Logarit

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Bi Gia Phong Gio vin Trng THPT Trng Vnh K Bn Tre.

http://giaphong.schools.officelive.com/ Trang 1

PHNG TRNH; H PHNG TRNH;BT PHNG TRNH M & LGARIT.

MC LC.

BI:

I/ PHNG TRNH CHA N S TRONG S M. Trang 02.

II/ PHNG TRNH CHA N S TRONG BIU THC LGARIT. Trang 03.

III/ MT S BI TP LIN QUAN N PHNG TRNHM & LGARIT (C CHA THAM S). Trang 04.

IV/ BT PHNG TRNH M & LGARIT. Trang 05.

V/ H PHNG TRNH M & LGARIT. Trang 06.

HNG DN GII:

I/ PHNG TRNH CHA N S TRONG S M. Trang 07.

II/ PHNG TRNH CHA N S TRONG BIU THC LGARIT. Trang 11.

III/ MT S BI TP LIN QUAN N PHNG TRNHM & LGARIT (C CHA THAM S). Trang 18.

IV/ BT PHNG TRNH M & LGARIT. Trang 23.

V/ H PHNG TRNH M & LGARIT. Trang 28.

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I/ PHNG TRNH CHA N S TRONG S M.

Gii cc phng trnh sau y:

1) x 3(0,04) 625 5.

2) 1 x8

0,125.16 .32

3)2 28 x 8 x 5 1 x2 .5 0,001.(10 ) .

4) 2x 1 3x 3x 33 .15 .5 9. 5) x x x x5.3 3.2 7.2 4.3 . 6) x x 1 x 2 x 1 x 1 x 25 5 5 3 3 3 . 7) 1 2x x2 15.2 8 0. 8) 2x 1 2 x5 5 124.

9) 2 2x 2 x3 2.3 27 0.

10) x x

3 35 2 6 5 2 6 10.

11) x x x25 15 2.9 . 12) x 1 x x 19 13.6 4 0. 13) x x 2x 149 2.35 7.5 0.

14)2 2sin x cos x9 9 6.

15)2 21 2sin x 2cos x4 9.4 5.

16)2cos x5 sinx.

17)2 2sin x cos x4 4 6 cos2x.

18) 3 3x 3 3x 4 x 4 x 33 3 3 3 10 . 19) x 1 x5 25 6.

20)x x5 33 5 .

21) x 5 2x3 2 . 22) x3 5 2x.

23)x

x 23 5 4. 24) x x4 (x 14)2 33 3x 0.

25) 2 x 1 x2 x 3 3x 1 4.3 1. 26) 2 x 1 x x 1 x 1 xx .5 3 3.5 x 2.5 3 0.

27) 3 3x x x x36 2 3 9.8 4.27 .

28) x 1 x 3(x 2) (x 2) .

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II/ PHNG TRNH CHA N S TRONG BIU THCLGARIT.

Gii cc phng trnh sau y:

1) x 13log (3 26) 2 x.

2) 3 3log (x 2) log (x 6) 2.

3) 33

5 0,2 25log x log x log x 7.

4) 4 2 4log (x 3) log x 1 2 log 8.

5) 22 0,5log (x 3) log (6x 10) log10 0.

6) 3 2log 3x 15x 15x 4 3log(x 2) 0.

7)

24

4 2

xlog log (4x) 10 0.4

8) 9

6 3

1 4log 5 x1.

log (2 x) log (2 x)

9)3( 2 log x 1,5 logx)x 10.

10) 3 13

10 log x 3 log x.

11) 5log(log x) log log(x ) 4 0.

12)2

5x 5

5log log x 1.x

13) 2 3x 16x 4x2

log x 14log x 40log x 0.

14)3 5

3 2 3 210log x.log x 15 2log x 5log x . 15) 22log x ln x 2ln x.log x log x.

16)2

5 5log 15 log (5x )25x 22.x 5.3 0. 17) 2log(x 3x 4) log(x 4) 2 x.

18) log( x 7 )3 x.

19) 2 2 2log 25 log x log 52x x .5 x . 20) x ln x x(3 x) 2(1 ln x).

21) 25 5(x 4) log (x 2) (2x 11)log (x 2) 6 0. 22) ex ln x 1. 23) 2 3 20,5 0,5log (x 1) log x 2x 3x .

24)2

22

x x 1ln x 4x 3.

2x 5x 4

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III/ MT S BI TP LIN QUAN N PHNG TRNHM & LGARIT (C CHA THAM S).

1) Tm x ba s: log2 , x

log 3 1 , x

log 5.3 7 theo th t lp thnh mt cp scng.

2) Chng minh rng phng trnh: 3ln(x 1) 2 c t nht mt nghim thc.3) Chng minh rng phng trnh: 2 x(x 2x 2)e 2 c ng mt nghim thc.4) Chng minh rng phng trnh: 525log x x c ng hai nghim thc phn bit.5) Chng minh rng phng trnh: 4x(4x2 + 1) = 1 c ng ba nghim thc phn bit.6) Tm m phng trnh: 22 ln(x 3) ln(mx 3x 9) c nghim (thc).

7) Tm m phng trnh: 2

25 0,24 log x log x m 0 c nghim thuc khong (0;1) .

8) Cho phng trnh:

x xx 2

7 3 m 7 3 2 (1).

a) Gii phng trnh (1) khi m = 4.b) Tm m phng trnh (1) c ng mt nghim thc.

9) Tm m phng trnh: 23 3 3

2 2 0,5log x log x 5log x 8 m 0 c ba nghim thc

phn bit thuc khong (0;64) .10) Cho phng trnh: x x3 9log (10 1).log (3.10 3) m (1).

a) Gii phng trnh (1) khi m = 6.b) Tm m phng trnh (1) c nghim x 1 .

11) Cho phng trnh: x x4 m.2 m 8 0 (1).

a) Gii phng trnh (1) khi m = 8.b) Tm m phng trnh (1) c hai nghim thc phn bit.c) Tm m phng trnh (1) c hai nghim thc x 1, x2 tha x1 < 2 < x2.

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IV/ BT PHNG TRNH M & LGARIT.

1/. Gii cc bt phng trnh sau y:

a)

29x 17x 11 7 5x

1 1 .2 2

b) 4 2log (x 8) log (x 2).

c)x 2x

x 11

39

.

d) 20,5log (x 4x) 5.

e) ln x 1 ln x 1 ln x ln x 13.5 13.7 7 5 .

f) 5 12

2x 3log log 0.

x 1

2/. Gii cc bt phng trnh sau y:

a) 2 x 2 x0,1 0,1log 1 25 1 log 11 9.5 .

b)1 1 1

x x x9.25 16.15 25.9 . c) 3 3xlog x log 27 3.

d)4

4 2 23 1 1 1

3 3 3

x4log x log 32log x 41 85log x.

81

3/. Gii cc bt phng trnh sau y:a) x x x2 4.5 4 10 . b) 2log x (x 3) log x 2 2x 0.

4/.Gii cc bt phng trnh sau y:a) x 2 x3 7log 2 1 log 5 2 3.

b)x x x3.2 7.5 49.10 2.

5/. Bt phng trnh c cha tham s:

a) Tm m bt phng trnh: log x ln x m nghim ng vi mi x thuc khong(0;e) .

b) Tm m bt phng trnh:2 2 2cos x sin x cos x3 5 m.5 c nghim.

c) Tm m bt phng trnh: x x9 (m 1)3 m 4 0 c nghim.

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V/ H PHNG TRNH M & LGARIT.

1) Gii cc h phng trnh sau y:

a) 2x y

5x y 4

15

25

b)3 3 3

x y 13

log x log y 2 log 4


a)x 3 x y 1

x x y

2 .3 1

2 3 11

b)

2 2

5 5

x 9y 125

log (x 3y).log (x 3y) 2


a)2 2

2

25x 16y 10

log(5x 4y) log (5x 4y) 1

b)

225 1

5

3 2

log x log y 0

2x 3y 0


a)x 1 y 1

x y 23 3 82

b)

log y 3log x log y 2x 10


a)3 3log y log x

1 13 3

x y 18

log x log y 1

b) x y

2 2

5log y log x

2log(x y ) 1 log3


a)x 2y

2

x 4

4(x 2y) log x 9

b)3

3

yx

yx

5 .2 200

25 4 689


a)x

y

log (4x 3y) 2

log (4y 3x) 2

b)

x

y

log (3x 4y) 3

log (3y 4x) 3


a)y x

2

4(x y).3

7293log (x y) x y

b)

2

2 9x y 6y x9 2

xy 1

x y

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PHNG PHP CHUNG: Vi 0 a 1 M Na a M N a alog M log N M N 0

a > 1:Hm s m y a ng bin trn R.

Hm s lrarit ay log x ng bin trn

(0; ) .

0 a 1 :Hm s m y a nghch bin trn R.

Hm s lrarit ay log x nghch bin trn

(0; ) .

I/ PHNG TRNH CHA N S TRONG S M.1) Phng php a v ly tha vi cng c s. Vi 0 a 1 : M Na a M N .

2 x 4 1/3(5 ) 5 .5 2x 13/35 5 13

2x3

13

x .6

2)3/2

3 4 1 x5

22 .(2 )

2 1 4x 7/22 2

71 4x

2

9x .

8

3)28 x 3 5 5x(2.5) 10 .(10)

28 x 2 5x10 10 2x 5x 6 0 x 1

x 6

.

