i n s . LE H O A N H P H O
Nha gido Uu tu

B

O

I

H

O

C

D

A

I

D

U

S

S

S

I

O

N

N

G

H

-

7

G

G

I

I

O

A

I

T

I

O

T

• 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 X U A 1 H N DAI HOC QUOC GIA HAJIOI

A

I

N

C

H

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. L E H O A N H P H O
Nhd gido Uu tu

B

O

I

D

H

O

C

U
S

I

d

N

N

H

G

,
G I O I

T

O

A

N

3>
D

A

I

S

O

-

G

I

1

2

A

I

T

I

- 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

C

H

NHA XUAT B A N D A I HOC QUOC GIA HA NOI
16 Hang Chudi - Hai Ba Triing Ha Npi
Dien thoai: Bien tap-Che ban: (04) 39714896;
Hanh chinh: (04) 39714899; Tong bien tap: (04) 39714897
Fax: (04) 39714899

Chiu

trach

nhiem

xudt

bdn:

Giam doc PHUNG QUOC BAO
Tong bien tap P H A M T H I T R A M
Bien tap noi
HAI NHU

dung

THUY NGAN

Sila bdi
LE HOA
Che

bdn

CONG T I A N P H A
Trinh bay bia
SON K Y
Doi tdc lien ket xudt ban
CONG T I A N P H A

SACH LIEN KET
BOI DUCfNG HQC SINH GIOI TOAN DAI SO GIAI TICH 12 -TAP 2
Ma so: 1L-181DH2010
In 2.000 cuon, kho 16 x 24 cm tai cong t i TNHH In Bao bi Hung Phu
So'xua't ban: 89-2010/CXB/11-03/DHQGHN, ngay 15/01/2010
Quyet dinh xua't ban so: 181LK-TN/XB
In xong va nop lu'u chieu quy I I nam 2010.

L d i

N 6 I

D

X

U

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 tdm sdch gido due ANPHA xin
trdn trong giai thieu 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. Mdt sd'bai tdp cd the khd nhung cdch
gidi duoc dua tren nen tdng kien thuc va ki nang co ban. Hoc sinh can 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 sin
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
tdi rat mong nhdn duoc gdp y ciia ban doc 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 Nguyen Tri Phuong, P.9, Q.5, Tp. HCM.
- Cong ti sach - thiet bj giao due Anpha
50 Nguyen Van Sang, Q. Tan Phii, Tp. HCM.
DT: 08. 62676463, 38547464 .
Email: alphabookcenter@yahoo.com
Xin chan thanh cam on!
3

M U C

L U C

Chuong I I : Ham so luy thira ham so mu va ham so logarit
§2. Phucmg trinh, he phuomg trinh, bat phuong trinh mu va logarit
Dang 1: Phuong trinh mu va logarit

5
5

Dang 2: Bat phuong trinh mu, logarit

21

Dang 3: He phuong trinh mu, logarit

31

Chuong H I : 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 I V . 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

C H U O N G II: H A M S O L U Y T H U A , H A M S O M U V A H A M
SO LOGARIT
§2. P H U O N G T R i N H , H £ P H U O N G T R I N H , B A T
PHUONG TRINH M U V A LOGARIT
A. K I E N T H U G C O B A N
Phuong phap chung:
- Dua ve cung mot co so
- Dat an phu
- Logarit hoa, mu hoa
- Su dung tinh chat cua ham so
B. P H A N D A N G T O A N
DANG 1: PHUONG TRlNH M U VA LOGARIT
- Phuong trinh mu co ban: a = b (a > 0, a * 1)
Neu b < 0, phuong trinh vo nghiem
Neu b > 0, phuong trinh co nghiem duy nhat x = log b.
a = l
(a>0)
, ^ = a
a * 1 , f ( x ) = g(x)
x

a

g(x)

a

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

b

a

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,...
V i du 1: Giai cac phuong trinh sau:
a) 2 '
x 2

c) 2
a) PT o

=4

b) (2 + V3 )

5 = 200

d) 0,125.4
Giai

3 x + 2

x + 1

X

2* 2

3x+2

2 x

2x-3

= 2 - V3
= (4V2)

x

= 2 o x - 3 x + 2 = 2 » x - 3 x = 0 o x = 0 hoac x = 3.
2

b) P T < ^ ( 2 + V 3 )

