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8/3/2019 phuong_trinh_mu_va_logarit_6938_5085
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PHNG TRNH M V LOGARITA.PHNG TRNH M
VN 1: Cc phng php gii phng trnh m.I.Cng thc ly tha v cn thc.
.
.
.
. .
m n m n
m n m n
m
n m n
n n n
n m n m
m n m n
a a a
a a a
a a
a b a b
a a
a a
II. Cc phng php gii phng trnh m.1) a v dng c bn.
( )0
(0 1) ( ) log
f x
a
b
a b a f x b
2)Phng php a v cng c s.Bin i phng trnh v dng :
( )( ) ( )
0 1
g x f x af x g x
a
Nu c s a khng ph thuc x ( a=a(x))
( ) ( ) 0( ) ( )( ( ) 1)( ( ) ( )) 0
g x f xa x
a x a xa x f x g x
3)Phng php dng n s ph.t t= ( )f xa chn c s a thch hpiu kin t >0Bin i phng trnh m v phng trnh bc 2 , bc3 theo tGii phng trnh ny v chn nghim t >0Gii tip suy ra x4)Phng phng php a v phng trnh tch.-Nhm cc s hng ri t tha s chung suy ra phng trnh tch5)Phng php ly logarit thch hp 2 v.
Dng ( ) ( )0 1
0 1
f x g xa
a bb
Ly logarit c s a 2 v( ).log ( ) log
( ) ( ).log
a a
a
f x a g x b
f x g x b
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PHNG TRNH M V LOGARIT6)Phng php dng tnh n iu.Bin i phng trnh v dng f(x)=g(x)Trong f(x) v g(x) l 2 hm s n iuon nhn 1 nghim x= 0x
Suy ra phng thnh c nghim duy nht x= 0x
III.Mt s v d.
VD1:Gii phng trnh0,5
1(0,2) 5.(0,04)5
xx
Gii:1
1 12
1
2
1 12( 1)2 2
2 3
5 1(1) 5.
255
5 5.5
5 5
2 3
3
xx
xx
x x
x x
x
VD2: Gii phng trnh:
2
2 2 442 5. 2 6 0x x
x x
Gii:iu kin 2 4 0 2x x hoc 2x
2
2 2 44(1) 2 5. 2 2 6 0x x
x x
t t=2 4( 2)x x . iu kin t>0
24
56 3
22
t
t tt
2 4
2 2
2 2
3( ai)
2
t=4 ( 2) 4
4 4 4 4
0 4
4 16 8
4
5
2
x x
t lo
x x x x
x
x x x
x
x
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PHNG TRNH M V LOGARIT
S:5
2x
VD3.Gii phng trnh8.3 3.2 24 6 x x x (1)
Gii:
(1) 8.(3 3) 2 (3 3) (3 3)(2 8) 0
3 3 1
2 8 3
x x x x x
x
x
x
x
S: x=1;x=3VD4.Gii phng trnh
2 4 23 5x x (1)Gii:Ly logarit c s 3 hai v
2 2
3 3 3
23
( 4) log 3 2 .log 5 4 2 log 5
2 log 5 4 0
x x x x
x x
2
3 3
2
3 3
log 5 log 5 4
log 5 log 5 4
x
x
VD5.Gii phng trnh
3 72
5 5
x
x
Gii: Ta thy x=1 l mt nghimca phng trnh
t3 7
( )5 5
x
f x
l hm s gim trn R
( ) 2xg x l hm s tng trn RM f(1)=g(1)
Vy phng trnh c nghim duy nht x=1VD6. Gii phng trnh:
1 1 12 3 5 2 3 5 x x x x x x Gii:
t1
( ) 2 3 5 x x x
f x
l hm s tng trn R1 1( ) 2 3 5 x x xg x l hm s gim trn R
M1 1
2 2f g
nn phng trnh c nghim x=