4) 2x 1 3x 3x 3x 2/33 .3 (5 .5 ) 3 5x 1 2/33 3 2

5x 13

1

x3

.

5) x x3 (5 4) 2 (7 3) x x3 .9 2 .4 23 3

2 2

x 2.

6) x 2 2 x 2 35 (5 5 1) 3 (3 3 1) 2 0

5 51

3 3

x 2.

7) Phng php t n ph: 1 2x x2 15.2 8 0 x 2 x2.(2 ) 15.2 8 0 .

t xt 2 0 22t 15t 8 0 ( t 8 loi)1

t2

12 2 x 1.

Ch : Khi t n ph cn xc nh iu kin ca n ph (da vo tnh cht ca cc hm s hchoc tm gi tr ln nht, gi tr nh nht ca hm s) v chn cc gi tr ph hp vi iu kin can ph (hn ch c nhng li gii di dng, khng cn thit).

8) Phng php t n ph: x 1 2 x5 5 124 2

xx

55.5 124.

5

t xt 5 0 25

5t 124t

25t 124t 25 0 (1

t5

loi) t 25 x 2.

9) Phng php t n ph: 2 2x 2 x3 2.3 27 0 21 x 1 x3 2.3. 3 27 0.

t 1 xt 3 0 2t 6t 27 0 ( t 3 loi) t 9 1 x 23 3 x 1. 10)Phng php t n ph:

Nhn xt: 5 2 6 5 2 6 1

x/ 3

/3

15 2 6

5 2 6

.

t x/ 3t 5 2 6 0 1t 10t

2t 10t 1 0 t 5 2 6 x 3.

11)Phng php t n ph: x x x25 15 2.9 (1) x x

x x

25 152

9 9

2x x5 5

2 03 3

. t

x5

t 0.3

2t t 2 0 (t = 2 loi) t 1

x5

13

x 0.

Nhn xt: a v ly tha ca cng mt c s, ta c th chia c hai v ca (1) cho 9 , hoc15 , hoc 25 . Trong cc bi tp gii bt phng trnh sau ny, khi chia hai v cho 9 , s dng

tnh ng bin ca hm s m5

y

3

s hn ch c nhng sai lm trong qu trnh gii bi tp.

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12)Phng php t n ph: x 1 x x 19 13.6 4 0 2x x

3 39. 13. 4 0

2 2

.

tx

3t 0.

2

29t 13t 4 0 t 1

4t

9

x 0

x 2

.

13)Phng php t n ph: x x 2x 149 2.35 7.5 0 x x x49 2.35 7.5.25 0

2x x

7 72. 35 05 5

. t

x

7t 05

2t 2t 35 0 (t = 5 loi) t 7 75

x log 7.

14)Cch 1: Phng php c lng hai v (Bt ng thc Csi):2 2 2 2sin x cos x sin x cos x9 9 2 9 2 9 6 . ng thc xy ra

2 2sin x cos x9 9

2 2sin x cos x sin x cosx t anx 1 x k (k Z).4

Cch 2: Phng php t n ph (Mt n ph):2 2sin x cos x9 9 6

2 2sin x 1 sin x9 9 6 . t2sin xt 9 (iu kin: 1 t 9 )

2

( x R, 0 sin x 1

20 sin x 1

9 9 9 ) 9

t 6t 2

t 6t 9 0 t 3

2sin x9 3

22sin x 13 3 21

sin x2

2

sinx2

x k (k Z).4 2

Cch 3: Phng php t n ph (Hai n ph dn n h phng trnh tng, tch):

t2sin xu 9 v

2cos xv 9 u v 6

u.v 9

. iu kin

1 u 9

1 v 9

.

Khi u, v l nghim ca phng trnh: 2X SX P 0 .Ch : iu kin ca n ph v chn cc gi tr ph hp vi iu kin ca n ph (hn ch c

nhng li gii di dng, khng cn thit).C th biu din trn ng trn lng gic hiu thm:

x k (k Z)4 x k (k Z)

4 2

15)Phng php t n ph:2 21 2sin x 2cos x4 9.4 5

2 21 2cos x 2cos x4 9.4 5 .

t22co s xt 4 (iu kin: 1 t 16 ) 2( x R, 0 2cos x 2

20 2cos x 24 4 4 )

1 9

t 54 t

2t 20t 36 0 (t = 16 loi) t 2 22cos x4 2

21

2co s x 24 4

21

2 cos x2

1

cosx2

x k2

3 (k Z)2

x k23

x k (k Z).3

Ch : iu kin ca n ph v chn cc gi tr ph hp vi iu kin ca n ph (hn ch cnhng li gii di dng, khng cn thit).

C th biu din trn ng trn lng gic hiu thm:

x k23 (k Z)2

x k23

x k (k Z).3

16)Phng php c lng hai v:2cos x5 sinx . 2x R, cos x 0

2cos x 05 5 1 s inx .

ng thc xy ra2cos x 0

sinx 1

cosx 0

s inx 1

x k (k Z)

2

.

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17)Phng php c lng hai v Tm gi tr ln nht, gi tr nh nht ca hm s.2 2sin x cos x4 4 6 cos2x (1) .

* x R, 1 cos2x 1 x R, 5 6 cos2x 7 .

*2 2 2

2

sin x cos x sin x

sin x

4y 4 4 4

4 . TX: D = R.

t2sin xt 4 0 . 2( x R, 0 sin x 1

20 sin x 14 4 4 )

4

y f (t) t t trn on [1; 4].

2

2 2

4 t 4

f '(t) 1 t t

f '(t) 0 (t = 2 loi) t = 2.

x R t [1; 4]

m ax y m ax f (t) f (1) f (4) 5

;x R t [1; 4]

min y min f(t) f (2) 4

4 y 5 .

(1)2 2sin x cos x4 4 5

6 cos2x 5

2sin x 0

cos2x 1

x k22 (k Z)

x k2

x k (k Z)2

18)Phng php t n ph:3 3x 3 3x 4 x 4 x 33 3 3 3 10 3x 3x x x 327.3 27.3 81.3 81.3 10

3x x 33x x1 127 3 81 3 103 3

. t x x1t 3 2.3 x xx x1 1t 3 2 3 . 23 3

3

3 x 3x 2x xx 2x 3x

1 1 1 1t 3 3 3.3 . 3.3 .

3 3 3 3

3x 33x

13 t 3t

3 .

3 327 t 3t 81t 10 3

3 10t27

10

t 23

.

10t

3 x

x

1 103

3 3 . t xy 3 0 2

10y y 1 0

3

y 3

1y

3

x 1

x 1

.

Ch : iu kin ca n ph.19)Phng php t n ph:

x 1 x5 25 6 x2x

255 6

5 . t xt 5 0 3 2t 6t 25 0 2(t 5)(t t 5) 0

t 5

1 21t

2

1 21

(t2

loi)

5

x 1

1 21x log

2

20)Phng php lgarit ha: M N 0 a alog M log N (0 a 1) .x x5 33 5

x x5 33 3log 3 log 5

x x 35 3 log 5 x

3x5 log 53

x

35 log 53

353

x log log 5

21)Phng php lgarit ha: M N 0 a alog M log N (0 a 1) .5 2x3 2 5 2x3 3log 3 log 2 3x (5 2x) log 2

3 3x(1 2 log 2) 5 log 2 3

3

5 log 2x

1 2log 2

.

22)Phng php hm s: x3 5 2x (1) Dng phng trnh c nghim duy nht.* x = 1 l nghim ca phng trnh (1).* f (x) 3 ng bin trn R v g(x) 5 2x nghch bin trn R. Phng trnh (1) c nghim duy nht x = 1.

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23)Phng php hm s:x

x 23 5 4 x

x3 5 4 x x

3 11 4 (1)

5 5

Phng trnh (1) c nghim duy nht x = 2 (tng t bi tp 22).24) Phng php hm s: x x4 (x 14)2 33 3x 0 (1)

t xt 2 0 2t (x 14)t 33 3x 0 . 2(x 8) t 3

t 11 x

* t = 3 x2 3 2

x log 3 .

* t = 11 x x2 11 x (2) Phng trnh (2) c nghim duy nht x = 1(tng t bi tp 22). Phng trnh (1) c hai nghim 2x log 3 v x = 1.

25) 2 x 1 x2 x 3 3x 1 4.3 1(1) Cch 1: Phng php phn tch thnh nhn t:

(1) 2 x2x 3x 1 6.3 (2x 1) 0 x1

2(x 1) x 6.3 (2x 1) 02

x(2x 1) x 1 6.3 0 x2x 1 0

x 1 6.3 0

Cch 2: Phng php tm nghim ca phng trnh bc hai (theo x):(1) 2 x x2x 3(1 4.3 )x 1 6.3 0 2 x x 29(1 4.3 ) 8(1 6.3 ) (1 12.3 )

1

x2

x 1 6.3

x

1x

21 x

36

1

x2

x 1

Ch : Phng trnh x1 x

36

c nghim duy nht x = 1(tng t bi tp 22).

26) 2 x 1 x x 1 x 1 xx .5 3 3.5 x 2.5 3 0 (1) . Tng t bi tp 25 (cng c phng php gii).Cch 1: Phng php phn tch thnh nhn t:

(1)

2 x x

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

x x

(x 1) (x 2)5 5.3 0 Cch 2: Phng php tm nghim ca phng trnh bc hai (theo x):(1) 2 x x x x xx .5 (5.3 3.5 )x 2.5 5.3 0

x 2 x x x x x 2(5.3 3.5 ) 4.5 (2.5 5.3 ) (5.3 5 )

x 1

3x 2 5

5

xx 1

3 x 25 5

x 1

x 1

Phng trnhx

3 x 25 5

c nghim duy nht x = 1(tng t bi tp 22).