2

2

<=>2x = - l o x = - i .
2
c) PT o 2 . 10 = 200 o 10 = 100 o x * 2
5x
5x
d) PT <=> 2" . 2 " = 2 o 2 " = 2
5x
<=> 4x - 9 = — <=> 8x - 18 = 5x o x = 6.
2
-BDHSG DSGT12/2- c
2 x

= (2+ S ) '

x

3

1

x

4x

6

T

4x

9

Y

V i du 2: Giai cac phuong trinh sau:
/o\

a) (1,5) 5x-7

x + 1

b) 7 -' = 2
X

1
3
c) 9 - 2 x+— = 2x+—
X

^ ylogx _ ^logx+1 _ ^ jlogx-1 _ J3 j^ogx-\

->2x-l

2

X

CO

5x-7

u ,
b)PT»f =2

-x-l

,2j

X

x

=2

- l o x = l

O 5x - 7

7<=>

x

1
9 +i.9
3

c) P T o

_

CO

Giai

7 <=> x = log, 7
i

+2.2

X + 2

X+2

A
-.9
3

x+= 3.2"

X

+2

= — O X - 1 = lOgg — O X = 1 - lOgg 2 .
2 2
Z

d) PT »

7

logx

1

+13.7

logx

.- = 5

logx

. 5 + 3.5

logx

.- o 7

28
5

^ ylogx | 20 | _ glogx

logx

1+

13

-logx 5 ?
.
5,
+

1_
20

<=> logx = 2 <=> x = 100.
V i du 3: Giai cac phuong trinh sau:
b) 3 + 1 + 18.3" = 29

a) 4 - 2 - 6 = 0
X

c) e

2 x

X

X

d) 2 7 + 12 = 2.8

- 3 e - 4 + 12e" = 0
x

x

x

x

X

X

Giai
a) Dat t = 2 , (t > 0) thi PT <=> t - t - 6 = 0
X

Chon nghiem t = 3 <^> 2

=3 o x = log 3
1R
b) Dat t = 3 \ t > 0 thi PT o 3t + — = 29
X

2

<=> 3t - 29t + 18 = 0 o t = 9 hoac t = 2

Giai ra nghiem x = 2 hoac c = log32 - 1.
c) Dat t = e\ (t > 0) thi PT <=> t - 3t - 4 + —
2

=0

O t - 3t - 4t + 12 = 0 <=> (t - 2)(t + 2)(t - 3) = 0.
3

2

Chon nghiem t = 2 hoac t = 3 nen x = ln2 hoac x = ln3.
6

-BDHSG DSGT12I2-

d) Chia 2 ve cho 8 > 0 thi PT:
X

27V

(12\*

( T
i' 3 ^ - 2 = 0 . Dat t =
+ {2
\ 2)
3

-2

,t>0.

PT <=> t + t - 2 = 0 <=> (t - l ) ( t + t + 2) = 0 o
3

t = 1 <=> x = 0.

2

Vi du 4: Giai cac phuong trinh:
a)2.25 + 5.4 = 7.10
x

x

c) U/2-V3

+U2+S)

b) 4

x

=4

d) 4

x

+6

X+Nx2

x

=9

" -5.2 "
2

x

1+VxI:2

=6

Giai
a) P T « 5 | | ]

-7

2 = 0. Dat t =

,t>0.

PT <=> 5t - 7t + 2 = 0 o t = 1 hoac t = |
2

(thoa man)

Suy nghiem x = 0 hoac x = 1.
b) Dieu kien x * 0, dat y = — va chia hai ve cho 4 , ta co:
x
y

PT«

1 + 75
.
1 + 75
— - — es> y = log — - —

- 1 = 0<=>

3

o
X

1 ,
1 + 75
1 ,
— = l o g — — c=>- = log
2 2
3

1 + 75

« x = log^_ 2
i

3

2X

c) Ta co 7 2 - 7 3 . 7 2 + 73 = 1, dat t = (72 + 73") , t >
PT<^t+-=4«t -4t+l=0
t
o t = 2 + 73 hoac t = 2- 73»x = 2 hoac x = -2.
2

d) Dat t = 2 - , t > 0 thi PT <=> t - -1 = 6
x+vx2

2

2

2
<=> 2t - 5t - 12 = 0. Chon nghiem t = 4.

