Bồi dưỡng học sinh giỏi toán lớp 12 Môn Đại số và giải tích tập 2

ins.
LE
HOANH
PHO
Nha gido
Uu tu
BOIDUSNG
7
HOC SINH GIOI TOAN
DAI
SO -
GIAI
TICH
•
Danh cho
HS
lap 12 on tap
&
nang cao
Id"
nang lam bdi.
Chudn
bi cho cdc
ki thi
qudc gia do
Bo GD&OT
to choc.
a
NHA XUA1HN DAI HOC
QUOC
GIA HAJIOI

B6I
DUBNG
HQC
SINH
QOI
TOAN
DAI
SO-GIAI
TICH
Boi
duOng hoc sinh
gioi
Toan Dai so 10-1.
Boi
duQng hoc sinh
gioi
Toan Dai so 10-2.
Boi
dti8ng hoc sinh
gioi
Toan
Hinh
hoc 10.
Boi
duQng hoc sinh
gioi
Toan Dai so 11.
-
Boi duQng hoc sinh
gioi

Toan
Hinh
hoc 11.
Bo de thi tii luan Toan
hoc.
Phan
dang
va phi/dng
phap
giai
Toan So phtic.
Phan
dang
va phirong
phap
giai
Toan To hop va
Xac
suat.
1234 Bai tap ttj luan
dien hinh Dai so
giai
tich
1234 Bai tap tu luan
dien hinh
Hinh
hpc
luting
giac
ThS.

LE
HOANH
PHO
Nhd gido
Uu tu
BOIDUdNG
,
HOC SINH GIOI TOAN
3>
DAI
SO-GIAI TICH
12
-
Ddnh cho
HS
ldp 12 on tdp
&
ndng cao
ki
ndng Idm bdi.
Chudn bi cho cdc
ki thi
qudc gia do Bo
GD&DT
to chut:.
NHA XUAT BAN DAI HQC
QUOC
GIA HA
NQI
NHA

XUAT
BAN DAI HOC
QUOC
GIA HA NOI
16
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dinh
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In
xong
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chieu
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nam 2010.
Ldi
N6I DXU
Be giup cho hoc sinh
ldp
12 cb them tai liiu
tu
boi
duong, ndng
cao
va
ren luyen
ki
nang gidi todn theo chuong trinh phdn ban mai. Trung
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sdch gido
due
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quy

ban dong nghiep
va
cdc
em hoc sinh cuon: "Boi dudng hoc sinh
gidi todn Dai so' Gidi tich 12" nay.
Cuon
sdch
nay nam trong
bo sdch
6
cuon gom:
- Boi duong
hoc
sinh gioi
todn
Hinh
hoc
10.
- Boi duong
hoc
sinh gidi
todn
Dai
so'
10.
- Boi duong
hoc
sinh
gioi todn
Hinh

hoc
11.
- Boi dudng
hoc
sinh
gidi todn Dqi so'- Gidi
tich 11.
- Boi dudng
hoc
sinh gidi
todn
Hinh
hoc
12.
- Boi dudng
hoc
sinh gidi
todn Gidi
tich 12.
do nha gido
uu tu,
Thac
si
Le
Hoanh Phd
to'chiec bien soqn.
Noi
dung
sdch
duoc bien

soqn theo chuong trinh phdn
ban:
co
bdn
vd ndng
cao
mdi cua
bo
GD &
DT, trong dd mot
so"van deduac
md
rong
vdi
cdc
dang
bdi
tap hay vd khd dephuc
vu
cho
cdc
em
yeu
thich
mud'n
ndng
cao
todn
hoc, cd
dieu kien phdt trien tot nhat

khd
ndng
ciia
minh.
Cud'n sdch
la
sir ke
thira nhirng hieu biet chuyen mdn vd kinh
nghiem
gidng day
ciia
chinh
tdc gid
trong
qua trinh true tiep dirng
ldp bdi
dudng
cho hoc
sinh
gidi cdc ldp
chuyen todn.
Vdi ndi dung
sue
tich, tdc gid da cd'gdng sap xep, chon loc cdc bdi todn tieu bieu cho
tirng the
loai
khdc nhau
ung
vdi noi dung
ciia

SGK.
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nen
tdng
kien
thuc va
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nang co
ban.
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tu
minh
hoan
thien
cdc
ki
ndng cung
nhu
phat trien
tu duy
qua viec gidi bai tap cd trong
sdch
trudc khi ddi
chieu
vdi
ldi

gidi
cd
trong
sdch
nay,
cd the
mdt soldi
gidi cd
trong
sdch con cd dong, hoc
sinh
cd
the'tu minh lam
ro
hon, chi tiet hon, ciing
nhu tu
minh dua ra nhirng
cdch lap
ludn mdi
han.
Cluing tdi
hy
vong bd
sdch
nay
se la
mdt
tai lieu
thie't
thuc,

bd' ich
cho ngudi day
va
hoc,
dqc biet
cdc
em hoc sinh
yeu
thich mdn todn vd
hoc
sinh chuan bi
cho cdc
ky
thi quoc
gia (tot nghiep THPT,
tuye'n
sinh
DH
-
CD) do
bd
GD &
DT
to chirc sap
tdi.
Trong qua trinh
bien soqn,
cudn
sdch
nay khdng the tranh khdi nhung thieu sdt, chiing

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y
ciia
ban
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gan xa debb
sdch
hoan thien
hon
trong ldn
tdi
ban.
Moi
y
kien
dong gop
xin lien
he:
Trung
tam
sach
giao due Anpha
225C
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Tri
Phuong, P.9,
Q.5, Tp. HCM.
-
Cong
ti

sach
- thiet
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Xin
chan thanh cam on!
3
MUC
LUC
Chuong
II:
Ham so luy thira ham so mu va ham so logarit
§2.
Phucmg
trinh,
he phuomg
trinh,
bat phuong

trinh
mu va logarit 5
Dang 1: Phuong
trinh
mu va logarit 5
Dang 2: Bat phuong
trinh
mu, logarit 21
Dang 3: He phuong
trinh
mu, logarit 31
Chuong
HI:
Nguyen ham,
tich
phan va ung dung
§
1.
Nguyen ham 46
Dang 1:
Dinh
nghia va
tinh
chat
47
Dang 2: Phuong
phap
bien doi bien so 55
Dang 3: Nguyen ham
tirng

phan 62
§2.
Tich
phan 70
Dang 1:
Dinh
nghia va
tinh
chat
71
Dang 2:
Tich
phan da thuc, phan thuc 81
Dang 3:
Tich
phan luong giac 89
Dang 4:
Tich
phan can
thiic
100
Dang 5:
Tich
phan mu - logarit 108
§3.
Ung dung cua
tich
phan 124
Dang 1:
Tinh

dien
tich
hinh phang 124
Dang 2:
Tinh
the
tich
vat thS 130
Chirong
IV.
So
phuc
§1.
S6
phuc
139
Dang 1:
Phep
toan ve so
phuc
140
Dang 2: Bieu dien va tap hop diem 143
§2.
Can bac hai va phucmg
trinh
151
Dang 1: Can bac hai
ciia
so
phuc