1
2
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PHNG TRNH M V LOGARITVD7 Gii phng trnh:
2 23.25 (3 10).5 3 0(1)x xx x
Gii :t t= 25x (t>0)
(1) 23 (3 10) 3 0(2)t x t x
1
3
3
t
t x
Vi
2
5
5
1 1 15 2 log
3 3 3
2 log 3
xt x
x
Vi23 5 3 (3)xt x x
(3) c 1 nghim x=2t 2( ) 5xf x l hm s tng trn R( ) 3g x x l hm s gim trn R
Vy (3) c nghim duy nht x=2Vy (1) c nghim : x=2 ; 52 log 3x
IV.Mt s bi tp:Bi 1: Gii phng trnh: 1 4 24 2 2 16 x x x
Bi 2: Gii phng trnh: 12log 9 5.3 4x x
Bi 3: Gii phng trnh: 2 3 2 3 4x x
Bi 4: Gii phng trnh: 2 1 24 .3 3 2 .3 2 6 x x x x x x x
Bi 5: Gii phng trnh:1 1 1
9 6 4 0 x x x
VN 2: Tm m phng trnh m c nghim, c nghim duy nht.I. Tm m phng trnh m:F(x,m)=0 (1) c nghim xD.Cch gii:-t n ph: t:=q(t), tm iu kin cho n ph t.-Chuyn iu kin xD thnh iu kin tT.-Bin i phng trnh (1) thnh phng trnh bc 2 theo t f(t,m)=0 (2).*Cch 1.-Bin i (2) tng ng vi f(t)=m (2) vi tT.-Tnh f(t), lp bng bin thin.- (1) c nghim xD khi v ch khi (2) c nghim tT iu ny cng tng ng
vi ng thng y=m c im chung vi th y=f(t)-Da vo bng bin thin tm iu kin ca m.
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PHNG TRNH M V LOGARIT*Cch 2.-Ta c (1) f(t,m)=0 (2) (bc 2 theo t)- (1) c nghim xD khi v ch khi (2) c nghim t T
Tc l (2) c 1 trong 2 nghim thuc T hoc c hai nghim u thuc T.II. Tm m phng trnh c nghim duy nht
*Cch 1.iu kin cn.-Gi s phng trnh c nghim x0. Da vo tnh i xng, hm s chn, gi tr
tuyt i phng trnh c nghim x1.-T phng trnh c nghim duy nht khi v ch khi x0=x1.-Thay vo phng trnh tm gi tr m.
iu kin .-Thay gi tr m va tm c vo phng trnh. -Gii phng trnh v chn m sao cho tha mn iu kin phng trnh c nghim
duy nht.T a ra kt lun cc gi tr m tha mn.
*Cch 2.-Bng cch t n ph t=q(x) a phng trnh cho v dng f(t)=m.-t y=f(t) vi tT-Tnh f(t), lp bng bin thin trn T. -T phng trnh (2) c nghim duy nht khi v ch khi ng thng y=m ch
c duy nht mt im chung vi th y=f(t).-Da vo bng bin thin c c gi tr m cn tm.
III.Mt s v d :VD1: nh m phng trnh: 1 4 2 3 2 3 0 1x xm m m c nghim
Gii:
t: t=2x
(t>0)
2
2 2
2 2
2
2
1 1 2 3 3 0
2 6 3
2 1 6 3
6 32 0
2 1
m t m t m
mt m m t t
m t t t t
t tm t
t t
t 2
2
6 30
2 1
t t f t t
t t
2
22
2
4 8 12
2 1
10 4 8 12 0
3
t tf t
t t
t f t t t
t
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PHNG TRNH M V LOGARITBng bin thin:
(1) c nghim 2x R c nghim t>0 ng thng y=m c im
chung vi th y f x .