27)Phng php hm s (vn dng tnh n iu ca hm s):

3 3x x x36 2 3 9.8 4.27 3 3

xx x 8 272 3

4 9

3 3x 3x 2 3x 22 3 2 3

Hm s tf (t) 2 3 ng bin trn R. 3f (x ) f (3x 2) 3x 3x 2 3x 3x 2 0

2(x 1)(x x 2) 0 x 1

x 2

Ch : Cho hm s y = f(x) ng bin (hoc nghch bin) trn K.

1 2 1 2x ,x K, f (x ) f (x ) 1 2x x .

28) x 1 x 3(x 2) (x 2) (1) .Phng php: Gi s cc hm s f, g c tp xc nh ln lt l Df, Dg v D = Df Dg.

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(x) g(x)x x D

1

hoc

x D

0 x 1

f (x) g(x)

(1)x 1 0

x 2 1

hoc

x 1 0

0 x 2 1

x 1 x 3

x 1

x 1

hoc

2

x 1

x 3 0

x 1 (x 3)

2

x 3

x 7x 10 0

x 3x 2

x 5

x = 5.

Ch : Hc sinh thng gii sai: A B 2

2A B (!)

Cn phi gii ng: A B 2

B 0

A B

II/ PHNG TRNH CHA N S TRONG BIU THC LGARIT.1) Phng php vn dng nh ngha lgarit v gii phng trnh m.

x 13log (3 26) 2 x

x 1

1 2 x

3 26 0

3 26 3

x 1 2 x3 26 3 0 x3.3 26 9.3

t xt 3 0 23t 26t 9 0 (1

t3

loi) t = 9 x = 2.

2) Phng php vn dng nh ngha v cc tnh cht ca lgarit.

3 3log (x 2) log (x 6) 2

3

x 2 0

x 6 0

log (x 2)(x 6) 2

2

x 6

(x 2)(x 6) 3

2

x 6x 4x 21 0

x 6

x 3

x 7

x = 7.

Nhn xt:Nu khng ch n iu kin xc nh ca phng trnh th thng dn n nhng php bin

i sai: 3 3log (x 2) log (x 6) 2 3log (x 2)(x 6) 2 , v tp xc nh ca

3 3f (x) log (x 2) log (x 6) khc vi tp xc nh ca 3g(x) log (x 2)(x 6) .

Vi 0 a 1 v M, N l hai s dng: a a alog M log N log (M.N)

Vi 0 a 1 v M, N l hai s cng du: a a alog M log N log (M.N)

3) Phng php a v lgarit vi cng c s. Vi 0 a 1 : a alog M log N M N 0 .

33

5 0,2 25log x log x log x 7 1 2/335 5 5log x log x log x 7

35 5 53

3log x 1log x log x 72

53

3 1 log x 72

5log x 2 x = 25.

Ch : Vi 0 a 1 : alog f (x) b f (x) a .

4) Phng php a v lgarit vi cng c s.

4 2 4log (x 3) log x 1 2 log 8

1/21/2

4 4 44

x 3 0

x 1 0

log (x 3) log x 1 2log 4 log 8

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2

4 4

x 1

x 3 4log log

x 1 8

x 1

x 32

x 1

x = 5.

Ch :

Khi gii phng trnh, c th tch thnh hai phn: iu kin (xc nh):x 3 0

x 1 0

v gii

phng trnh: 1/21/2

4 4 44og (x 3) log x 1 2log 4 log 8 . Khi gii c x = 5, ta kim trax 5 tha iu kin.

Nu iu kin khng qu phc tp th nn dng php bin i tng ng. Php bin i tng ng ng:

4 2 4log (x 3) log x 1 2 log 8 1/21/2

4 4 44log (x 3) log x 1 2log 4 log 8 .

Nu khng ch n iu kin xc nh ca phng trnh th thng dn n nhng php bini sai:

4 4 4 4log (x 3) log x 1 2log 4 log 8 4 4 4x 3

log 2 log 4 log 8x 1

, v tp xc nh

ca 4 4f (x) log (x 3) log x 1 khc vi tp xc nh ca 4x 3

g(x) log

x 1

.

5) Phng php a v lgarit vi cng c s: 22 0,5log (x 3) log (6x 10) log10 0 (1)

K:2x 3 0

6x 10 0

. (1) 22 2 2log (x 3) log (6x 10) log 2 0

2

2

(x 3)2log 0

6x 10

2

2 2

x 3log log 1

3x 5

2x 31

3x 5

2x 3x 2 0 (x = 1 loi) x = 2 tha K.

Ch : Khi gii phng trnh khng nht thit phi gii phn iu kin (nu iu kin phc tp).

xc nh nghim tha K, ta kim tra bng cch th gi tr tm c vo K. Vi 0 a 1 :

a alog f (x) log g(x)

f (x) 0

g(x) 0

f (x) g(x)

f (x) 0

(I)f (x) g(x)

g(x) 0(II)

f (x) g(x)

.

C th trong mi bi tp, cn chn h (I) hoc h (II) mt cch kho lo hn ch bt iu kin(m vn bo m php bin i tng ng).

i vi bi tp 3, nn trnh by li gii nh sau:2

2 0,5log (x 3) log (6x 10) log10 0 2

2 2 2log (x 3) log (6x 10) log 2

22 2log (x 3) log (3x 5) 23x 5 0

x 3 3x 5

2

x 5/3

x 3x 2 0

x 5/3

x 2

x 1

x 2 .

6) Phng php: Vi 0 a 1 : a alog f (x) log g(x) f (x) 0

(I)f (x) g(x)

g(x) 0(II)

f (x) g(x)

.

3 2log 3x 15x 15x 4 3log(x 2) 0 3 2log 3x 15x 15x 4 3log(x 2)

3 2 3

x 2

3x 15x 15x 4 (x 2)

3 2

x 2

2x 9x 3x 4 0

2

x 2

(x 1)(2x 7x 4) 0

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x 2

x 1

x 4

x 1/2

x 1

x 1/2

.

Ch : Hc sinh c th bin i phng trnh v dng :3 2

3

3x 15x 15x 4log 0

(x 2)

.

Tuy nhin cch gii ny khng chnh xc, thng thng phngtrnh:

3log f (x) log g(x) 0 (a) khng tng ng vi phng trnh:

3

f(x)log 0 (b)

g (x)

, v

tp xc nh ca (a) v (b) thng khc nhau.

7)2

44 2

xlog log (4x) 10 0 (1)

4

. K: x 0.

(1) 2 2 4 44 4 2 2log x log 4 log 4 log x 10 0 2 2log x 2 8 4log x 10 0

2log x 0 0x 2 x 1 (tha K).

Hoc trnh by li gii nh sau:2

44 2xlog log (4x) 10 04

2 2 4 44 4 2 2log x log 4 log 4 log x 10 0

2 2log x 2 8 4log x 10 0 2log x 0 0x 2 x 1.

Ch n iu kin ca cc cng thc s dng: Vi N > 0 v 0 a 1 : a alog N log N ( R)

Vi N 0 v 0 a 1 : 2a alog N 2 log N ( R)

Hc sinh thng gii sai nh sau:2

44 2

xlog log (4x) 10 0

4

4 2

x2log 4log (4x) 10 0

4

(!)

8) Phng php a v lgarit vi cng c s (vn dng cc cng thc i c s):

9

6 3

4log 5 x11(1)

log (2 x) log (2 x)

. K:

0 2 x 15 x 0

x 15 x 2

.

(1) 3 3

3 3

log 6 log (5 x)1

log (2 x) log (2 x)

3 3log 6(5 x) log (2 x)

2x 3x 4 0 (x = 1 loi) x = 4 tha K.9) Phng php lgarit ha: Vi 0 a 1 : M N 0 a alog M log N .

3(2 log x 1,5 logx)x 10 3(2 log x 1,5log x) log x log 10 4 24 log x 3log x 1 0

( 21

log x4

loi) 2log x 1 log x 1 x 10

x 0,1

.

(C th t n ph t log x hoc 2y log x ).10)Phng php t n ph: 3 1

3

10 log x 3 log x (1)

K:13

x 0

log x 0

1 13 3

x 0

log x log 1

x 0

x 1

0 x 1 .

t 13

t log x 0 2 13

t log x 23log x t

(1) 2t 3t 10 0 (t = 5 loi) t = 2 13

log x 4 4

1 1x

3 81

tha K.

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11)Phng php t n ph:5log(log x) log log(x ) 4 0 log(log x) log 5log x 4 0 (1)

K:

x 0

log x 0

5log x 4 0

x 0

4logx

5

. t4

t log x5

(1) log t log 5t 4 0 log t 5t 4 log1 25t 4t 1 0

( 1t3

loi) t = 1 x = 10 tha K.

12)Phng php bin i v cng c s, t n ph.

25x 5

5log log x 1 (1)

x

. K: 0 5x 1 0 x 1/5 .

(1)5

25

5

5log

xlog x 1

log (5x)

25 5

5

1 log xlog x 1 0

1 log x

25 5

5

1 log x(1 log x) 0

1 log x

25 51 log x 1 (1 log x) 0 5 5 51 log x log x 2 log x 0

5

5

5

log x 1

log x 0

log x 2

x 5

x 1

x 1/25

(tha iu kin 0 x 1/5 ).