151
Dang 2: Phuong
trinh
nghiem
phuc
156
§3.
Dang luong giac 165
Dang 1:
Viet
dang luong giac 165
Dang 2: Toan ung dung 171
CHUONG
II:
HAM SO LUY THUA, HAM SO MU VA HAM
SO LOGARIT
§2.
PHUONG TRiNH, H£ PHUONG TRINH, BAT
PHUONG TRINH MU VA LOGARIT
A.
KIEN
THUG CO BAN
Phuong
phap
chung:
- Dua ve
cung
mot co so
- Dat an phu
- Logarit hoa, mu hoa

- Su
dung
tinh
chat
cua ham so
B.
PHAN DANG TOAN
DANG
1: PHUONG TRlNH MU VA LOGARIT
,
a
^ = a
g(x)
(a>0)
log
a
f(x)
= logag(x), (a > 0, a * 1) o
-
Phuong
trinh mu co ban: a
x
= b (a > 0, a * 1)
Neu b < 0,
phuong
trinh vo
nghiem
Neu b > 0,
phuong
trinh co

nghiem
duy
nhat
x = log
a
b.
a = l
a *
1
, f(x) = g(x)
-
Phuong
trinh logarit co ban: log
a
x =
b(a>0,
a^l)
Phuong
trinh logarit co ban luon co
nghiem
duy
nhat
x = a
b
.
f(x)>0
hayg(x)>0
f(x)
= g(x)
Chii

y: Ngoai 4
phuong
phap
chinh de giai
phuong
trinh mu, lograrit, ta co
the
dung
dinh
nghTa,
bien
doi
thanh
phuong
trinh tich so,
dung
do thi, bit
dang
thuc,
Vi
du 1: Giai cac
phuong
trinh sau:
a) 2
x2
'
3x+2
=4 b) (2 + V3 )
2x
= 2 - V3

c) 2
x+1
5
X
= 200 d)
0,125.4
2x-3
= (4V2)
x
Giai
a) PT o 2*
2
-
3x+2
=2
2
ox
2
-3x
+ 2 = 2»x
2
-3x = 0ox = 0
hoac
x = 3.
b)
PT<^(2+
V3)
2x
= (2+ S)'
1

<=>2x = -l ox =
-i.
2
c) PT o 2 . 10
x
= 200 o 10
x
= 100 o x * 2
5x 5x
d) PT <=> 2"
3
. 2
4x
"
6
= 2
T
o 2
4x
"
9
= 2
Y
5x
<=>
4x - 9 = — <=> 8x - 18 = 5x o x = 6.
2
-BDHSG DSGT12/2- c
Vi
du 2: Giai cac

phuong
trinh sau:
a) (1,5)
5x-7
/o\
x+1
b) 7
X
-' = 2
X
1 3
x+—
x+—
c) 9
X
- 2
2
=2
->2x-l
^ ylogx _ ^logx+1 _ ^ jlogx-1 _ J3 j^ogx-\
Giai
CO
5x-7
_
CO
-x-l
u,
,2j
O 5x - 7
-lox = l

b)PT»f = 2
x
7<=> 7
<=>
x = log, 7
1 i - A x+-
c) PTo 9
X
+i.9
x
=2
X+2
+2.2
X+2
9
X
=
3.2"
+2
3 3
=
— O
X
-
1
= lOgg — O
X
= 1 - lOgg 2 .
Z
2

1
2
d) PT »
7
logx
+13.7
logx

=
5
logx
.
5 +
3.5
logx

o
7
logx
1 +
13
-logx
5
+
?
. 5,
20
^ ylogx | | _ glogx
<=> logx = 2
<=>

x = 100.
Vi
du 3: Giai cac
phuong
trinh sau:
a) 4
X
- 2
X
- 6 = 0
c)
e
2x
-3e
x
-4+
12e"
x
= 0
28 1_
5 20
b) 3
X
+ 1 +
18.3"
x
= 29
d)
27
x

+
12
X
= 2.8
X
Giai
a) Dat t = 2
X
, (t > 0) thi PT
<=>
t - t - 6 = 0
Chon
nghiem
t = 3 <^> 2
X
=3 ox =
log
2
3
1R
b) Dat t = 3\ t > 0 thi PT o 3t + — = 29
<=> 3t
2
- 29t + 18 = 0 o t = 9 hoac t = -
Giai ra
nghiem
x = 2
hoac
c =
log32

- 1.
c) Dat t = e\ (t > 0) thi PT
<=>
t
2
- 3t - 4 + — =0
O t
3
- 3t
2
- 4t + 12 = 0 <=> (t - 2)(t + 2)(t - 3) = 0.
Chon
nghiem
t = 2
hoac
t = 3 nen x = ln2
hoac
x = ln3.
6
-BDHSG DSGT12I2-
d)
Chia 2 ve cho 8
X
> 0 thi PT:
27V
(12\*
(
3
T
i

'3^
-
+
-
{2
\
2)
-
2 = 0
.
Dat t =
-2
,t>0.
PT
<=>
t
3
+ t - 2 = 0
<=>
(t - l)(t
2
+ t + 2) = 0o t = 1
<=>
x = 0.
Vi du 4: Giai cac phuong trinh:
a)2.25
x
+ 5.4
x
= 7.10

x
b)
4
x
+ 6
x
= 9
c)
U/2-V3
+U2+S)
=4 d) 4
X+Nx2
"
2
-5.2
x
"
1+VxI:2
=6
a)
PT«5||]
-7
Giai
2 = 0. Dat t =
,t>0.
PT
<=>
5t
2
- 7t + 2 = 0 o t = 1

hoac
t = | (thoa man)
Suy nghiem x = 0 hoac x = 1.
b) Dieu kien x * 0, dat y = — va chia hai ve cho 4
y
, ta co:
x
PT«
-1
=
0<=>
1
+ 75 .
1
+ 75
—-—
es> y =
log
3
—-—
1
,
1
+ 75 1 ,
o — =
log
3
——
c=>- =
log

3
X
2
2 X
2
1
+ 75
«x
=
log^_
i
-
2
c)
Ta co 72-73.72 + 73 = 1, dat t = (72 + 73") , t >
PT<^t+-=4«t
2
-4t+l=0
t
o t = 2 + 73 hoac t = 2- 73»x = 2 hoac x = -2.
d) Dat t = 2
x+vx2
-
2
, t > 0 thi PT <=> t
2
- -1 = 6
2
<=>
2t - 5t - 12 = 0. Chon nghiem t = 4.