nen x + 7 x - 2 = 2 <=>7 x - 2 =2x
2

2

<=> 2 - x > 0 va x - 2 = 4 - 4x + x <=> x < 2 va x
2

2

2
-BDHSG DSGT12/2-

<=> x = 2
1

V i du 5: Giai cac phuong trinh:
a) x + ( x - 2 ) = 0

b) V2 V4 (0,125) = 4^2

c) Ul6-x

d) / x T l - / x ^ T = v x - 1

3

6

X

+ $fx~+~l=3

X

3

X

3

/

2

Giai
a) B K : x > 0 , x - 2 > 0 < = > x > 2 .
V o i x > 2 thi V T > 0 nen PT vo nghiem.
b) D K : x # 0, PT o

1 6
7
2 (2~ )
= 2 o
2x

2

3

x

3

7
£ 2x— 6
2 2
=2
k
>
2

x

3

1 7
3 » - +- - — = 2 3 2x 3
,
1
» 5x - 14x - 3 = 0 <=> x = — hoac x = 3.
5
c) D K : - 1 < x < 16. D a t u = # L 6 - x , v = 7 x + l thi u, v > 0.
lu + v = 3
Ta co he:
Dat S = u + v, P = uv
u + v =17
»

_
22.23"^

2

= 2

4

4

thi u + v = ( u + v ) = ((u + v ) - 2uv) - 2u v
4

4

2

2

2

2

17 = ( 9 - 2 P ) - 2 P
2

2

2

2

2

= 2 P - 3 6 P + 81.
2

Do do P = 2 hoac P = 16. V i S - 4P > 0 nen chon P = 2 suy ra S = 3 nen
nghiem x = 1 hoac x = 15.
2

d) D K : x < - 1 hoac x > 1. V i 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.
P T < » 7(x + l ) - 7 ( x - l )
2

2

= v x ^lci>6
/

Y

x+1
x-l
x - l Vx+1

e p ^ i . t > 0 thi PT t - - = 1 < = > t - t - l = 0
x-l
t
1 + V5
Chon nghiem t
suy ra nghiem ciia PT cho la:
2

x=±

't + r

6

VOI t =

t-1

1 + V5

V i du 6: Giai cac phuong trinh sau:
b) 3 .8

a) 3 * = 4 "
4

X

3

c) S"" ^ =8.4
1

2

X+1

=36

X

5 15
-BDHSG DSGT12/2-

Giai
a) Hai ve deu duong, logarit hoa theo co so 10 ta co :
log 4
« x = log4(log 4)
log 3
3
3x
x-2
2
=
2 o^ 3TX-2
~ . 2* = 1
_ 3
o2 . n,2
,x-2
1 <=> x - 2 = 0 hoac 3.2
=1

4 log3 = 3 log4 <=>
x

x

b) PT o 3

X

X+1

3.2

2

3

l

X

2

+1

X+1

s

» x = 2 hoac 2

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

log (3 - .2 ) = log (8.4 - )
x

1

x2

x

2

2

2

<=> (x - l)log 3 + x = log 8 + (x - 2)log 4
<=>x -(2-log 3)x + 1 -log 3 = 0 o x = 1 hoacx= 1 -log 3
d) Hai ve deu duong, logarit hoa hai ve theo co so 5 ta co:
2

2

2

2

2

2

2

2

<
IM
/ ^^ i
r ^(x-l)l0g (-) -l0g (-)-—
3

1

5

3

+

3

x

4

1

- -

5

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

5

<=> X

5

4 log 3 - 7 ^ - 4 + l o g ^ ^
5

_ 2(log 3-4)

x

5

2

4log 3-7
5

V i du 7: Giai cac phuong trinh:
a) 2

log3x2

.5

c) x

l o g 4

+4

log3X

l o g x

=400

b) 4

=32

d) 3

l n x + 1

-6

l n x

-2.3

l n x 2 + 2

I

^ + 3 ' i x—^ V ^
Giai

, U B 4

U 6 4

=

a) Dieu kien x > 0, phuong trinh <=> 4

log3 x

.5

400

= 20 <=> log x = 2 x = 9 (thoa man)
<=> 2 0
b) D K : x > 0, PT « 4 . 2
- 6 - 18.3
=0
lnx
thi duoc PT <=> 4t - t - 18 = 0.
Chia ca hai ve cho 3 , dat t log3X