Ch : Hc sinh thng khng ch n nhn t chung, thng bin i v phng trnh bc ba. Khi

li phi tch bc ba thnh bc nht nhn bc hai. C th t n ph 5t log x.

0 x 1/5 5log (5x) 0 51 log x 0 .13)Phng php bin i v cng c s, t n ph.

2 3x 16x 4x

2log x 14log x 40log x 0 (1). K:

x 0

x 2

x 1/16

x 1/4

.

(1)2 3

2 2 2

2 22

log x log x log x14 40 0

x log (16x) log (4x)log2

2

2 2

2 2 2

1log x2log x 3log x 214 40 0

log x 1 log x 4 log x 2

. t 2t log x

2t 42t 20t0

t 1 t 4 t 2

t(t 4)(t 2) 21t(t 1)(t 2) 10t(t 1)(t 4) 0 2t( 10t 15t 10) 0

* t = 0 x = 1. * t = 2 x = 4. * 1t2

2x2

.

Ch : Hc sinh thng chia hai v ca phng trnh cho t, dn n lm mt nghim t = 0. x 2 2log x 1 2log x 1 0 t 1 0 (tng t vi x 1/16 , x 1/4 ).

C th a v xlog 2 nh sau:

2x/2 x/2

x x

2 2log x 2log x

log (x/2) 1 log 2

(tng t vi 316xlog x , 4xlog x )

Tuy nhin khi x = 1 th 1log 2 khng xc nh, trong khi phng trnh (1) c nghim x = 1.

Nh vy, nu mun bin i v xlog 2 th cn xt cc trng hp x = 1 v x 1 .

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14) 3 53 2 3 210 log x.log x 15 2log x 5log x (1) Cch 1: Phng php phn tch thnh nhn t.

3 53 2 3 210log x.log x 15 2log x 5log x 3 2 3 210 log x.log x 6 log x 15 25log x

3 2 22 log x(5log x 3) 5(3 5log x) 2 3(5 log x 3)(2log x 5) 0

2

3

3log x

55

log x 2

35

5

2

x 2

x 3

Cch 2: Kt hp cc phng php: a v lgarit cng c s, t n ph t, tm nghim caphng trnh bc hai (theo t):

(1) 3 2 2 3 2 210(log 2.log x) log x 15 6(log 2.log x) 25log x

t 2t log x ; 3a log 2 (a 0,63) 210at (6a 25)t 15 0

2 2(6a 25) 4.10a.15 (6a 25)

3t

55

t2a

2

2 2

3log x

55

log x log 32

2

2 3 2

3log x 5

5log 3.log x log 3

2

2

3

3log x 5

5log x

2

3

5

52

x 2

x 3

15) 22 log x ln x 2 ln x.log x log x (1) Cch 1: Phn tch thnh nhn t: (1) (log x ln x)2log x log x ln x 0

(log x ln x)(2log x 1) 0 1

logx2

log x ln x

1x

10x 1

Cch 2: t n ph (mt n ph t), tm nghim ca phng trnh bc hai (theo t):

(1)

2

2 log x (2 ln x 1) log x ln x 0 t t log x 22t (2ln x 1)t ln x 0 2 2(2 ln x 1) 8ln x (2ln x 1)

1

t2

t ln x

1

logx2

log x ln x

1x

10x 1

.

Cch 3: t n ph (hai n ph):u log x

v ln x

22u u 2uv v 0

u(2u 1) v(2u 1) 0 (2u 1)(u v) 0

16)Phng php t n ph:2

5 5log 15 log (5x )25x 22.x 5.3 0 (1) . K: x > 0.

(1)

25 5 5log x log x 1 2log x

5.5 22.15 5.3 0

5 5 5log x log x log x

5.25 22.15 15.9 0

5 52log x log x5 5

5. 22. 15 03 3

. t

5log x5t 0

3

25t 22t 15 0

(t = 5 loi)3

t5

5log x 15 5

3 3

5log x 1 1

x 05

Ch : Trong qu trnh bin i phng trnh, nu khng bit r php bin i dn n phng trnh

tng ng hay phng trnh h qu th u tin nn t iu kin. Vi 0 a 1 v b > 0: alog bb a . Vi a, c l hai s dng v 0 b 1 : b blog c log aa c .

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(v b blog c log aa c b blog c log ab blog a log c b b blog c.log a log a.log c )17)Phng php hm s (Dng phng trnh c nghim duy nht):

2log(x 3x 4) log(x 4) 2 x (1) . K:2x 3x 4 0

x 4 0

x > 1.

(1)2x 3x 4

log 2 xx 4

(x 1)(x 4)log 2 x

x 4

log(x 1) 2 x (2)

* x = 2 l nghim ca (2).* f (x) log(x 1) ng bin trn (1; )

1f '(x) 0, x (1; )

(x 1) ln 10

.

g(x) 2 x nghch bin trn R ( g(x) 2 x nghch bin trn (1; ) ). (2) c nghim duy nht x = 2 (1) c nghim duy nht x = 2.

Nhn xt: Thng thng phng trnh: log f (x) log g(x) 2 x (a) khng tng ng vi

phng trnh:f(x)

log 2 x (b)g(x)

, v tp xc nh ca (a) v (b) thng khc nhau.

Nhng: 2log(x 3x 4) log(x 4) 2 x 2x 3x 4

log 2 xx 4

(trn (1; ) ) l mt

trng hp c bit.18)Phng php t n ph, phng php hm s (Dng phng trnh c nghim duy nht):

log( x 7 )3 x (1) . K: x > 0.

t t log(x 7) x 7 10 tx 10 7 .

log (x 7 )3 x t t3 10 7 t3 7 10 t t

3 17 1(2)

10 10

Phng trnh (2) c nghim duy nht t = 1 (tng t bi 17) Phng trnh (1) c nghim duy nht 1x 10 7 3 0. Nhn xt: iu kin xc nh ca phng trnh (1) l: x + 7 > 0 x > 7; v hm s log( x 7 )f (x) 3 c

TX: fD ( 7; ) , hm s g(x) = x c TX: Dg = R, khi f gD D ( 7; ) .

iu kin x > 0 l iu kin phng trnh (1) c nghim, v nu x l nghim ca (1) thlog( x 7 )x 3 0 .

19)Phng php t n ph, phng php hm s (Dng phng trnh c nghim duy nht):2 2 2log 25 log x log 52x x .5 x 2 2 2log x log x log x225 x .5 5 2 2 2log x log x log x25 x 5 5

2log x 25 x 1 . t 2t log x x 2 2t t5 2 1 t t5 4 1

t

4 11

5 5

c nghim duy nht (tng t bi 17) t = 1 2log x 1 x = 2.

20)Phng php tm nghim ca phng trnh bc hai (theo x), dn n dng phng trnh c nghimduy nht:x ln x x(3 x) 2(1 ln x) 2x (3 ln x)x 2(1 ln x) 0

2 2(3 ln x) 84(1 ln x) (1 ln x)

x 2

x 1 ln x

x 2

ln x 1 x

x 2

x 1

(Phng trnh: ln x 1 x c nghim duy nht x = 1, tng t bi 17).

Nhn xt: Vi kt qux 2

x 1 ln x

, ta cng on c nu gii bng phng php phn tch

thnh nhn t th: (x 2) x 1 ln x 0 .

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+

-1

0

0

x + e-1

f ' (x)

f(x)

0

-1

0

0

+

2 +

f ' (x)

f(x)

x

0

21)Kt hp nhiu phng php: t n ph t, tm nghim ca phng trnh bc hai (theo t), dngphng trnh c nghim duy nht):

25 5(x 4) log (x 2) (2x 11) log (x 2) 6 0 . t 5t log (x 2) .

2(x 4)t (2x 11)t 6 0 2 2(2x 11) 24(x 4) (2x 5)

t 2

3t

x 4

5

5

log (x 2) 2

3log (x 2)

x 4

51

x25

x 7

Trong : 5 3log (x 2) x 4

l dng phng trnh c nghim duy nht (tng t bi 17).

Nhn xt: x = 4 khng tha phng trnh 2(x 4)t (2x 11)t 6 0 ,

v x = 4 5t log (4 2)

(2.4 11)t 6 0

5

t log 2

t 2

(v l).

Khi c kt qu5

5

log (x 2) 2

3log (x 2)

x 4

, ta cng on c nu gii bng phng php phn tch

thnh nhn t th: 5 5log (x 2) 2 x 4 log (x 2) 3 0

.22)Phng php c lng hai v, s dng hm s tm gi tr ln nht, gi tr nh nht.Xt hm s f (x) ex ln x trn khong (0; ) f '(x) eln x e e(1 ln x)

f '(x) 0 1x e

1x (0; )min f (x) f (e ) 1

f (x) 1 1x e Ch : C th xt du ca f '(x) bng cch gii bt phng trnh f '(x) 0 1 ln x 0 hoc s dng

my tnh cm tay ( f '(0,1) 1,3 suy ra du ca f '(x) trong khong 1(0; e ) ).

xl im(xlnx)

.

x 0lim(xlnx)

(dng 0.) =

x 0

ln xlim

1/x(dng

) =

2x 0

1/xlim

1/x (quy tc LHospitale)

=x 0lim ( x) 0

.