x nen x + 7x
2
-2 = 2
<=>
7x
2
- 2 =2-
<=>
2 - x > 0 va x
2
- 2 = 4 - 4x + x
2
<=>
x < 2 va x
<=>
x = -
2 2
-BDHSG DSGT12/2-
1
Vi
du 5:
Giai
cac
phuong
trinh:
a)
x
3
+(x-2)
6

=0 b)
V2
X
V4
X
(0,125)
X
=
4^2
c)
Ul6-x + $fx~+~l=3
d)
3
/xTl-
3
/x^T
=
v
/
x
2
-1
Giai
a)
BK:x>0,
x-2>0<=>x>2.
Voi
x > 2
thi VT
> 0 nen PT vo

nghiem.
b)
DK:
x # 0, PT o 2
2
_
17
»
22.23"^
=2
3 »- +
—=
-
2
3 2x 3
,
1
»
5x - 14x - 3 = 0
<=>
x = — hoac x = 3.
5
c)
DK:-1
<x< 16.
Datu= #L6-x,
v= 7x + l thi u, v > 0.
lu
+
v = 3

1
6
7 £
2x—
6
7
2
2x
(2~
3
)
x
=
2
3
o 2
2
2
x
=
2
3
k
>
Ta
co he:
u
4
+v
4

=17
Dat
S = u + v, P = uv
thi
u
4
+ v
4
= (u
2
+
v
2
)
2
= ((u + v)
2
-
2uv)
2
-
2u
2
v
2
17
=
(9-2P)
2
-2P

2
= 2P
2
-36P + 81.
Do
do P = 2 hoac P = 16. Vi S
2
- 4P > 0 nen
chon
P = 2 suy ra S = 3 nen
nghiem
x =
1
hoac x = 15.
d)
DK:
x < -1 hoac x > 1.
Vi
x =
±1 khong
la
nghiem
nen
dieu kien:
x < -1
hoac x > 1. Ta co x la
nghiem thi
-x
cung
la

nghiem,
do do xet x > 1.
PT<»
7(x +
l)
2
-7(x-l)
2
=
v
/
x
Y
^lci>6
x
+
1
x-l
x-l
Vx+1
ep^i
. t > 0 thi PT
<i>
t - - = 1
<=>t
2
-t-l
= 0
x-l
t

Chon nghiem
t
1
+ V5
suy
ra
nghiem ciia
PT cho la:
x
= ±
't
+
r
6
t-1
VOI
t =
1
+ V5
Vi
du 6:
Giai
cac
phuong
trinh
sau:
a)
3
4
*

=
4
3
"
c) S""
1
^
2
=8.4
X
b)
3
X
.8
X+1
=36
5 15
-BDHSG DSGT12/2-
Giai
a) Hai ve deu duong, logarit hoa
theo
co so 10 ta co :
4
x
log3
= 3
x
log4
<=>
log

4
log
3
«x
= log4(log
3
4)
3
3x x-2
_
o2
n,2
^
TX-2
b) PT o 3
X
2
X+1
= 3
2
. 2
l
o 3
X
~
2
. 2*
+1
= 1
,x-2

3.2
s
1
<=>
x - 2 = 0
hoac
3.2
X+1
=1
»
x = 2
hoac
2
X+1
= -
<=>
x = 2
hoac
x = -1 - log32.
3
c) Hai vd deu duong, logarit hoa hai ve
theo
co so 2 ta co:
log
2
(3
x
-
1
.2

x2
) = log
2
(8.4
x
-
2
)
<=> (x - l)log
2
3 + x
2
= log
2
8 + (x - 2)log
2
4
<=>x
2
-(2-log
2
3)x +
1
-log
2
3 = 0ox=
1
hoacx=
1
-log

2
3
d) Hai ve deu duong, logarit hoa hai ve
theo
co so 5 ta co:
<
IM /
3
^^
1
i r
3
^-
3x
-
4 1
(x-l)l0g
5
(-)
+
-l0g
5
(-)-—

O x(log
5
3 - 1) - log
5
3 + 1 - I + I log
5

3 = - 1 - \
<=>
X
4 log
5
3-7^-4
+ log^ ^
x
_ 2(log
5
3-4)
2 4log
5
3-7
Vi
du 7: Giai cac
phuong
trinh:
a) 2
log3x2
.5
log3X
=400 b) 4
lnx+1
-6
lnx
-2.3
lnx2+2
c) x
log4

+ 4
logx
=32
I
i
x—
d) 3
,UB4
^+3'
U64
^
=
V^
Giai
a) Dieu kien x > 0,
phuong
trinh
<=>
4
log3 x
.5 400
<=>
20
log3X
= 20
2
<=>
log
3
x = 2 <s> x = 9 (thoa man)

b) DK: x > 0, PT « 4.2
2lnx
-
6
lnx
- 18.3
21nx
= 0
Chia ca hai ve cho
3
21nx
,
dat t -
9 _
2
Chon nghiem t =
—
<=>
x = e
•
&
• 2
lnx
thi
duoc
PT <=> 4t - t - 18 = 0.
c) DK: x > 0, ta co: x
log4
=
4

Iog
^'°
e4
= 4
log4 lo
^
x
= 4
logx
nen PT « 2.4
logx
= 32 o 4
logx
= 16 o logx = 2 o x = 100.
-BDHSG
DSGT1Z/2-
d)
DK: x > 0, dat t = log
3
x thi x = 4
l
PT o V3 . 3' + -L . 3
1
= 2
1
o 4.3' = V3 . 2'
V3
log;
3
V

73 . _ _ , V3
l-j
=^ ot= log^
Vayx=
4
Vi du 8: Giai cac phuong trinh:
a) (4 - Vl5)
tanx
+ (4 + Vl5)
tanx
=8 b) 81
sin2 x
+ 81
c
30
c) (cos72°)
x
+
(cos36°)
x
= 3.2~
x
sin(x—)
d)
e
4
= tan x
Giai
a) Vi (4 - y/l5 )(4 + 7l5) = 1 nen dat (4 - JTE )
tanx

= t, t > 0 thi phuong
trinh ot+ ^=8«>t
2
-8t+ 1 = 0«t = 4±Vl5.
Do
do tanx = -1
hoac
tanx =
1
nen nghiem x =
±—
+
leir,
k e Z.
4
b) Datt= 81
smZx
, 1 <t<81 thiPT
o 81
sm2x
+ si
1
"
51
"
2
" = 30«t + — = 30
t
o t
2