2

3

2lnx

lnx

21nx

21nx

9
_
Chon nghiem t = — <=> x = e
•
•
2
c) D K : x > 0, ta co: x
= 4 ^'°
2

&

l o g 4

nen PT « 2.4

logx

-BDHSG DSGT1Z/2-

Iog

= 32 o 4

l o g x

e4

=4

l o g 4 l o

^ = 4
x

l o g x

= 16 o logx = 2 o x = 100.

d) D K : x > 0, dat t = log x thi x = 4

l

3

PT o V3 . 3' + -L . 3 = 2 o 4.3' = V3 . 2'
V3
1

1

log;
._ _ ,
V3
o t = l o g ^ Vayx= 4

3 V 73
l-j = ^

Vi du 8: Giai cac phuong trinh:
a) (4 - Vl5) + (4 + Vl5) =8 b) 81
tanx

tanx

sin2 x

+ 81

c) (cos72°) + (cos36°) = 3.2~
x

30

c

x

x

d) e

Giai

sin(x—)
4

= tan x

a) Vi (4 - y/l5 )(4 + 7l5) = 1 nen dat (4 - JTE ) = t, t > 0 thi phuong
tanx

trinh ot+ ^=8«>t -8t+ 1 = 0«t = 4±Vl5.
2

Do do tanx = - 1 hoac tanx = 1 nen nghiem x = ±— + leir, k e Z.
4
b) Datt= 81 , 1 smZx

o 81 + si " " " = 30«t + — = 30
sm2x

1

51

2

t
o t - 30t + 81 = 0 <=> t = 27 hoac t = 3 (chon)
2

Do do 3

4s,n2 x

= 27 hoac S ^ =3 <=> 4sin x = 3 hoac 4sin x = 1.
4

2

2

^V3, „ . 1
o sinx = ± — h o a c sinx = ±—
2
•
2
it
<=> x = ±— + kn hoac x
kn, k e Z.
6
•
3
c) Phuong trinh: (2cos72°) + (2cos36°) = 3
2sin36°.cos36°.cos72
Vi:2cos72° 2cos36°
=1
sin 36°
x

x

0

Dat t = (2cos72°) , t > 0 thi PT o t + ± = 3
x

^

t

2 _

3

t

+

1

=

0 « t =

N/5 + 1

Ta co: 2cos72° = 2sinl8° = — — - suy ra nghiem x = ±2.
d) Dieu kien cosx * 0, v i sinx = 0 khong thoa man phuong trinh nen PT
,_ %/2sinx 72 cosx
72(stnx-cosx) s m x
e *
e
<=>
=
cosx
sinx
cosx
A

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 ~2~
0

U
72t

V

Xet ham s6 y = f(t) = —, vai t e (-1; 0) u (0; 1).
V2t

^

72t
e 2

V2t
v
y' = = ———— < 0 suy ra ham so nghich bien tren
t
2t
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
2

2

2

<=> f(u) = ftVr'o u = v «tanx = 1 <=>x= — + kn (chon).
4
Vi du 9: Giai cac phuong trinh:
a)(sin-) + (cos-) = 1 b) 4 - 3 = 7
5
5
x

x

X

X

c)(-) = x + 4 d)2 = x+l.
3
Giai
71 71
a) V i 0 < sin— < 1 va 0 < cos— < 1 do do:
5
5
Neu x > 2 thi ta co (sin- ) < (sin- ) va (cos- ) < (sin- f
5
5
5
=> VT < 1 (loai).
x

x

x

2

x

5

Neu x < 2 thi ta co (sin- ) > (sin- ) va (cos - ) > (sin- )
5
5
5
5
=> VT> 1 (loai).
Neu x = 2 thi PT nghiem dung, do la nghiem duy nhat
13
b) PT <=> ( — ) + ( — ) = 1 va ta co x = 2 thoa man PT. V i 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 - 1 = 0
Xet f(x) = 2 - x - 1, D = R. Ta co:
f '(x) = 2 ln2 - 1, f "(x)= 2 .ln 2 > 0, Vx
Do do f '(x) dong bien tren R, f '(x) = 0 <=> x = -log (ln2)
x

x

2

x

2

x

X

X

X

x

2

2

-BDHSG DSGT12/2- 11

BBT
X

-loKln2)
0
•+

f

+ 0

+00

f

°

+ o c

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

V i du 10: Giai cac phuong trinh:
a) (^6 + V l 5 ) + ( ^ 7 - v / l 5 )
x

c) 6 + 15 = 3
X

X+1

+ 5.2

x

b) (2 - V3) + (2 + V 3 ) = 4

= 13

x

x

X

d) x.2 = x(3 - x ) + 2 ( 2 - 1).