Nu khng tm c cc gii hn nh trn, da vo bng bin thin ta cng suy ra c1

x ( 0; )min f (x) f (e ) 1

23)Phng php c lng hai v; s dng bt ng thc, s dng hm s tm gi tr ln nht, gitr nh nht.

2 3 20,5 0,5log (x 1) log x 2x 3x (1). K: x > 0.

*2

20,5 0,5 0,5 0,5 0,5

x 1 1log (x 1) log x log log x log 2 1

x x

.

ng thc xy ra: 20,5 0,5log (x 1) log x 1 x = 1 (v x > 0).

* Xt hm s 3 2f (x) 2x 3x trn khong (0; ) .2f '(x) 6x 6x

f '(x) 0 x 0 (0; )

x 1 (0; )

x (0; )min f (x) f (1) 1

.

3 2

f (x) 2x 3x 1 .

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3 22x 3x 1 x = 1.Phng trnh (1) c nghim duy nht x = 1.Ch :

Bt ng thc Csi: Vi x > 0,1 1

x 2 x. 2x x

. Trng hp khng vn dng c

bt ng thc th tm gi tr ln nht, gi tr nh nht ca hm s (nh v phi).

Hm s 0,5y log x nghch bin trn khong (0; ) nn khi1

x 2

x

th

0,5 0,5

1log x log 2

x

.

24)Phng php hm s (vn dng tnh n iu ca hm s).2

22

x x 1ln x 4x 3

2x 5x 4

2 2 2ln x x 1 ln 2x 5x 4 x 4x 3

2 2 2ln x x 1 ln 2x 5x 4 x 4x 3

2 2 2 2ln x x 1 ln 2x 5x 4 2x 5x 4 x x 1

2 2 2 2ln x x 1 x x 1 ln 2x 5x 4 2x 5x 4

Xt hm s f (t) ln t t trn (0; ) .1

f '(t) 1 0, t (0; )t

Hm f ng bin trn (0; ) .

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

x 3

Ch : Cho hm s y = f(x) ng bin (hoc nghch bin) trn K.

1 2 1 2x ,x K, f (x ) f (x ) 1 2x x .

Nhn xt: 2 2x R, x x 1 0, 2x 5x 4 0 (v < 0).

(Tng t bi tp 27 I/ PHNG TRNH CHA N S TRONG S M)

III/ MT S BI TP LIN QUAN N PHNG TRNH M &LGARIT (C CHA THAM S).

1) K: x3 1 0 x > 0.Theo tnh cht cp s cng: x x2log 3 1 log 2 log 5.3 7

2x xlog 3 1 log 2 5.3 7

2x x3 1 2 5.3 7 2x x3 12.3 13 0

( x3 1 loi) x3 13 3x log 13 .

2) Phng php p dng tnh cht ca hm s lin tc.3ln(x 1) 2 x 3ln(x 1) 2 0 (1) . K: x > 1.

3f (x) ln(x 1) 2 trn TX: D ( 1; ) .

Hm f lin tc trn D v1

f (0).f (3) (ln 4 1) 08

0 0x (0; 3) : f (x ) 0 .

Phng trnh (1) c nghim 0x (0; 3) ( 1; ).

3) Phng php p dng tnh lin tc v s bin thin ca hm s.2 x(x 2x 2)e 2 2 x(x 2x 2)e 2 0 (1)

Xt hm s 2 xf (x) (x 2x 2)e 2 trn TX: D = R.2 xf '(x) x e 0, x R.

f '(x) 0 x 0 Hm f lin tc v ng bin trn R.

f (0) 0

Phng trnh (1) c mt nghim duy nht x = 0.

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25

ln5 +

f '(x)

f(x)

x

0

0

f '(x)

f(x)

x1x - +

0

x2

0

Ch : Sch gio khoa cng nhn tnh lin tc ca hm s m, hm s lgarit, hm s ly tha (trn

tp xc nh ca n). Hc sinh thng chng minh f '(x) 0, x R Hm f ng bin trn R. y l chng

minh sai. Theo sch gio khoa: Nu f '(x) 0, x D v f '(x) 0 ti mt s hu hn im th hm f

ng bin trn D.4) Phng php p dng tnh lin tc v s bin thin ca hm s.

525log x x 5x 25log x 0 (1) . K: x > 0.Xt hm s 5f (x) x 25log x trn (0; ) .

25 x ln 5 25f '(x) 1

x ln 5 x ln 5

f '(x) 0 25

xn 5

Da vo bng bin thin Phng trnh: f(x) = 0 c khng qu 2 nghim thc phn bit.f (1) 1 0 , f (5) 5 25 20 0 , 3f (5 ) 125 25.3 0 .Hm f lin tc trn khong (0; ) .

Phng trnh (1) c ng hai nghim 1x (1; 5) v 2x (5;125) .

5) Phng php p dng tnh lin tc v s bin thin ca hm s.f(x) = 4x(4x2 + 1) 1. TX: D = R.

x 2f '(x) 4 (4ln 4)x 8x ln 4

f '(x) 0 2(4 ln 4)x 8x ln 4 0

> 0 f '(x) 0 c 2 nghim phn bit.Da vo bng bin thin:Phng trnh: f(x) = 0 c khng qu 3 nghim thc phn bit.f(1/2) = 0; f(0) = 0; f(3).f(2) < 0 pcm.

6) Phng php gii phng trnh lgarit, xt du nh thc bc nht, tam thc bc hai.

22 ln(x 3) ln(mx 3x 9) (1) 2 2

x 3 0

(x 3) mx 3x 9

2

x 3

(1 m)x 9x 0

x 3

x (1 m)x 9 0

x 3

(1 m)x 9 0

x 3

9x

1 m

(1) c nghim9

x 31 m

3

1 01 m

2 m

01 m

(2 m)(1 m) 0 2 < m < 1.

Ch :

Vi 0 a 1 : a alog f (x) log g(x) f (x) 0

(I)f (x) g(x)

g(x) 0(II)

f (x) g(x)

.

Hc sinh thng gii sai:

9

31 m 9 3(1 m) (!)

a

a,b R, 0b

ab 0 . Xt du ca: (2 m)(1 m) tng ng vi xt du tam thc

bc hai.f(m) 00

m

+

1-2

7) Phng php t n ph, p dng s bin thin ca hm s; phng php tam thc bc hai.

2

25 0,24 log x log x m 0

25 5log x 2 log x m 0 (1)

t 5t log x . Phng trnh (1) tr thnh:2t 2t m 0 2m t 2t (2)

(1) c nghim x (0;1) (2) c nghim t ( ; 0) .

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+

f '(t)

f(t)

t

0

0 2

0

-

4

-

-

f '(t)

f(t)

t

0

0

0

-1

1

f '(t)

f(t)

t

-10

-17

1-

-

0

5

0

15

6

Cch 1: 2f (t) t 2t trn khong ( ; 0) .f '(t) 2t 2 f '(t) 0 t 1 (2) c nghim t ( ; 0) m 1 .

Cch 2: 2t 2t m 0 (2 ')

Nu (2) c hai nghim t1, t2 th t1 + t2 = 1 (2) c nghim t ( ; 0) .

V vy, (2) c nghim t ( ; 0) ' 1 m 0 m 1 .Nhn xt: i vi bi tp ny th gii cch 2 nhanh v gn hn cch 1 (bit nhn xt phng trnh

(2) nu c nghim t th t < 0).Cch 1 s c li th khi bi tp yu cu bin lun; da vo bng bin thin ta thy:

(2) c ng mt nghim t ( ; 0) m = 1 hoc m 0.

(2) c hai nghim m phn bit thuc khong ( ; 0) 0 < m < 1.8) Phng php t n ph, p dng s bin thin ca hm s; phng php tam thc bc hai.

(1)

x x

7 3 7 3m 4

2 2

. t

x

7 3t 0

2

x

7 3 1

2 t

Khi phng trnh (1) tr thnh: 2t 4t m 0 (2)

Khi m = 4 t = 2 ax log 2 vi7 3

a2

.

(1) c ng mt nghim x (2) c ng mt nghim t > 0.Cch 1: ' 4 m

* ' 0 m = 4 t = 2 ax log 2 vi7 3

a2

(m = 4 tha)

* ' 0 m < 4: Phng trnh (2) c hai nghim phn bit t1, t2 v 1 2t t 4 0 . (2)

c ng mt nghim t > 0 1 2t 0 t m 0 .

Kt lun: (2) c ng mt nghim t > 0 m = 4 hoc m 0 .Cch 2: 2t 4t m (2 ')

Xt hm s 2f (t) t 4t trn khong (0; ) .

f '(t) 2t 4 f '(t) 0 t 2

Da vo bng bin thin:(2) c ng mt nghim t > 0 m = 4 hoc m 0 .

Ghi ch:Hc sinh xem thm th pha di y, tuy nhin trong phn

bi lm khng nht thit phi v th.9) Phng php t n ph, p dng s bin thin ca hm s.

23 3 32 2 0,5log x log x 5log x 8 m 0 3 22 2 2log x 9 log x 15log x 8 m 0 (1)

t 2t log x . Khi (1) tr thnh:3 2t 9t 15t 8 m (2)

0 x 64 2 2log x log 64 6 t 6

(1) c ba nghim phn bit thuc khong (0;64) (2) c ba nghim phn bit thuc khong

( ; 6)

Xt hm s 3 2f (t) t 9t 15t 8 trn ( ; 6) .2f '(t) 3t 18t 15

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+

-1/2 +

f '(t)

f(t)

t

0

6

2

f '(t) 0 t 1 D

t 5 D

Da vo bng bin thin:(2) c ba nghim phn bit thuc khong ( ; 6) 17 m 10 .Ghi ch:Hc sinh xem thm th pha di y, tuy nhin trong

phn bi lm khng nht thit phi v th (tr trng hpbi ton c phn kho st v v th).