- 30t + 81 = 0 <=> t = 27 hoac t = 3 (chon)
Do do 3
4s,n2 x
= 27 hoac S
4
^ =3 <=> 4sin
2
x = 3 hoac 4sin
2
x = 1.
^V3, „ . 1
o sinx =
±—hoac
sinx = ±—
2 • 2
it
kn,
k e Z.
<=>
x = ±— + kn
hoac
x
6 • 3
c) Phuong trinh: (2cos72°)
x
+ (2cos36°)
x
= 3
Vi:2cos72° 2cos36°
2sin36°.cos36°.cos72

0
sin 36°
=
1
Dat t =
(2cos72°)
x
,
t > 0 thi PT o t + ± = 3
^
t
2_
3t+1=
0«t=
N/5
+ 1
Ta co:
2cos72°
=
2sinl8°
= ——- suy ra nghiem x = ±2.
d) Dieu kien
cosx
* 0, vi sinx = 0 khong thoa man phuong
trinh
nen PT
,_ %/2sinx 72 cosx
72(stnx-cosx)
smx e * e
A

<=>
=
cosx
sinx
cosx
10
-BDHSG DSGT12/2-
Dat
u = sinx, v = cosx, u, v e (-1; 1), u.v > 0 nen ta co phuong
trinh
72u Vgv
o 2
0
~2~
U
V
72t
Xet ham s6 y = f(t) = —, vai t e (-1; 0) u (0; 1).
V2t
^
v
2
72t
e 2 V2t
y' = = ———— < 0 suy ra ham so nghich bien tren
t
2
2t
2
moi khoang (-1; 0) va (0; 1).

Vi u, v cung dau nen u, v cung thuoc mot khoang (-1; 0) hoac (0; 1) do do PT
<=> f(u) = ftVr'o u = v «tanx = 1 <=>x= — + kn (chon).
4
Vi du 9: Giai cac phuong trinh:
a)(sin-)
x
+ (cos-)
x
= 1 b) 4
X
- 3
X
= 7
5 5
c)(-)
x
= x + 4 d)2
x
= x+l.
3
Giai
71 71
a) Vi 0 <
sin—
< 1 va 0 < cos— < 1 do do:
5 5
Neu x > 2 thi ta co (sin- )
x
< (sin- )
2

va (cos- )
x
< (sin- f
5 5 5 5
=> VT < 1 (loai).
Neu x < 2 thi ta co (sin- )
x
> (sin- )
2
va (cos - )
x
> (sin- )
2
5 5 5 5
=> VT> 1 (loai).
Neu x = 2 thi PT nghiem dung, do la nghiem duy nhat
1 3
b)
PT
<=>
(
—
)
x
+
(
—
)
x
= 1 va ta co x = 2 thoa man PT. Vi ve

trai
la ham so
4 4
nghich bien tren R nen co nghiem duy nhat x = 2.
c) Vi 0 < - < 1 nen khi x > -1 thi VT < 3, VP > 3 (loai), khi x < -1 thi VT > 3.
3
VP < 3 (loai), con khi x = -1 thi PT nghiem diing. Vay nghiem duy nhat
lax = -l.
d) PT <=> 2
X
- x - 1 = 0
Xet f(x) = 2
X
- x - 1, D = R. Ta co:
f '(x) = 2
X
ln2 - 1, f "(x)= 2
x
.ln
2
2 > 0, Vx
Do do f '(x) dong bien tren R, f '(x) = 0 <=> x = -log
2
(ln2)
-BDHSG DSGT12/2- 11
BBT
X
-loKln2)
+0
°

f
0
•
+
f
+00
+oc
Vay
f(x) = 0 co toi da 2 nghiem ma f(0) =
f(l)
= 0 nen tap nghiem la:
S =
{0;
1}.
Minh
hoa
bang
do thi cau c) va cau d)
y
t y
Vi
du 10:
Giai
cac phuong
trinh:
a) (^6
+Vl5)
x
+
(^7-v/l5)

x
= 13
c)
6
X
+ 15 = 3
X+1
+ 5.2
X
b)
(2 -
V3)
x
+ (2 + V3)
x
=4
X
d)
x.2
x
= x(3 -x) + 2(2
x
- 1).
Giai
a) Ta co x = 3 la nghiem
ciia
phuong
trinh,
vi ham so
f(x)

= (\o + Vl5 )
x
+ (^7 -
Vl5)
x
la tong cua hai ham so mu voi co so
Ion
hon 1 nen f(x) dong bien tren R. Vay x = 3 la nghiem duy
nhat
ciia
phuong
trinh.
b)
Ta co x = 1 la nghiem
ciia
phuong
trinh.
Bien doi PT
2-V3V
(z
+
S
=
1
thi ve
trai
la ham
f(x)
nghich bien, vay x = 1
la

nghiem duy
nhat
ciia
phuong
trinh.
c) PT«6
X
-3.3
X
+ 15 - 5.2
X
= 0
o
(2
X
- 3)(3
X
- 5) = 0 ci> 2
X
= 3
hoac
3
X
= 5.
<=>
x = log23
hoac
x = log35.
d)
PT » x.2

x
- x(3 - x) - 2.2
X
= 0 o 2
x
(x - 2)
2
x
(x - 2) + (x - l)(x - 2) = 0 o (x - 2)(2
<=>
x - 2 = 0
hoac
2
x
+ x=
l<=>x
= 2
hoac
x = 1.
(VI
f(x) = 2
X
+ x dong bien tren R va f(0) = 1).
x
- 3x + 2 = 0
-x-
1) = 0
12
-BDHSG DSGTi?/?.
Vi

du 11: Chiing minh rang phuong
trinh:
a) 4
x
(4x
2
+ 1) = 1 co
diing
ba nghiem phan biet
b)
x
x+1
= (x + l)
x
co mot nghiem duong duy
nhat.
Giai
a) PT «• 4
x
(4x
2
+ 1) - 1 = 0. Xet ham s6 f(x) = 4
x
(4x
2
+ 1) - 1, D = R.
Ta co
f'(x)
= 4
x

ln4.(4x
2
+ 1) + 8x .4
X
=
4
x
[ln4.(4x
2
+ 1) +
8x].
f
(x)
= 0 o ln4.(4x
2
+ 1) + 8x = 0 O (41n4)x
2
+ 8x + ln4 = 0 (*).
PT (*) nay co biet thiic A > 0 nen co
diing
2 nghiem phan biet. Tir
bang
bien thien ciia f(x) suy ra phuong
trinh
f(x)= 0 co khong qua ba nghiem
phan biet.
Mat
khac: f(-) = 0, f(0) = 0; f(-3) . f(-2) < 0
Do
do phuong