X

x

x

Giai
a) Ta co x = 3 la nghiem ciia phuong trinh, v i ham so
f(x) = ( \ o + V l 5 ) + (^7 - V l 5 )
x

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) P T « 6 - 3 . 3 + 15 - 5.2 = 0
o (2 - 3)(3 - 5) = 0 ci> 2 = 3 hoac 3 = 5.
<=> x = log23 hoac x = log35.
d) PT » x.2 - x(3 - x) - 2.2 = 0 o 2 (x - 2) x - 3x + 2 = 0
2 (x - 2) + (x - l ) ( x - 2) = 0 o (x - 2)(2 - x - 1) = 0
<=> x - 2 = 0 hoac 2 + x = l<=>x = 2 hoac x = 1.
(VI f(x) = 2 + x dong bien tren R va f(0) = 1).
X

X

X

X

X

X

x

X

X

x

x

x

X

12

-BDHSG

DSGTi?/?.

V i du 11: Chiing minh rang phuong trinh:
a) 4 (4x + 1) = 1 co diing ba nghiem phan biet
x

b) x

2

x + 1

= (x + l ) co mot nghiem duong duy nhat.
x

Giai
a) PT « • 4 (4x + 1) - 1 = 0. Xet ham s6 f(x) = 4 (4x + 1) - 1, D = R.
x

2

x

2

Ta co f'(x) = 4 ln4.(4x + 1) + 8x .4 = 4 [ln4.(4x + 1) + 8x].
x

2

X

x

2

f (x) = 0 o ln4.(4x + 1) + 8x = 0 O (41n4)x + 8x + ln4 = 0 (*).
2

2

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 = — - , x e (-3; - 2 ) .
2

3

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 ) = l n x + ^ i - l n ( x + l ) - — = hi-?+- +x
x+1
x + l x x + 1
f "(x) = [ —~) ^-j < 0, Vx > 0 nen f' nghich bien tren (0; +°o),
^ X + x" x ) (x + 1)
vi l i m 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

c) 3 ^

-4

x

=x-1

- 2 ^

b) 4

= 4c~^

b g 3 X

+2

I o g 3 A

=2x

d) cot2 = tan2 + 2tan2
x

x

x+l

Giai
a) P T « 2

X + I

+ (x+ l) = 2

2x

+ 2x.

Xet ham sd f(t) = 2 + 1.1 e R thi f '(t) = 2'.ln2 + 1.
l

V i f ' (t) > 0, Vt nen f dong bien tren R.
PT f(x + 1) = f(2x) o x + 1 = 2 x o x = 1.
-BDHSG DSGT1Z/2-

13

b) D K : x > 0, dat t = log x => x = 3't
t
(4) t '2\
=2
+ <3)
3

4V
(2^
V i f(t) = | J J + | ± 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 - 2' = t <=> [ -1 + \ = 1.
l

, y
2 .ln- + l-t.ln3
Xet f(t) = ( | J + - L . _ i < < 1 thi f ' ( t ) =
S__
t

9

t

Xet g(t) = 2'.ln- + 1 - tln3 voi 0 < t < 1 thi g'(t) = 2 .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.
t

<=> cosx = 0 hoac cosx = 1 o x = k27i hoac x = — + kn.
2
d) DK: 2 * k- . Dat t = tan2 thi PT o - = t + o t - 6t + 1 = 0
4
t
1 -1
ot = 3 +2^2 =(V2 ± l) ot = ±(^ + 1)
X

x

4

2

2

2

2

Vay tan2 = ±{42 ± 1), tu do suy ra nghiem x.
Vi du 13: Giai cac phuong trinh sau:
a) log [x(x - 1)] = 1 b) log (9 - 2 ) = 10 x

X

2

log(3 x)