10)Phng php t n ph, p dng s bin thin ca hm s; phng php tam thc bc hai.x x

3 9log (10 1).log (3.10 3) m (1). K:x10 1 0 x > 0.

(1)

x x

3 3 3

1

log (10 1) log 3 log (10 1) m2

x x

3 3log (10 1) 1 log (10 1) 2m

t x3t log (10 1) . Phng trnh (1) tr thnh:

2t t 2m 0 (2)

Khi m = 6 2t t 12 0 t 4

t 3

x

3

x3

log (10 1) 4

log (10 1) 3

82x log

81x log 28

x 1 x10 10 x10 1 9 x3t log (10 1) 2 .

Cch 1: (1) c nghim x 1 (2) c nghim t 2 1 2

1 2

2 t t

t 2 t

1 2t 2 t

( 1 22 t t loi v 1 2t t 1 ) 1 2t 2 0 t 2 .

t y t 2 t y 2 . Khi phng trnh (2) tr thnh 2y 5y 6 2m 0 (3) (2) c nghim 1 2t 2 t (3) c nghim 1 2y 0 y 1.(5 2m) 0 m 3 .

Cch 2: 2t t 2m (2 ') . (1) c nghim x 1 (2) c nghim t 2 .2f (t) t t trn [2; ) .

f '(t) 2t 1

f '(t) 0 1

t [2; )2

Da vo bng bin thin:(2) c nghim t 2 2m 6 m 3 .Ch : Cch gii: (2) c nghim 1 2t 2 t a.f (2) 0 vi

2f (t) t t 2m khng ph hp

vi chng trnh ci cch gio dc v sch gio khoa hin hnh (Sch gio khoa hin hnhkhng trnh by kin thc so snh s vi hai nghim ca phng trnh bc hai).

11)Phng php t n ph, p dng s bin thin ca hm s; phng php tam thc bc hai.x x4 m.2 m 8 0 (1). t xt 2 0 .

Khi phng trnh (1) tr thnh: 2t mt m 8 0 (2)

Khi m = 8 2t 8t 16 0 t 4 x 2 . (1) c hai nghim phn bit (2) c hai nghim dng phn bit.

2m 4m 32 0

S m 0

P m 8 0

m > 8.

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f '(t)

f(t)

t

+ +

8

-8

-

-2 1- + 0

0

4

0

(1) c hai nghim x1, x2 tha x1 < 2 < x2 (2) c hai nghim t1, t2 tha t1 < 4 < t2.Cch 1: t y t 4 t y 4 .

Khi phng trnh (2) tr thnh: 2y (8 m)y 3(8 m) 0 (3)

(2) c hai nghim t1, t2 tha 1 2t 4 t (3) c hai nghim y1, y2 tha 1 2y 0 y

1.3(8 m) 0 m 8 .

Cch 2: (2) 2t 8 m(t 1)

(phng trnh ny khng c nghim t = 1)

2t 8

m (2')t 1

.

Xt hm s2t 8

f(t)t 1

trn D (0; ) \{1}.

2

2

t 2t 8f '(t)

(t 1)

f '(t) 0 t 4 D

t 2 D

Da vo bng bin thin:(2) c hai nghim t1, t2 tha t1 < 4 < t2 m 8 .

Nhn xt: Da vo bng bin thin ta cng tm c cc kt qu sau y:

(1) v nghim (2) v nghim hoc (2) ch c nghim m 8 m 8 . (1) c ng mt nghim (2) c mt nghim kp dng hoc (2) c hai nghim tri du m = 8 hoc m < 8.

Ch : Cch gii: (2) c nghim 1 2t 4 t a.f (4) 0 vi2f (t) t mt m 8 khng ph

hp vi chng trnh ci cch gio dc v sch gio khoa hin hnh (Sch gio khoa hin hnhkhng trnh by kin thc so snh s vi hai nghim ca phng trnh bc hai).

Ghi ch:Hc sinh xem thm th pha di y, tuy nhin trong phn bi lm khng nht thit phi v th (tr trng hp bi ton c phn kho st v v th).

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IV/ BT PHNG TRNH M & LGARIT.1/.

a) Vi 0 a 1 : Hm s m y a nghch bin trn R.29x 17 x 11 7 5x

1 12 2

29x 17x 11 7 5x 29x 12x 4 0

2(3x 2) 0 2

x3

.

Ch :Bt phng trnh: 29x 12x 4 0 l dng bt phng trnh bc hai; phng php chung gii

l xt du tam thc bc hai 2f (x) 9x 12x 4 .

Cch gii: 29x 12x 4 0 2(3x 2) 0 ch l mt trng hp c bit.b) Phng php a v lrarit cng c s.

4 2log (x 8) log (x 2) 2 21

log (x 8) log (x 2)2

2 2log x 8 log (x 2)

2

x 2

x 8 (x 2)

2

x 2

x 3x 4 0

x 2

4 x 1

2 x 1.

Ch :

Cn trnh cc php bin i lm thay i iu kin xc nh ca bt phng trnh.

iu kin xc nh ca bt phng trnh: 2 2log (x 8) 2 log (x 2) l:x 8 0

x 2 0

.

iu kin xc nh ca bt phng trnh: 22 2log (x 8) log (x 2) lx 8 0

x 2 0

.

C th tch ring phn iu kinx 8 0

x 2 0

x 2

v phn gii 4 2log (x 8) log (x 2) . . . 4 x 1 .Sau chn nghim tha iu kin 2 x 1.

c) Phng php a v ly tha cng c s.x 2x1

13

9

2x2x x 13 3

2x2x

x 1

2x2x 0

x 1

12x 1 0

x 1

2x(x 2)

0x 1

2x(x 2)(x 1) 0 x 2 hoc 1 x 0 .

Ch : Trc khi gii c th t iu kin (xc nh): x 1 0 x 1 . Xt du f (x) 2x(x 2)(x 1) :

000f(x)

0x -2 -1

Hc sinh thng bin i sai: 2x2xx 1

2x(x 1) 2x (!)

hoc:2x

2xx 1

1

1x 1

(!)

Bt phng trnh:2x(x 2)

0 (1)x 1

c tp xc nh l R \{ 1} .

Bt phng trnh: 2x(x 2)(x 1) 0 (2) c tp xc nh l R.(1) (2) trn R \{ 1} v x 1 khng l nghim ca c (1) v (2).

a,b R :a

0b

a.b 0 .a

0b

b 0

a.b 0

.

d) Phng php a v lrarit cng c s.

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03x - 5

x -1 0

0

1 5/33/2 +-

2x - 3

x

20,5log (x 4x) 5

20,5 0,5log (x 4x) 5log (0,5)

2 50,5 0,5log (x 4x) log (0,5)

2 50,5 0,5log (x 4x) log 2 2

2 5

x 4x 0

x 4x 2

2

2

x 4x 0 (1)

x 4x 32 0 (2)

Tp nghim: S ( 4; 0) (4; 8) .Ch : Hc sinh thng xt du v tm nghim ca tng bt phng trnh (1), (2). Sau tm giao

hai tp nghim ca (1) v (2). Tuy nhin ta nn xt du chung ca c (1) v (2) ri tm giao

trn bng xt du chung :

x2 - 4x - 32 0

00

0

- + 84-4 0x

x2 - 4x

Nu bi tp l gii bt phng trnh: 20,5log (x 4x) 5 th tp nghim l

S [ 4; 0) (4; 8] .

Nu gii ring l tng bt phng trnh trong h m khng thy c mi lin quan gia ccbt phng trnh trong h th i khi lm cho cch gii di dng. Th d nu bi tp l giibt phng trnh: 20,5log (x 4x) 5 (*) .

Khi (*)2

2

x 4x 0

x 4x 32

2x 4x 32 0 x < 4 hoc x > 8.

e) Phng php a v ly tha cng c s.

ln x 1 ln x 1 ln x ln x 13.5 13.7 7 5 ln x 1 ln x 15 (3.5 5) 7 (1 13.7 ) ln x ln x28 20

5 75 7

ln x 2

5 57 7

ln x 2 2ln x ln e

2

x 0

x e

2x e .

Ch : Trc khi gii c th t iu kin: x > 0.

C th gii:ln x 1 ln x 1 ln x ln x 1

3.5 13.7 7 5

ln x 1 2 ln x 1

5 (3 5 ) 7 (7 13)

.f) Phng php a v lrarit cng c s.

5 1

2

2x 3log log 0

x 1

5 1 5

2

2x 3log log log 1

x 1

1 1

2 2

2x 3 1log log

x 1 2

2x 30

x 12x 3 1

x 1 2

2x 30

x 13x 5

0x 1

3 5

x .2 3

Ch :

Nhn xt:a

0b

ab 0 a v b cng du.a

0b

ab 0 a v b tri du.