Da vo bng bin thin, ta c:3
3
2
m
S:3
32
m
V d 2: Cho phng trnh: 3 16 2 1 4 1 0 1x x x m m Tm m phng trnh c 2 nghim tri du.Gii:t: 4 0xt t phng trnh (1) tr thnh 23 2 1 1 0 2 f t m t m t m Phng trnh (1) c 2 nghim tri du
1 20
1 2 1 20 4 4 4 1x x
x x t t
(2) c nghim t1, t2 tha 0 < t1 < 1 < t2
. 1 0
. 0 0
3 4 3 0
3 1 0
33
4 31
3 4
1
a f
a f
m m
m m
m
mm
m
Vy phng trnh c 2 nghim tri du khi: .3
14
m
V d 3: Tm m phng trnh sau c nghim duy nht: 1
13 2 1
2x
m
Gii:Phng trnh (1) c nghim khi v ch khi:
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PHNG TRNH M V LOGARIT
1
2
2
2
23 2 0
3
1 11 2 1 log
3 2 3 2
1 log 3 21 log 3 2
x
m m
xm m
x mx m
Phng trnh c nghim duy nht
2 2
2
1 log 3 2 1 log 3 2
log 3 2 0 3 2 1 1
m m
m m m
IV.Mt s bi tp:Bi 1: Tm m phng trnh 4 9 2 2 3 1 0x xm m m c nghim.
Bi 2: Tm m phng trnh .2 2 5 0x xm c 1 nghim duy nht.
Bi 3: nh m phng trnh:
3 2 2 3 2 2
tgx tgx
m
C ng 2 nghim trong ,2 2
Bi 4:Tm k phng trnh 11 4 3 2 .2 3 1 0x xk k k
c 2 nghim tri du.Bi 5:Gii v bin lun phng trnh .3 .3 8x xm m
B.PHNG TRNH LOGARITVN 1: Cc phng php gii phng trnh logarit.I.Dng c bn:
log
log 0, 1
log , ; ; 0a
N
a
xx
a
x N x a a a
a x x a x x
Cng thc i c s:log
log log log loglog
aa a b b
a
x x b x x
b
1log
logx
a
ax
; log logb bc aa c
3
1log log
3log log
aa
aa
x x
x x
II.Cc phng php gii phng trnh logarit.1.Phng php a v cng c s-Bin i phng trnh v dng:
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PHNG TRNH M V LOGARIT
log log 0 1
0
0
a a f x g x a
f x
g x
f x g x
2.Phng php t n ph:-Chn n s ph thch hp, bin i phng trnh cho thnh mt phng trnh i s.3.Phng php a v dng phng trnh tch:-Nhm cc s hng, t tha s chung suy ra phng trnh tch. 4.Phng php dng tnh n iu.-Suy on 1 nghim c bit v chng minh nghim duy nht.
5.Dng: 0 1
log log0 1
m
a b
a f x a g x
b
-Suy on nghim x0v chng minh nghim duy nht.
-Nghim duy nht x0 tha:
0
0
m
n
f x a
g x a
6.Dng phng php i lp.
A BA m
A mB m
B m
7.Dng: log loga x a x f x g x
0
1
0
a x
a x
f x
f x g x
III.Mt s v d:
V d 1: Gii phng trnh: 421 1
log 3 log 4 12 4
x x
Gii:
K:0
1
x
x
2 2 2
2
1 11 log 3 . .8log 1 log 4
4 2
log 3 1 log 4
3 1 4 2
x x x
x x x
x x x
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PHNG TRNH M V LOGARIT Nu 0< x 1
S: 3; 3 2 3x x V d 2: Gii phng trnh:
1 2
1 14 lg 2 lgx x
Gii:
K: 400
lg 4 10
lg 2 1
100
xx
x x
xx
t: lg 4 2t x t t
2
2
1 21 1
4 2
2 2 4 4 2
10 8 4 2
3 2 0
1
2
t t
t t t t
t t t t
t t
t
t
1 lg 1 10t x x 22 lg 2 10 100t x x
S: x=10; x=100V d 3: Gii phng trnh:
3 2log log 1 1x x
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PHNG TRNH M V LOGARITGii: iu kin: 0x t: 2log 3
tt x x
21 log 1 3
2 1 3
1 31 2
2 2
t
tt
tt
t
Nhn xt: t=2 l nghim ca (2) V tri l hm s gim.V phi l hm s hng.Nn phng trnh c 1 nghim duy nht l 232 log 2 3 9t x x
S: x=9IV.Mt s bi tp:Bi 1: Gii phng trnh
2 2 4 2 4 22 2 2 2log 1 log 1 log 1 log 1 x x x x x x x x
Bi 2: Gii phng trnh:4
2 1
2log 1
2 1x
x
x
Bi 3: Gii phng trnh: 2 23 2 3log 2 9 9 log 4 12 9 4 0x x x x x x
Bi 4: Gii phng trnh: 9
log 1 lg 02
x x
Bi 5: Gii phng trnh: 2 23
1log 3 1 2 log 1
log 2x
x x
VN 2: nh m phng trnh logarit c nghim, c nghim duy nht:I.Tm m phng trnh: , 0 1F x m c nghim x D
-t n s ph: logat x thch hp.