2

c)——\ + —— = 3 d) 5 /log (-x)=log >^
5 - 4 log x 1 + log x
Giai
a) PT <=> x(x -l) = 2x -x-2 = 0<=>x = -l hoac x = 2.
b) Dieu kien x < 4. PT: 9 - 2 = 2 o 2 - 9.2 + 8 = 0
o2 =l hoac 2 = 8. Chon nghiem x = 0.
14 5
c) V o i x > 0, dat t = logx thi PT <=>
+
= 3,
5-4t
1+t
<=> 2t - 3t + 1 = 0 <=> t = 1 hoac t = - (chon).
>

2

2

2

X

x

3-3

2x

X

X

2

Suy ra nghiem x = 10 hoac x = vlO .
d) DK: x < 0, PT o 5Vlog (-x) = log (-x)
2

2

o >g (-x).(5-Vlog (-x)) = 0
2

2

<=> ^/log (-x) = 0 hoac /log (-x) =5ox = -l hoacx = -2
2

14 -BDHSG DSGT12/2-

N

2

25

t * - , t * - l
4

V j du 14: Giai cac phuong trinh:
a) log x + log (x - 1) = 1
c) l o g ( 3 - l ) . l o g ( 3 - 3 ) = 12
2

b) log x + log x + log4X = 1
d) logx-,4 = 1 + l o g ( x - 1).
2

2

x

x+,

3

3

3

2

Giai
a) D K : x > 1, PT o log 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 2 + log 2).log x = 1
2

3

4

<=> (3 + log 2)log x = 1 o log x =
3

2

2

2

3 +log, 2

Vay nghiem = 2
*
c) DK: x > 0 , P T o log (3 - 1)[1 + log (3 - 1)] = 12
Dat t = log (3 - t) thi PT <=> t ( l + t) = 12
o t + t - 1 2 = 0 o t = - 4 hoac t = 3.
o log (3 - 1) = - 4 hoac log (3 - 1) = 3
1
hoac 3 - 1 = 27
81
82
o 3 = - hoac 3 = 28 <=> x = log 82 - 4 hoac x = log 28
81
d) D K : x > 1, x ^ 2, PT <=>
1 + l o g ( x - 1)
log (x-l)
3+21og

2

x

x

x

3

3

x

3

2

x

x

3

3

X

x

X

3

3

2

2

Dat t = log (x - 1) thi PT: - = 1 + t « t + t - 2 = 0
2

2

5

o t = l hoac t = - 2 . Giai ra nghiem x = — hoac x = 3.
3
'
V i du 15: Giai cac phuong trinh:
a) log [(x + 2)(x + 3)] + i l o g 2 L Z ^ = 2
2
x+3
b) l o g ( x + 12).log 2= 1
4

2

4

x

c) - ^ - l o g ( x - 2 ) - ^ = log V 3 x - 5
b
d
log x
l o g 9x
d)
log 3x l o g 27x
2

1

3

27

9

81

Giai
(x + 2)(x + 3 ) > 0
a) D K :

x-2
>0
x +3

PT <=> log

4

<=>

(x + 2)(x + 3)

-BDHSG DSGT12/2:

x<-3
x>2

x-2
x+3

= log 16 <=> x - 4 = 16.
2

4

x = 20 o x = ±2 V5 (chpn).
2

b ) B K : x > 0 , x * l . P T o - l o g ( x + 12) — ! — = 1.
2
log x
<=>log (x + 12) = log x »x+12 = x ox -x-12 = 0.
Chon nghiem x = 4.
c) DK: x > 2, phuong trinh tro thanh:
2

2

2

2

2

2

2

^log (x - 2) + ilog (3x -5)=U log (x - 2)(3x - 5) = 2
D
b
d
2
cs> (x - 2 ) ( 3 x - 5 ) = 4 o x = 3 hoac x = - Chon nghiem x = 3.
3
d) DK: x > 0, x * i x * —, dat t = log x thi PT o—=
3
27
'
1 + t 3(3 +1)
ot + 3t-4 = 0ot=l hoac t = -4.
1
Suy ra nghiem x = 3 hoac x =
81
Vf du 16: Giai cac phuong trinh:
2

2

2

2(2 + t)