trinh
f(x)
= 0 co
diing
ba nghiem phan biet:
xi
= 0, x
2
= —-, x
3
e (-3; -2).
b) Voi x > 0, PT <=> (x + l)lnx = xln(x + 1) » (x+ l)lnx - xln(x + 1) = 0
Xet
ham so
f(x)
= (x +
l)lnx
-
xln(x
+ 1), x > 0.
f(x)
=
lnx+^i-ln(x+l)-—
=
hi-?-
+
-
+ -
x
x+1

x+lxx+1
f "(x) = [ —~) ^-j < 0, Vx > 0 nen f' nghich bien tren (0; +°o),
^
X
+
x"
x ) (x
+
1)
vi
lim f '(x) = 0 nen f '(x) < 0, Vx Do do f(x) nghich bien tren R nen
f(x)
= 0 co toi da 1 nghiem. Ma ham f(x)
lien
tuc tren khoang (0; +oo),
f(2)
= 31n2 - 21n3 = ln8 - ln9 > 0 va f(3) = 41n3 - 31n4 = ln81 - ln64 > 0
=>
dpcm.
Cach khac: Xet ham
f(t)
= —, t > 0.
Vi du 12: Giai cac phuong trinh:
a)2
x+l
-4
x
= x-1 b) 4
bg3X
+2

Iog3A
=2x
c)
3^ - 2^ = 4c~^ d) cot2
x
= tan2
x
+ 2tan2
x+l
Giai
a)
PT«2
X+I
+ (x+ l) = 2
2x
+ 2x.
Xet
ham sd f(t) = 2
l
+ 1.1 e R thi f
'(t)
= 2'.ln2 + 1.
Vi
f'
(t)
> 0, Vt nen f dong bien tren R.
PT f(x + 1) = f(2x) ox + 1 = 2xox= 1.
-BDHSG DSGT1Z/2-
13
b) DK: x > 0, dat t = log

3
x => x = 3'
= 2
t
t
(4)
t
'2\
-
+
-
<3)
4V
(2^
Vi
f(t) = |JJ + |±J ta co f '(t) > 0 va f(0) = f(l) = 0 nen chi co 2
nghiem t = 0 hoac t = 1 <=> x = 1 hoac x = 3.
c) Dat t = Vcosx , 0 < t < 1 thi PT o 3
l
- 2' = t <=> [ -1 + \ = 1.
,
9
y
2
t
.ln- + l-t.ln3
Xet f(t) = (|J +
-L.
_i
<

t
< 1 thi f'(t) = S__
Xet g(t) = 2'.ln- + 1 - tln3 voi 0 < t < 1 thi g'(t) = 2
t
.ln2.1n- - ln3 < 0
3 3
nen f '(t) nghich bien tren [0; 1]. Lap BBT thi f(t) = 1 co toi da 2 nghiem
ma f(0) = f(l) = 1 nen PT tuong duong t = 0 hoac t = 1.
<=> cosx = 0 hoac cosx = 1 o x = k27i hoac x = — + kn.
2
d) DK: 2
X
* k- . Dat t = tan2
x
thi PT o - = t + o t
4
- 6t
2
+ 1 = 0
4 t
1
-1
2
ot
2
= 3 +2^2 =(V2 ± l)
2
ot = ±(^ + 1)
Vay tan2
x

= ±{42 ± 1), tu do suy ra nghiem x.
Vi du 13: Giai cac phuong trinh sau:
a) log
2
[x(x - 1)] = 1 b) log
2
(9 - 2
X
) = 10
log(3
-
x)
c)——\ + —— = 3 d) 5
>
/log
2
(-x)=log
2
>^
5-4 log x
1
+ log x
Giai
a) PT <=> x(x -l) = 2<i=>x
2
-x-2 = 0<=>x = -l hoac x = 2.
b) Dieu kien x < 4. PT: 9 - 2
X
= 2
3-3

o 2
2x
- 9.2
X
+ 8 = 0
o2
x
=l hoac 2
X
= 8. Chon nghiem x = 0.
14 5
c) Voi x > 0, dat t = logx thi PT
<=>
+ = 3,
t*-,t*-l
5-4t
1
+ t 4
<=> 2t
2
- 3t + 1 = 0 <=> t = 1 hoac t = - (chon).
Suy ra nghiem x = 10
hoac
x = vlO .
d) DK: x < 0, PT o 5Vlog
2
(-x) = log
2
(-x)
o >g

2
(-x).(5-Vlog
2
(-x)) = 0
<=> ^/log
2
(-x) = 0 hoac
N
/log
2
(-x) =5ox = -l hoacx = -2
25
14 -BDHSG DSGT12/2-
Vj
du 14:
Giai
cac
phuong
trinh:
a) log
2
x
+
log
2
(x
- 1) = 1
c) log
3
(3

x
-l)
.
log
3
(3
x+,
-3)
= 12
b) log
2
x
+
log
3
x
+
log4X
= 1
d) logx-,4
=
1 +log
2
(x-
1).
Giai
a) DK:
x > 1, PT o
log
2

x(x
- 1) =1 o
x(x
- 1) = 2
<=>
x - x - 2 = 0.
Chon nghiem
x = 2.
b) DK:
x > 0, PT o
(1
+
log
3
2
+
log
4
2).log
2
x
= 1
<=>
(3 +
log
3
2)log
2
x
=

1
o
log
2
x
=
3
+log,
2
Vay nghiem
x
=
2
3+21og
*
2
c) DK:
x>0, PTo
log
3
(3
x
-
1)[1
+
log
3
(3
x
- 1)] = 12

Dat
t =
log
3
(3
x
- t) thi PT
<=>
t(l
+ t) = 12
ot
2
+ t-12 = 0ot = -4 hoac t = 3.
o log
3
(3
x
- 1) = -4 hoac
log
3
(3
x
- 1) = 3
1
81
hoac 3
X
-
1
= 27

82
o3
x
= - hoac 3
X
= 28
<=>
x =
log
3
82
- 4 hoac x =
log
3
28
81
d)
DK:x>
1, x ^ 2, PT
<=>
log
2
(x-l)
1
+log
2
(x-
1)
Dat
t =

log
2
(x
- 1) thi PT: - =
1
+ t « t
2
+ t - 2 = 0
5
ot=l
hoac t = -2.
Giai
ra
nghiem
x =
—
hoac x = 3.
3 '
Vi
du 15:
Giai
cac
phuong
trinh:
a) log
4
[(x
+ 2)(x +
3)] +ilog
2

2LZ^
= 2
2
x
+
3
b) log
4
(x+ 12).log
x
2=
1
c) -^-log
2
(x-2)-^
=
log
1
V3x-5
b
d -
d)
log
3
x
log
27
9x
log
9