-Chuyn iu kin x D t T -Bin i (1) thnh phng trnh bc 2 theo t. Bin i phng trnh ny v dng:
2 f t m
-Tnh , f t t T . Lp bng bin thin- (1) c nghim trn D (2) c nghim trn T.-Da vo bng bin thin iu kin ca mII. nh m phng trnh logarit c nghim duy nht:Cho phng trnh ( cha logarit )
, 0 1F x m
-t: t p x
-Tm iu kin ca t T -Bin i phng trnh (1) v dng:
2 f t m
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PHNG TRNH M V LOGARIT
-Tnh f t vi t T -Lp bng bin thin trn T-Phng trnh (1) c nghim duy nht
(2) c nghim duy nht trn T.-Da vo bng bin thin k ca m.
Cch khc:Phng trnh (1) (2) l phng trnh bc hai vi x (1) c nghim duy nht 2 c 1 nghim kp
1 22
bx x
a hoc c 2 nghim 1 2x x
0
2
b
a
hoc af 0
III.Mt s v d:
V d 1: Tm m phng trnh: 2lg 2 lg 1 0 1 x mx x c nghim.Gii:
Ta c: 21 lg 2 lg 1 x mx x
2
2
1 0
2 1
1
12
2
x
x mx x
x
x xm
x
t: 2 1
12
x x f x x
x
2
2
2 20
4
xf x
x
v x>1
Bng bin thin:
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PHNG TRNH M V LOGARIT
(1)c nghim khi v ch khi (2) c nghim x>1 12
m
V d 2: Tm m phng trnh: 21 1
2 2
1 log 4 2 1 log 4 2 0 1m x m x m
C 2 nghim x1, x2 tha mn: 4 < x1 < x2 < 6Gii:t: 1
2
log 4t x
iu kin:
1 12 2
4 6 0 4 2
log 4 log 2 1
x x
t x
21 1 . 2 1 . 2 0 2 f t m t m t m
(1)c 2 nghim tha mn : 1 24 6x x 2 c 2 nghim 1 2,t t tha 1 21 t t
0 9 0
af 1 0 1 4 2 0
4 11 0 0
2 2 2
11
121
1 21
4
m m
S m
m
m m
m m
m m
Vy:1
12
m m
IV.Mt s bi tp
Bi 1: Tm m phng trnh 2
2 1
2
4 log log 0 x x m
c nghim thuc khong 0,1 Bi 2: Gii v bin lun phng trnh theo m
3 3 32log log 1 log 0 x x m
Bi 3: Tm m phng trnh 2
2 2lg lg 3 0 x mx x c nghim.
Bi 4: Cho phng trnh: 3 22 2log 5 6 log 3 1 1mmx mx x x Tm cc gi tr ca x nghim ng phng trnh (1) vi mi 0m
Bi 5: Vi gi tr no ca a th phng trnh: 2loglog
a
x
a a xa
C nghim duy nht.