdngtnhniu



x

x Phng php th

Cc hot ng



2 4) + x = 3 + log2 (x + 2).

Hng gii. iu kin

x2 4 > 0x + 2 > 0

x > 2.

Phng trnh tng ng vi

log2 (x 2) = 3 x.


Vi x > 3 th log2 (x 2) > log2 1 = 0, v 3 x < 0. Do phng trnh v nghim.


19/61

Phngphpsdngtnhniu



x

x Phng php th

Cc hot ng



2 4) + x = 3 + log2 (x + 2).

Hng gii. iu kin

x2 4 > 0x + 2 > 0

x > 2.

Phng trnh tng ng vi

log2 (x 2) = 3 x.


Vi x > 3 th log2 (x 2) > log2 1 = 0, v 3 x < 0. Do phng trnh v nghim.Vi x < 3 th log2 (x 2) < log2 1 = 0, v 3 x > 0. Do phng trnh v nghim.


20/61

Phngphpsdngtnhniu



x

x Phng php th

Cc hot ng



2 4) + x = 3 + log2 (x + 2).

Hng gii. iu kin

x2 4 > 0x + 2 > 0

x > 2.

Phng trnh tng ng vi

log2 (x 2) = 3 x.


Vi x > 3 th log2 (x 2) > log2 1 = 0, v 3 x < 0. Do phng trnh v nghim.Vi x < 3 th log2 (x 2) < log2 1 = 0, v 3 x > 0. Do phng trnh v nghim.

Vy phng trnh c nghim duy nht x = 3.


21/61

Phng php nh gi



x

x Phng php th

Cc hot ng


S dng:s Tam thc bc hai.


22/61

Phng php nh gi



x

x Phng php th

Cc hot ng



s

Cc BT c bn nh Cauchy, Bunhiacopsky.


23/61

Phng php nh gi



x

x Phng php th

Cc hot ng



s

Cc BT c bn nh Cauchy, Bunhiacopsky.s Tnh cht ca gi tr tuyt i.


24/61



x

x Phng php th

Cc hot ng


Vd4. Gii phng trnhlog3

2 (

4 x +

x + 5) = 1.


25/61



x

x Phng php th

Cc hot ng



2 (

4 x +

x + 5) = 1.

Hng gii. iu kin5 x 4.


26/61



x

x Phng php th

Cc hot ng



2 (

4 x +

x + 5) = 1.

Hng gii. iu kin5 x 4.Theo BT Bunhiacopsky ta c


27/61



x

x Phng php th

Cc hot ng



2 (

4 x +

x + 5) = 1.


4

x +

x + 5

(12 + 12)(4

x + x + 5) = 3

2.


28/61



x

x Phng php th

Cc hot ng



2 (

4 x +

x + 5) = 1.


4

x +

x + 5

(12 + 12)(4

x + x + 5) = 3

2.

Suy ra log32 (

4 x +x + 5) < log32 3

2 = 1.


29/61



x

x Phng php th

Cc hot ng



2 (

4 x +

x + 5) = 1.


4

x +

x + 5

(12 + 12)(4

x + x + 5) = 3

2.

Suy ra log32 (

4 x +x + 5) < log32 3

2 = 1.

Du = xy ra khi v ch khi4

x

1 =

x

+ 51 x = 12 .


30/61



x

x Phng php th

Cc hot ng



2 (

4 x +

x + 5) = 1.


4

x +

x + 5

(12 + 12)(4

x + x + 5) = 3

2.

Suy ra log32 (

4 x +x + 5) < log32 3

2 = 1.

Du = xy ra khi v ch khi4

x

1 =

x

+ 51 x = 12 . Vyx = 12 l nghimduy nht ca bi ton.


31/61

Phng php th



x

x Phng php th

Cc hot ng


Vd5. Hy bin lun theo m s nghim ca phngtrnhlg (m x2) = lg (x2 3x + 2).


32/61

Phng php th



x

x Phng php th

Cc hot ng



Hng gii. Phng trnh tng ng vix2 3x + 2 > 0m

x2 = x2

3x + 2

x > 2 x < 12x2

3x + 2 = m

2


33/61

Phng php th



x

x Phng php th

Cc hot ng



Hng gii. Phng trnh tng ng vix2 3x + 2 > 0m

x2 = x2

3x + 2

x > 2 x < 12x2

3x + 2 = m

2

S nghim ca phng trnh cho l s honh giao im ca parabol y = 2x2 3x + 2 vi ng

thng y = m.


34/61

Cc hot ng

LUYN TP GIIPHNG TRNHLNG GIC

Cc hot ngxHot ng 1xHot ng 2

xHot ng 3xHot ng 3



35/61

Hotng1



xHot ng 3xHot ng 3


Tm phng php gii cc phng trnh sau:


36/61

Hotng1



xHot ng 3xHot ng 3


Tm phng php gii cc phng trnh sau:1)

log2 x

log4 2x=

log8 9x

log16 8x.


37/61

Hotng1



xHot ng 3xHot ng 3



log2 x

log4 2x=

log8 9x

log16 8x.

2) log5(x2 + 1) + log1/5 5 = log5 (x + 2)

2log1/25 x

2.


38/61

Hotng1



xHot ng 3xHot ng 3



log2 x

log4 2x=

log8 9x

log16 8x.

2) log5(x2 + 1) + log1/5 5 = log5 (x + 2)

2log1/25 x

2.

3) (x + 2) log23(x + 1) + 4(x + 1) log3 (x + 1) 16 = 0.


39/61

Hotng1



xHot ng 3xHot ng 3



log2 x

log4 2x=

log8 9x

log16 8x.

2) log5(x2 + 1) + log1/5 5 = log5 (x + 2)

2log1/25 x

2.