3x
log
81
27x
Giai
a) DK:
(x
+ 2)(x +
3)>0
x-2
x +
3
>0
<=>
x<-3
x>2
PT
<=>
log
4
(x
+ 2)(x +
3)
x-2
x +
3
=
log
4
16

<=>
x
2
- 4 = 16.
-BDHSG
DSGT12/2:
x
2
= 20 o x =
±2 V5 (chpn).
b)BK:x>0,x*
l.PTo -log
2
(x+
12)
—!—=
1.
2 log
2
x
<=>log
2
(x + 12) = log
2
x
2
»x+12 = x
2
ox
2

-x-12 = 0.
Chon nghiem x = 4.
c) DK: x > 2, phuong trinh tro thanh:
^log
2
(x - 2) + ilog
2
(3x -5)=U log
2
(x - 2)(3x - 5) = 2
D
b d
2
cs>
(x -
2)(3x-5)
= 4ox =
3
hoac x = -
Chon nghiem
x = 3.
3
d) DK: x > 0, x * i x * —, dat t = log
3
x thi PT o—=
2(2 + t)
3 27 '
&
1
+

t
3(3
+1)
ot
2
+ 3t-4 = 0ot=l hoac t = -4.
1
Suy
ra
nghiem
x = 3 hoac x =
Vf du 16: Giai cac phuong trinh:
a) l
0
g
2
(4x) + log
2
^- = 8
2
8
81
b) log
2
^ x+31og
2
x
+
log
1

x
=
2
c) log
4
xlog
2
x
+
log
2
log
4
x
= 2 d)
log
x2
16
+
log
2x
64 = 3
Giai
a) DK: x > 0, ta co log
2
— = log, x
2
- log
2
8 = 2 log

2
x-3
log
2
(4x)
=
2
\2
logj
4 +
logi
x
V 2 2 J
=
(-2-log
2
x)
2
=(2
+
log
2
x)
2
Dat
t =
log
2
x thi
PT

<=>
(2 + t)
2
+ 2t - 3 = 8
ot
2
+ 6t-7 = 0<=>t = l hoac t = -7.
Suy ra nghiem x = 2~
7
hoac x = 2.
b) DK: x > 0, dat t = -log ^ x thi PT » t
2
+ -t - it = 2 c> t
2
+ t - 2 = 0
V2
2 2
<=> t = 1 hoac t = -2. Giai ra nghiem x = —, x = 72
'
- 2
c) DK:
x > 1,
phuong
trinh
tro
thanh
log
22
log
2

x + log
2
log
22
x = 2- o ilog
2
log
2
x + log
2
f^log
2
x j = 2
11 3
o -log
2
log
2
x
+
log
2
-
+
log
2
log
2
x
= 2

<=>
-log
2
log
2
x
= 3.
o log
2
log
2
x = 2 <=> log
2
x = 4 <=> x = 16 (chon).
16
-BDHSG
DSGT12/2-
d)
DK:x>0,x*l,x*i
thi
2
PT 02102,2 + = 3o + —- = 3
l
+ log
2
x log
2
x l + log
2
x

Dat t = log
2
x thi PT: - + —- = 3 o 3t
2
- 5t - 2 = 0
t
1
+ t
o t = 2
hoac
t = -— Suy ra nghiem x = 4, x = JJ=
3 %/2
Vi
du 17: Giai cac
phuong
trinh
a) log
5
x . log
3
x = log
5
x + log
3
x
b) 21og
2
x . log
5
x + log

2
x -
101og
5
x
= 5
c) log
2
x log
3
x log
5
x = log
2
x log
3
x + log
2
x log
5
x + log
3
x log
5
x
Giai
a) DK: x > 0, ta co x = 1 la mot nghiem.
Neu x * 1 thi PT o - = —-— + —-—
log
x

51og
x
3
log
x
5 log
x
3
o log
x
5 + log
x
3 = 1 o log
x
15 = 1 o x = 15 (chon)
b) DK: x > 0, PT o log
2
x (21og
5
x + 1) - 5(21og
5
x + 1) = 0
o (log
2
x - 5)(21og
5
x + 1) = 0 o log
2
x = 5
hoac

21og
5
x = -1
o x = 32 hoac x = —T= (chon).
V5
c) DK: x > 0,
phuong
trinh o
(lgx)
3
= (lgx)
2
(lg2 + lg3 + lg5)
o (lgx)
2
(lgx - lg30) = 0 o lgx = 0
hoac
lgx = lg30
O x = 1
hoac
x = 30 (chon).
Vi
du 18: Giai cac
phuong
trinh
a) log
2
x = 3 - x b) log
3
x + log

4
(2x - 2) = 2
c)
log
2
(l
+ Vx) = log
3
x d)
log
3
(l
+ Vx + Vx) = -log
2
Vx
Giai
a) DK: x > 0, vi ham so ve trai dong bien, ham so ve phai nghich bien va
x
= 2 la nghiem nen do la nghiem duy
nhat.
b) DK: x > 1. Ta co f(x) = log
3
x + log
4
(2x - 2) la ham dong bien nen
f(x)
> f(3) = 2 voi x > 3 va
f(x)
<
f(3)

= 2 voi 1< x < 3.
Vay
x = 3 la nghiem duy
nhat.
c) DK: x > 0, dat log
3
x = y thi x = 3
y
r—
r— flY ( R
V
PTo
log
2
(l
+ V3
y
) = y«l + V3
y
= 2
y
o|±| +
-BDHSG
DSGT12/2- 17
Ta co y = 2 thoa man phuong
trinh,
vi ve trai la ham nghich bien nen PT
co nghiem duy
nhat
y = 2 nen x = 2.

d)
DK:x>0,datx = 2
r2y
thiPT
o log
3
(l + 2
6y
+ 2
4y
) = -log
2
2
6y
<=> log
3
(l + 2
6y
+ 2
4y
) = 4y
3
'
1
~
f
+
r64
T
+

'16^
,81.
J
V8lj
,8lJ
ol+2
6y
+ 2
4y
= 3
4y
<=>
Ta co y = 1 thoa man va vi ham so f(y) = —
+\
— + — nghich
1
81J \ 81 J I 81
1
=
1.
lY T64Y fiej
-
bien tren R nen y = 1 la nghiem duy
nhat,
do do PT cho co nghiem x = 2
Vi
du 19: Giai cac phuong
trinh:
a) 21og
2

x = x
b)
log
2
[31og
2
(3x-l)-l ] = x
c)
log
2
x + log
3
(x + 1) = log
4
(x + 2) + log
5
(x + 3)
d)
log
3
x2+x
+ 3
^
x
2
+3x + 2
2x
2
+ 4x + 5
Giai

a) DK: x > 0, PT: log
2
x =-t> — = — .
6
2 x 2
Xet ham s6 f(x) = —. x > 0 thi f(x) =
1-1
°
x
X
X
-
f'(x)
= 0ox = e, lap BBT thi
r(2) = f(4)=i
1/2 + 1
In
9
f(x)
= 0 co toi da 2 nghiem ma
f(2)
=
f(4)
= — nen S =
{2;
4}.
b)
DK: 3x - 1 > 0, 31og
2
(3x - 1) > 1 o x >