C th gii:

2x 30

x 12x 3 1

x 1 2

bng bng xt du:

C th gii:

2x 30

x 12x 3 1

x 1 2

2x 3 0

x 1 0

3x 5 0

hoc

2x 3 0

x 1 0

3x 5 0

3 5

x .2 3

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000f(t)

2t -1 0

Nu bi tp l gii bt phng trnh: 5 12

2x 3log log 0 (2)

x 1

Khi (2) 1 12 2

2x 3 1log log

x 1 2

2x 30

x 12x 3 1

x 1 2

2x 3 1

x 1 2

3x 50

x 1

x 1 hoc

5

x 3 .2/.

a) Phng php t n ph:

2 x 2 x0,1 0,1log 1 25 1 log 11 9.5 2 x

2 x 11 9.51 25 (1)10

t 2 xt 5 0 . Khi (1) c dng: 210(1 t ) 11 9t 210t 9t 1 0

1

t 110

. Kt hp vi K: t > 0 0 t 1 2 x5 1 2 x 0 x > 2.

Ch : 2 xx R : 1 25 0 ; 2 x11 9.5 0.

Hm s 0,1y log x nghch bin trn khong (0; ).

b) Phng php t n ph:1 1 1

x x9.25 16.15 25.9

2 1

x x5 59 16 25 0 (1)

3 3

.

t

1

x5t 0

3

. Khi (1) c dng: 29t 16t 25 0 ( t 1 loi)25

t9

12

x5 5

3 3

1

2x

1

2 0x

1 2x

0x

x 0

x(1 2x) 0

10 x

2 .

Ch : Trc khi gii c th t iu kin: x

0.

Hc sinh thng gii sai:1

2x

1 2x (!)

Nu chia hai v ca (1) cho1

25 v t

1

x3t 0

5

. Gii tng t, ta c:

12

x3 35 5

12

x (Hm s

3y

5

nghch bin trn R)

c) Phng php t n ph: 3 3xlog x log 27 3 (1) . K:1

0 x .3

(1)3

33

log 27

log x 3log (3x) 3 3

3

log x 3 (1')1 log x

t 3t log x . Khi (1) c dng:3

t 31 t

2t 2t

01 t

2

t 1

(t 2t)(1 t) 0

t 1 hoc 0 t 2 3log x 1 hoc 30 log x 2 1

x3

hoc 1 x 9 .

Kt hp vi iu kin:1

0 x3

(1) c nghim:1

0 x3

hoc 1 x 9 .

Ch : Xt du 2f (t) (t 2t)(1 t) :

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d) Phng php t n ph:4

4 2 23 1 1 1

3 3 3

x4log x log 32 log x 41 85log x (1).

81

K: x > 0.

t 3t log x . Khi (1) c dng:4 2 24t (4 4t) 32t 41 85t 4 24t 101t 25 0

21

t 254

1

t 52

1t 5

21

5 t

2

3

3

1log x 5

21

5 log x

2

3 x 243

1 1x

243 3

.

3/.a) Phng php phn tch thnh nhn t: x x2 4.5 4 10 x x x2 10 4.5 4 0 x x x2 (1 5 ) 4(1 5 ) 0 x x(1 5 )(2 4) 0

x

x

1 5 0

2 4 0

hoc

x

x

1 5 0

2 4 0

x

x

5 1

2 4

hoc

x

x

5 1

2 4

x > 2 hoc x < 0.

Ch :

a

0b

ab 0 a v b cng du.

a

0b ab 0 a v b tri du.b) Phng php phn tch thnh nhn t da vo nghim ca tam thc bc hai.

2log x (x 3) log x 2 2x 0

Xt tam thc bc hai (theo logx): 2f (log x) log x (x 3) log x 2 2x 2 2(x 3) 4(2 2x) (x 1)

Tam thc bc hai c hai nghim:

x 3 x 1log x 2

2x 3 x 1

log x 1 x2

Xt phng trnh: log x 1 x (1)

* Hm y = logx ng bin trn khong (0; ) .* Hm y = 1 x nghch bin trn R nn nghch bin trn khong (0; ) .

* x = 1 l nghim ca (1) (v log1 = 1 1).* x > 1 logx > log 1 = 0 v 1 x < 1 1 = 0 1 x < 0 < logx.* 0 < x < 1 logx < log 1 = 0 v 1 x > 1 1 = 0 1 x > 0 > logx.Khi : 2log x (x 3) log x 2 2x 0 (log x 2)(log x x 1) 0

log 2 0

log x x 1 0

hoc

log x 2 0

log x x 1 0

log x 2

log x 1 x

hoc

log x 2

log x 1 x

2x 10

x 1

hoc2x 10

0 x 1

1 x 100 .

Ch : Trc khi gii c th t iu kin: x > 0. C th t n ph t log x .

Phng trnh: log x 1 x c nghim duy nht x = 1. C th bin i:

2 2f (log x) log x (x 3) log x 2 2x log x 3log x 2 x log x 2x f (log x) (log x 1)(log x 2) x(log x 2) (log x 2)(log x 1 x)

4/.

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+

f(x)

+

f '(x)

x 0

loge - 1

e

a) Phng php vn dng s bin thin ca hm s: x 2 x3 7log 2 1 log 5 2 3.

Hm s x 2 x3 7f (x) log 2 1 log 5 2 xc nh trn R.

x 2 x

x 2 x

2 .ln 2 5 .ln 5f '(x) 0, x R.

2 1 ln 3 5 2 ln 7

x 2 x3 7f (x) log 2 1 log 5 2 ng bin trn R.f (1) 3 x 1, f (x) f (1) 3 .

(1) c nghim x 1. b) Phng php vn dng s bin thin ca hm s:

x x x3.2 7.5 49.10 2 x x3.2 7.5 2 49.10

x x

x

3.2 7.5 249

10

x x x1 1 1

3 7 2 495 2 10

.

Hm sx x

1 1 1f (x) 3 7 2

5 2 10

xc nh trn R.

x x x1 1 1 1 1 1

f '(x) 3 ln 7 ln 2 ln 0, x R.5 5 2 2 10 10

x x1 1 1f (x) 3 7 2

5 2 10

nghch bin trn R.

f ( 1) 49 x 1, f (x) f ( 1) 49 .(1) c nghim x 1.

5/.a) Phng php vn dng s bin thin ca hm s: log x ln x m

Xt hm s f (x) log x ln x trn khong (0;e) .1 1 1 ln10

f '(x)x ln10 x x ln10

f '(x) 0, x (0; e).

(v e < 10 lne = 1 < ln10)

x 0lim f (x)

,

x elim f (x) loge 1

.

Da vo bng bin thin:f (x) m nghim ng vi mi x thuc khong (0;e) m log e 1 .Ch :

x 0 x 0 x 0

ln x 1 ln10lim f (x) lim ln x lim ln x

ln10 ln10

(vx 0lim ln x

v

1 ln100

ln10

)

x elim f (x) loge ln e log e 1 (do tnh lin tc ca hm y log x v y ln x )b) Phng php t n ph v vn dng s bin thin ca hm s:

2 2 2cos x sin x cos x3 5 m.5 2 2 2cos x 1 cos x cos x3 5 m.5

2 2cos x cos x3 1

5 m (1)5 25

t 2t cos x 2( x R, 0 t cos x 1) . Khi (1) c dng:t t

3 15 m

5 25

.

Hm st

3 1f (t) 5

5 25

xc nh trn on [0; 1].

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3

4

f '(t)

f(t)

t

+

-3 -1- + 0

0

1

0

t t3 3 1 1

f '(t) ln 5 ln 0, t [0;1].5 5 25 25

t

3 1f (t) 5

5 25

nghch bin trn [0; 1].

f (0) 6 ;4

f (1)5

4

f (x) 6, t [0;1].5

Bt phng trnh: f (t) m m 6 .

c) Phng php t n ph v vn dng s bin thin ca hm s:x x9 (m 1)3 m 4 0

2x x3 (m 1)3 m 4 0 (1)

t xt 3 0 . Khi (1) c dng: 2t (m 1)t m 4 0 2t t 4 m(t 1)

2t t 4

mt 1

(v t > 0 nn t + 1 > 0)

Hm s2t t 4 4

f (t) tt 1 t 1

xc nh trn khong (0; ) .

2

2 2

4 (t 1) 4f '(t) 1

(t 1) (t 1)

f '(t) 0 t 1 2t 1 2

t 1 (0; )t 3 (0; )

t (0; ), 3 f (t) m 3 m m 3 .

V/ H PHNG TRNH M & LGARIT.

1) Phng php th kt hp vi vic vn dng cc cng thc v ly tha, lgarit.

g) 2x y

5x y 4

15

25

2 5x 4 2y 4 5x5 5

2y 4 5xx 5x 6 0

x 2y 6

hoc x 3y 11

.

h) (I)3 3 3

x y 13

log x log y 2 log 4

. K:

x 0

y 0

Cch 1: (I) 3 3

y 13 x

log x(13 x) log (9.4)

2

y 13 x

x 13x 36 0

x 4

y 9

hoc

x 9

y 4

(tha iu kin

x 0

y 0

).

Cch 2: (I) 3 3

x y 13

log (x.y) log (9.4)

x y 13

xy 36

Khi x, y l nghim ca phng trnh: 2X 13X 36 0 X 4

X 9

H phng trnh (I) c hai nghim:x 4

y 9

v

x 9

y 4

(tha iu kin

x 0

y 0

).