3) (x + 2) log23(x + 1) + 4(x + 1) log3 (x + 1) 16 = 0.4) log2(x + 3log6 x) = log6 x.


40/61

Hotng1



xHot ng 3xHot ng 3



log2 x

log4 2x=

log8 9x

log16 8x.

2) log5(x2 + 1) + log1/5 5 = log5 (x + 2)

2log1/25 x

2.

3) (x + 2) log23(x + 1) + 4(x + 1) log3 (x + 1) 16 = 0.4) log2(x + 3log6 x) = log6 x.5) Tu theo m hy bin lun s nghim ca phngtrnh


41/61

Hotng1



xHot ng 3xHot ng 3



log2 x

log4 2x=

log8 9x

log16 8x.

2) log5(x2 + 1) + log1/5 5 = log5 (x + 2)

2log1/25 x

2.

3) (x + 2) log23(x + 1) + 4(x + 1) log3 (x + 1) 16 = 0.4) log2(x + 3log6 x) = log6 x.5) Tu theo m hy bin lun s nghim ca phngtrnh

logm2+2(1 x2 + m2) = logm2+2[2(1 + x + m)].


42/61

Hotng2



xHot ng 3xHot ng 3


Thi ua gii cc phng trnh sau:


43/61

Hotng2



xHot ng 3xHot ng 3


Thi ua gii cc phng trnh sau:log5

5

x

log5 5x

+ log25 x = 1.


44/61

Hotng2



xHot ng 3xHot ng 3



5

x

log5 5x

+ log25 x = 1.

Li gii.


45/61

Hotng2



xHot ng 3xHot ng 3



5

x

log5 5x

+ log25 x = 1.

Li gii.iu kin x > 0.


46/61

Hotng2



xHot ng 3xHot ng 3



5

x

log5 5x

+ log25 x = 1.

Li gii.iu kin x > 0. P T 1 log5 x1 + log5 x

+ log25 x = 1.


47/61

Hotng2



xHot ng 3xHot ng 3



5

x

log5 5x

+ log25 x = 1.


+ log25 x = 1.

t t = log5

x th phng trnh tr thnh


48/61

Hotng2



xHot ng 3xHot ng 3



5

x

log5 5x

+ log25 x = 1.


+ log25 x = 1.

t t = log5


1 t1 + t

+ t2 = 1 t3 + t2 2t = 0 t = 0 t = 1 t = 2.


49/61

Hotng2



xHot ng 3xHot ng 3



5

x

log5 5x

+ log25 x = 1.


+ log25 x = 1.

t t = log5


1 t1 + t

+ t2 = 1 t3 + t2 2t = 0 t = 0 t = 1 t = 2.

T tm c x = 1, x = 5, x = 125

.

H 3


50/61

Hotng3



xHot ng 3xHot ng 3


Tm sai lm trong li gii bi ton sau v a ra li giing:

H 3


51/61

Hotng3



xHot ng 3xHot ng 3


Tm sai lm trong li gii bi ton sau v a ra li giing:Gii phng trnh

H 3


52/61

Hotng3



xHot ng 3xHot ng 3



32

log 14

(x + 2)2 3 = log 14

(4 x)3 + log 14

(x + 6)3

H t 3


53/61

Hotng3



xHot ng 3xHot ng 3



32

log 14

(x + 2)2 3 = log 14

(4 x)3 + log 14

(x + 6)3

Li gii.

H t 3


54/61

Hotng3



xHot ng 3xHot ng 3



32

log 14

(x + 2)2 3 = log 14

(4 x)3 + log 14

(x + 6)3

Li gii.iu kin

(x + 2)2 > 04 x > 0x + 6 > 0

H t 3


55/61

Hotng3



xHot ng 3xHot ng 3



32

log 14

(x + 2)2 3 = log 14

(4 x)3 + log 14

(x + 6)3

Li gii.iu kin

(x + 2)2 > 04 x > 0x + 6 > 0

6 < x < 4, x = 2.

H t 3


56/61

Hotng3



xHot ng 3xHot ng 3



32

log 14

(x + 2)2 3 = log 14

(4 x)3 + log 14

(x + 6)3

Li gii.iu kin

(x + 2)2 > 04 x > 0x + 6 > 0

6 < x < 4, x = 2.

Phng trnh tng ng vi

Hotng3


57/61

Hotng3



xHot ng 3xHot ng 3


3log1/4(x + 2) 3 = 3 log1/4(4 x) + 3 log1/4(x + 6) log1/4

x + 2

(4

x)(6 + x)= 1

x + 2(4 x)(6 + x) =

1

4

x2 + 6x

16 = 0

x =

8 (loi)

x = 2.

Vy phng trnh cho c nghim x = 2.

Hotng3


58/61

Hotng3



xHot ng 3xHot ng 3



x + 2

(4

x)(6 + x)= 1

x + 2(4 x)(6 + x) =

1

4

x2 + 6x

16 = 0

x =

8 (loi)

x = 2.


Hotng3


59/61

Hotng3



xHot ng 3xHot ng 3



x + 2

(4

x)(6 + x)= 1

x + 2(4 x)(6 + x) =

1

4

x2 + 6x

16 = 0

x =

8 (loi)

x = 2.


Hotng3


60/61

Hotng3



xHot ng 3xHot ng 3



x + 2

(4

x)(6 + x)= 1

x + 2(4 x)(6 + x) =

1

4

x2 + 6x

16 = 0

x =

8 (loi)

x = 2.

Hotng3


61/61

Hotng3



xHot ng 3xHot ng 3


x + 2

(4

x)(6 + x)= 1

x + 2(4 x)(6 + x) =

1

4

x2 + 6x

16 = 0

x =

8 (loi)

x = 2.


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