Dat y = log
2
(3x - 1) thi co he:
3
x
= log
2
(3y-l)
y
= log
2
(3x-l)
Do
do log
2
(3x - 1) + x = log
2
(3y - 1) + y
Xet f(t) = log
2
(3t - 1) + t, t > - thi f' (t) = + 1 > 0 voi moi t > -
3 (3t-l)ln2 3
nen f la ham dong bien, do do phuong
trinh
f(x)
= f(y)
<=>
x = y o x = log
2
(3x - 1) cs> 3x - 1 = 2

X
o2
x
-3x+l=0. Xet g(x) = 2
X
- 3x - 1, x > -
3
Ta co g'(x) = 2
x
.ln2 - 3, g"(x) = 2
x
.ln
2
2 > 0 nen g* dong bien tren D. Do
do g(x)
—
0 co toi da 2 nghiem, ma g(l) = g(3) = 0 nen suy ra tan nghiem
S =
{1;3}.
18 -BDHSG DSGTW2-
c)
DK: x > 0. Xet x = 2 thi PT thoa man.
VM ^^i
+
u-
x x
+
2
, x + 1 x + 3 ,
Xet

x > 2 thi
—
> >
1
, > > 1
2 4 3 5
nen VT > VP (loai). Xet x < 2 thi VT < VP (loai)
Vay PT co nghiem duy nhat x = 2.
2 Q
d)
Phuong
trinh:
log
3
-
x +x + d
= (2x
2
+ 4x + 5)-(x
2
+ x + 3)
2x
+
4x
+
5
<=> log
3
(x
2

+ x + 3) + (x
2
+x + 3) = log
3
(2x
2
+4x + 5) + (2x
2
+4x + 5)
Xet ham so f(t) = log
3
1 +1, (t>0) thi f'(t)= —+ l>0,Vt>0
t.
In
3
Do do f(t) dong bien, nen phuong trinh f(x
2
+ x + 3) = f (2x
2
+ 4x + 5)
<=> x
2
+ x + 3 = 2x
2
+ 4x + 5 <=> x
2
+ 3x + 2 = 0
Vay phuong trinh co 2 nghiem x = -1 va x= -2
Vi du 20: Giai cac phuong trinh sau:
a) log

2
(cotx + tan3x) - 1 = log
2
(tan3x)
b) 21og3(cotx) = log
2
(cosx)
Giai
s r^r^ [cotx + tan3x > 0
a) DK { thi PT <» cotx + tan3x = 2tan3x
[tan3x>0
<=> cotx = tan3x <=> tan3x = tan(— - x)
o3x= x + k7i<=>x= — + — ,keZ.
2 8 4
Chon nghiem: x = — + kit vax= — + kn, keZ.
'
8 8
b) DK: f
cotx>0
«k2n<x<- +k2n,ke Z.
[cosx>0 2
Dat log
2
(cosx) =2t=> cosx = 4
l
=> cos
2
x = 16
l
.

Do do 21og
3
(cotx) = 2t => cotx = 3
l
=> cot
2
x = 9
l
nen 9t =
16
t
o9'= 144' + 16*<=>
1-16'
Suy ra PT co nghiem duy nhat t = —
2
TC
Chon nghiem x = — + k2n, k e Z.
"
• 3
-BDHSG DSGT12/2-
f
144
T+f
16\
I
9 J \
I
<=>
cosx
=

—
2 '
Vi
du 21: Tim dieu kien de phuong
trinh:
a) y*
+
QC°S
2
^
=m C
6
nghiem.
b)
logg x + ^logg x +
1
- 2m
—
1
= 0 co nghiem thuoc doan
Giai
a) Dat t =
9
s,n2
x
, vi 0 < sin
2
x < 1 nen 1 < t < 9.
PT
<=>

t +
—
= m. Xet
f(t)
= t + —. 1 < t < 9.
t
t
f'(t)
=
-^;f'(t)
= Okhit = 0.
t
BBT:
l;3
X
1
3
9
f
—
0
h
f
10 10
*•
6
Vay
dieu kien
f(t)
= m co nghiem thoa l<t<9 1a6<m<10.

}
>/3
b)
Dat t = ^/logg x + 1, x
1;
3
%
ol<t<2
PT
<=>
t
2
-
1
+ t - 2m -
1
= 0 «• t
2
+ t = 2m + 2
Xet
f(t) = t
2
+ t, 1 < t <
2,f'(t)
= 2t + 1 > 0 nen f dong bien tren
[1;
2]
Dieu
kien co nghiem:
f(l)

< 2m + 2 < f(2)
o2<2m+l<6o0<m<2
Vi
du 22: Tim dieu kien de phuong
trinh
co nghiem duy
nhat
a) (S + l)
x
+ 2m(V5 - l)
x
= 2x b)
log^(x
+ 3) = log
3
(ax)
Giai
a) PTo
S +
l
(
+
2m
Vs-i
rp ,
V5+lV5-l
,
Taco: = 1. Dat t
=
1

fV5+l
A
PT: t +
2
2m
,t>0.
lot-t
+ 2m = 0.
Xet
t = 0 => m = 0 thi PT: t
2
- t = 0
<=>
t = 0 hay t = 1 (thoa man).
Xet
t * 0, didu kien co nghiem t > 0: ti < 0 < t
2
hoac
0 < t, < t
2
O P < 0
hoac
(A > 0, P > 0, S > 0) o m < 0
hoac
m = -
Vay:
m < 0
hoac
m
1

20
-BDHSG DSGT12/2-
Cach
khac:
Xet ham so va lap
bang
bien
thien.
b) PT:
21og
3
(x
+ 3) = log
3
(ax)'
fx
+ 3 > 0 , „
n
<=> <=> (x + 3)
2
= ax, x + 3 > 0
|log
3
(x + 3)
=log
3
(ax)
o x
2
+ 6x + 9 = ax, (x > -3)

Xet x = 0 (Loai). Xet x * 0 thi co: a
:
x
2
+ 6x + 9
x
(x>-3)
Datfl;x)=^^,(x>-3vax^)
J
f
l
(x) =
^
!
f
,
(x)
=
Oix=3
L
:
X
0 3 +co
f
0 -
- 0 +
f
—a
+00 +00
)