2) Phng php t n ph a v h tng tch.

a) (I)x 3 x y 1

x x y

2 .3 1

2 3 11

. t:

x

x y

u 2 0

v 3 0

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Khi h (I) tr thnh:u v 11

u.v 24

u, v l nghim ca phng trnh: 2X 11X 24 0

X 8

X 3

u 8

v 3

hoc

u 3

v 8

(tha iu kin

u 0

v 0

)

3

x y

2 2

3 3

hoc

x

x y

2 3

3 8

x 3

y 2

hoc 2

2 3

x log 3

y log 3 log 8

b) (I)2 2

5 5

x 9y 125log (x 3y) log (x 3y) 2

. K: x 3y 0x 3y 0

.

(I) (I) 5 5

5 5

log (x 3y) log (x 3y) 3

log (x 3y).log (x 3y) 2

t: 5

5

u log (x 3y)

v log (x 3y)

. Khi (I) c dng:

u v 3

u.v 2

u, v l nghim ca phng trnh: 2X 11X 24 0 X 2

X 1

u 2

v 1

hoc

u 1

v 2

(I)

5

5

log (x 3y) 2

log (x 3y) 1

hoc

5

5

log (x 3y) 1

log (x 3y) 2

x 3y 25

x 3y 5

hoc

x 3y 5

x 3y 25

x 15

10y

3

hocx 15

10y

3

3) Phng php cng, phng php th.

a) (I)2 2

2

25x 16y 10 (1)


. K:

5x 4y 0

5x 4y 0

(I)2

log(5x 4y) log(5x 4y) 1


2



2

log(5x 4y) log(5x 4y) 1log(5x 4y) log 10.log(5x 4y) 0

2

log(5x 4y) log(5x 4y) 11 log 10 .log(5x 4y) 0


log(5x 4y) 0

(v 21 log 10 0 )

5x 4y 10

5x 4y 1

x 11/10

y 9/8

.

b) (I)

225 1

5

3 2

log x log y 0 (1)

y 2x 3y 0 (2)

K:x 0

y 0

(1) 5log x log y 0 5x

log 0y

x y th vo (2)

3 2y 2y 3y 0 2y 2y 3 0 (v y > 0) y = 3 (y = 1 loi) x 3 x 3 .

H (I) c hai nghim:x 3

y 3

v

x 3

y 3

Ch : Vi N > 0 v 0 a 1 : a alog N log N ( R)

Vi N 0 v 0 a 1 : 2a alog N 2 log N ( R)

Hc sinh thng gii sai nh sau: 225 15

log x log y 0 5 5log x log y 0 (!)

4) Phng php m ha, lgarit ha, a v h tng tch.

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a) (I)x 1 y 1

x y 2

3 3 82

Cch 1: (I)x 1 3 x

y 2 x

3 3 82

2x x

y 2 x

3 82.3 27 0

x

y 2 x

3 27

hoc

x

y 2 x

13

3

x 3

y 1

hocx 1

y 3

Cch 2: (I)x y 2

x y

3 3

3(3 3 ) 82

x y

x y

3 .3 9

3 3 82/3

3x, 3y l nghim ca phng trnh: 282

X X 9 03

X 27

1X

3

3

1

3 3

3 3

hoc

1

3

3 3

3 3

x 3

y 1

hoc

x 1

y 3

b) (I)log y 3

log x log y 2

x 10

Cch 1: (I)log x 2 log y

log y.log x 3

log x 2 log y

log y.(2 log y) 3

2

log x 2 log y

log y 2log y 3 0

log x 3

log y 1

hoc

log x 1

log y 3

3x 10

y 10

hoc

1

3

x 10

y 10

.

Cch 2: (I)

xlog 2

y

log y.log x 3

2x 10

y

log y.log x 3

2

2

x 10 y

logy.log(10 y) 3

. . .

Cch 3: (I)log x ( log y) 2

log x( log y) 3

.

Khi logx, (logy) l nghim ca phng trnh: 2X 2X 3 0 . . .

5) Kt hp cc phng php t n ph, phng php th.

a) (I)3 3log y log x

1 13 3

x y 18

log x log y 1

. K:x 0

y 0

Cch 1: Kt hp vi iu kin trn, (I)

3 3log x log x

13

y y 18

xlog 1

y

3log xy 9

x3

y

3 3 3log x.log y log 9

x 3y

3 3

log (3y).log y 2

x 3y

3 3

(1 log y) log y 2

x 3y

23 3log y log y 2 0

x 3y

3log y 1

x 3y

hoc 3log y 2

x 3y

x 9

y 3

hoc x 1/3y 1/9

.

Cch 2: Kt hp vi iu kin trn, (I)3 3log x log x

3 3

y y 18

log x log y 1

3 3

3 3

log x.log y 2

log x log y 1

3 3

3 3

log x.( log y) 2

log x ( log y) 1

Khi log3x, (log3y) l nghim ca phng trnh: 2X X 2 0 . . .Ch : Vi a, c l hai s dng v 0 b 1 : b blog c log aa c ( v b blog c log ab blog a log c ).

i vi phng trnh: 3 3(1 log y) log y 2 c th t n ph: 3t log y .

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b) (I) x y

2 2

5log y log x

2log(x y ) 1 log3

. K:0 x 1

0 y 1

Kt hp vi iu kin trn, (I)x

x

2 2

1 5log y

log y 2

log(x y ) log 30

2x x

2 2

2log y 5log y 2 0

x y 30

x2 2

log y 2x y 30

hoc x

2 2

1log y 2x y 30

2

2 2y xx y 30

hoc2

2 2x yx y 30

2

2

y x

y y 30 0

hoc

2

2

x y

x x 30 0

2y x

y 5

hoc

2x y

x 5

(v x 0, y 0 )

x 5

y 5

hoc

x 5

y 5

(v x 0, y 0 ).

Ch : i vi phng trnh: xx

1 5log y

log y 2 c th t n ph: xt log y .

6) Phng php t hai n ph.a)

x 2y

2

x 4

4(x 2y) log x 9

(I) 2

2

(x 2y) log x 2

4(x 2y) log x 9

. iu kin (xc nh): x > 0.

t2

u x 2y

v log x

. Khi h phng trnh (I) c dng:

u.v 2

4u v 9

2

v 9 4u

4u 9u 2 0

u 2

v 1

hoc

u 1/4

v 8

2

x 2y 2

log x 1

hoc

2

x 2y 1/4

log x 8

x 2

y 0

hoc

x 256

y 1023/8

(tha K x > 0)

c) (I)

3

3

yx

yx

5 .2 20025 4 689

. K: y 0 . t3

x

y

u 5 0v 2 1

(v y 0 nn y 02 2 1 ).

d) Khi h phng trnh (I) c dng:2 2

u.v 200

u v 689

2

u.v 200

(u v) 2u.v 689

2

u.v 200

(u v) 1089

u.v 200

u v 33

(v u + v > 0)

Khi u, v l nghim ca phng trnh: 2X 33X 200 0 X 25

X 8

u 25

v 8

hocu 8

v 25

x 8

y 9

hoc

3

522

x 27log 2

y 4log 5

.

7) Phng php cng, phng php th.

a) (I) x

y

log (4x 3y) 2

log (4y 3x) 2

. K:

0 x 1

0 y 1

(I)2

2

4x 3y x (1)

4y 3x y (2)

. (1) (2) 2 2x y x y (x y)(1 x y) 0

x y

x 1 y

* Th x = y vo (1) 27x x x 7 (v x > 0) x y 7 (tha K).

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* Th x = 1 y vo (2) 24y 3(1 y) y 2y y 3 0

1 13y

2

1 13y

2

1 13y

2

loi v

1 130

2

;

1 13y

2

loi v

1 13y 1

2

( x 1 y 0 )

H (I) c nghim duy nht x = y = 7.

b) (II) x

y

log (3x 4y) 3log (3y 4x) 3

. K:

0 x 10 y 1

(II)3

3

3x 4y x (1)

3y 4x y (2)

. (1) (2) 3 3x y x y 2 2(x y)(x xy y 1) 0

x = y (v2

2 2 21 3x xy y 1 x y y 1 0, x, y R 2 4

)

* Th x = y vo (1) 37x x x 7 (v x > 0) x y 7 (tha K).

H (II) c nghim duy nht x y 7 .Ch : Cc h phng trnh (I), (II) trong bi tp ny l cc h phng trnh i xng i vi x, y.

8) Phng php m ha, lgarit ha.

a) (I)y x

2

4(x y).3

7293log (x y) x y

Cch 1: (I)

y x

2

4(x y).3

729x y

log (x y)3

y x

x y3

4(x y).3

729

(x y) 2

x y 2

y x36

22 .3

3

x y23

y 6

2 2

3 3

y 63 32 23 3

x y 6 .

Khi : (I)x y 6

x y 4

x 5

y 1

.

Cch 2: (I)2 2 2 2

2

log (x y) (y x) log 3 log 4 log 729

x ylog (x y)

3

2 2x y

(x y) log 3 2 6 log 33

2 2

1 3log 3(x y) 2(1 3log 3)

3

x y 6 .

(Phn cn li tng t cch 1)

b) (II)

2

2 9x y 6y x9 2

xy 1

x y

* Khi x = 1 2

6y 1

x 1

y 1

1 y

x 1

y 1

.

* Khi 0 x 1

2

2 9x y2 6y x9 2

x y

y y

2x y

2 92 x y 6y x

9 2

2 22 9

2 y y 6y y9 2

25

y 15y9

31

y27

1

y3

2

1x 9

3

.

H (II) c hai ngimx 1

vx 9

1

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