^
A
12^"
Bieu
kifn
co
nghiem
duy
nhat:
a < 0 hay a = 12.
DANG
2: BAT PHUONG
TPJNH
MO, LOGARIT
- Bat
phuong
trinh mu:
Neu a > 1 : a
f(x)
< a
g(x)
» f(x) < g(x)
Niu
0 < a < 1: a
m
< a
g(x)
<=>
f(x) > g(x)
- Bat

phuong
trinh logarit:
'f(x)
> 0
Neu a > 1: log
a
f(x) <
logag(x)
<=> \ g(x) >0 <=> 0 < f(x) < g(x).
f(x)<g(x)
f(x)>0
Neu 0 < a < 1: log
a
f(x) <
logag(x)
<=> lg(x)>0 •» f(x) > g(x) > 0.
f(x)>g(x)
Phuong
phap
chung:
Bua ve
cung
co so, dat an phu, mu hoa, lograrit hoa
va tinh
chat
don
dieu
ciia ham so.
Chii
y: a

x
< m o x < log
a
m (vod m > 0 va a > 1)
a
x
< m o x > log
a
m (voi
m>0va0<a<l)
log
a
x < m <=> 0 < x < a
m
(voi a > 1)
log
a
x < m o x > a
m
(voi 0 < a < 1).
Vi
du 1: Giai cac bat
phuong
trinh sau:
a) (0,5)i>
0,0625
b) 3
|x
"
2

' >
9
|x+11
c)
I
x
I
<1
d) (3 +
2V2)
X
>(3
-2V2 )
2x-5
-BDHsn r>sn.Ti?/?-
21
Giai
2
11 1 - 4x i
o-<4o
4<flo < 0 o x < 0
hoac
x > -
xx x 4
x-2| .
T2|X+1I
- <=> - ^ -
)
16
UJ 1

2
b)
3
|x
-
2|
>3
2|x+1|
o
|x-2| >2|x+l|
(vicas63>l)
Ixl >1, x
2
-x-2<0
o x
2
- 4x + 4 > 4(x
2
+ 2x + 1) o 3x
2
+ 12x < 0 » -4 < x < 0.
c) BPT o
0<Ixl<
1, x
2
-x-2>0
|x|
>1, -l<x<2
0 < |x| < 1, x < -1 hay x > 2
O

I
x | > 1 (vai -l<x<2)ol<x<2.
b) BPT«(3 + 2V2)
x
>(3 + 2V2)
5
-
2x
(vi3 + 2V2 > 1)
<=>x>5-2xox>-
3
Vi du 2: Giai cac bat phuong trinh:
a) <
3
X
+ 5 3
X+1
-1
c) (2+ V3)
x
+ (2- V3)
x
>4
Giai
b) 2
X
+ 2"
x+1
- 3 < 0
d) (0,4)

X
-(2,5)
X+I
>1,5
1
a) DK:x^-l,datt = 3
x
, t>0 thi BPTo—!—<—!—
t
+ 5 3t -1
f3t
-1 < t + 5 l
~l3t-l>0 -3<t^o-Kx<l.
b) Dat t = 2
X
, t > 0 thi BPT <=> t + ~ - 3 < 0 <=> t
2
- 3t + 2 < 0 » 1< t < 2
ol<2"<2o0<x<l.
c) Ta co: (2 + V3 )(2 - V3 ) = 1 nen dat t = (2 + V3 )
x
, t > 0 thi BPT:
t+->4ot
2
-4t+l>0
t
<=> t < 2 - V3 hoac t>2+V3 »x<-l hoac x > 1.
d) BPT: (0,4)
x
- 2,5.(0,4)-

x
- 1,5 > 0. Dat t = (0,4)
x
, t > 0 thi BPT trd thanh
t
2
- l,5t - 2,5 > 0
<=> t < -1 hoac t > 2,5. Chpn nghiem t > 2,5 nen (0,4)
x
> 2,5
o(0,4)
x
>(0,4)
:
'ox<-l.
Vi du 3: Giai cac bat phuong trinh
a) 3
2x+1
- 2
2x+1
- 5.6
X
< 0 b) 2
2x2
"
4x
"
2
-
4

.2
2x
~
x2+1
-2 <0
o
4
X
-3
5
<4
qX
d) ^^<3
3
X
-2
22
-BDHSG DSGT12/2-
Giai
a) Chia 2 ve cho 2
2x
> 0, BPT
3
fsY
r
f3Y
Y3Y
~
of
3

X
-
-5 - -2<0<=>
-
-
2
3
-
+
1
UJ
[2
UJ
<0
f'3Y
.3
—
< 2 o x <
log
3
2 (vicoso
—
> 1)
b) Dat t = 2
x'-2x-l
t
> 0. Bat
phuong
trinh t -
—

- 2 < 0
ot
3
-2t-4<0o(t-2)(t
2
+ 2t + 2)<0ot<2.
Dooo0< 2
x2
-
2x
"
1
< 2<=>x
2
-2x-2<0 ol- V3 < x <1+V3
c) DK: x * 0, xet x < 0 thi VT < 0 < 4 diing
Xet
x > 0 thi 4
X
> 3
X
nen BPT: 4
X
< 4(4
X
- 3
X
)
o
4.3

X
< 3.4
X
o 3
X_1
< 4
X_1
ox-l>0ox>l.
VayS =
(-oo;0)
u(l;+oo).
d) DK: x * log
3
2, BPT:
3
X
3
X
-2
3<0o^<0o^>0o
3
X
>3
3
X
>2
x>l
x
<
log

3
2
Vi
du 4: Giai cac bat
phuong
trinh:
a) 4
X
< 3.2^
+x
+4
1+Vx
b)
v
/
8 + 2
1+x
-4
X
+ 2
1+x
> 5 .
c) 4x
2
+ 3.3^ + x.3
7
* < 2x
2
.3
7x

+ 2x + 6 .
d) 2
-x
z
-3x-2+4Vx
z
+3x
<x
2
-2x + 5
Giai
a) DK: x > 0, BPT: 2
2x
-
3.2
Vx
2
x
- 4.2
2Vx
< 0
4
Chia hai ve cho 2^ . 2
X
> 0. BPT: 2
X
~^ - 4.2
7x
"
x

-3 < 0
Dat t = 2
x_7x
, t > 0 thi BPT ot 3<0ot
2
-3t-4<0o-l <t<4
t
Chon 0<t<4ox-Vx<2ox-Vx-2<0
»0<>/x
<2o0<x<4.
b) Dat t = 2\ t > 0 thi BPT: V8 + 2t-t
2
> 5 - 2t
"5-2t<0, 8 + 2t-t
2
>0
5-2t>0, 8 + 2t-r >0
Dod61<t<4ol<2
x
<4«0<x<2.
5
-<t<4
2
l<t<-
2
-BDHSG
DSGT12/2-
23