3

&

2

a) l g (4x) + log ^- = 8
2
c) log xlog x + log log x = 2

b) log ^ x+31og x + log x = 2

2

0

2

2

2

1

8

4

2

2

d) l o g 1 6 + l o g 64 = 3

4

x2

2x

Giai
a) DK: x > 0, ta co log — = log, x - log 8 = 2 log x-3
2

2

2

2

\2
log (4x) = logj 4 + logi x
2
V22J
2

= ( - 2 - l o g x ) = ( 2 + log x )
2

2

2

2

Dat t = log x thi PT <=> (2 + t ) + 2t - 3 = 8
ot + 6t-7 = 0<=>t = l hoac t = -7.
Suy ra nghiem x = 2~ hoac x = 2.
2

2

2

7

b) DK: x > 0, dat t = -log ^ x thi PT » t + -t - it = 2 c> t + t - 2 = 0
2
2
2

2

V 2

<=> t = 1 hoac t = -2. Giai ra nghiem x = —, x = 72
' 2
c) DK: x > 1, phuong trinh tro thanh
log log x + log log x = 2- o ilog log x + log f^log x j = 2
22

2

11 3
o

2

22

2

2

2

2

- l o g l o g x + l o g - + log log x = 2 <=> - l o g l o g x = 3.
2

2

2

2

2

2

2

o log log x = 2 <=> log x = 4 <=> x = 16 (chon).
2

16

2

2

-BDHSG DSGT12/2-

d) D K : x > 0 , x * l , x * i thi
2
PT 02102,2 + = 3o + —- = 3
l + log x

log x

2

l + log x

2

2

Dat t = log x thi PT: - + —- = 3 o 3t - 5t - 2 = 0
t 1+t
2

2

o t = 2 hoac t = -— Suy ra nghiem x = 4, x = JJ=
3
%/2
V i du 17: Giai cac phuong trinh
a) log x . log x = log x + log x
b) 21og x . log x + log x - 101og x = 5
c) log x log x log x = log x log x + log x log x + log x log x
Giai
a) DK: x > 0, ta co x = 1 la mot nghiem.
5

3

2

5

3

5

2

3

2

5

5

2

3

2

5

3

5

Neu x * 1 thi PT o

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

x

x

x

x

x

x

2

2

5

5

5

2

5

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

2

2

2

3

c) l o g ( l + V x ) = log x
2

4

d) l o g ( l + Vx + Vx) = - l o g Vx

3

3

2

Giai
a) D K : x > 0, v i ham so ve trai dong bien, ham so ve phai nghich bien va
x = 2 la nghiem nen do la nghiem duy nhat.
b) D K : x > 1. Ta co f ( x ) = log x + log (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) D K : x > 0, dat log x = y thi x = 3
3

4

y

3

r—
r—
f l Y (
P T o l o g ( l + V 3 ) = y « l + V3 = 2 o | ± | +
y

y

R

V

y

2

-BDHSG DSGT12/2-

17

Ta co y = 2 thoa man phuong trinh, v i ve trai la ham nghich bien nen PT
co nghiem duy nhat y = 2 nen x = 2.
d) D K : x > 0 , d a t x = 2 t h i P T
r 2 y

o log (l + 2 + 2 ) = -log 2 <=> log (l + 2 + 2 ) = 4y
3
' 1 f~
'16^
T
ol+2
+ 2 = 3 <=>
= 1.
,8lJ
,81.J V 8 l j
6y

4y

6y

3

6y

2

6

y

4y

3

4 y

r

4y

6

4

+

+

lY T64Y fiej
—
+\ —
+ —
nghich
81J \ 81 J I 81
bien tren R nen y = 1 la nghiem duy nhat, do do- PT cho co nghiem x = 2
V i du 19: Giai cac phuong trinh:
a) 21og x = x
b) l o g [ 3 1 o g ( 3 x - l ) - l ] = x
c) log x + log (x + 1) = log (x + 2) + log (x + 3)
Ta co y = 1 thoa man va vi ham so f(y) =

1

1

2

2

2

2

d) log

3

3

4

^
2x + 4x + 5
x

2

+

x

+

3

x

2

+

3

5

x +

2

2

Giai
a) DK: x > 0, PT: log x =-t> — = — .
2
x
2

2

6

Xet ham s6 f(x) = —. x > 0 thi f(x) = °
X
f'(x) = 0 o x = e, lap BBT thi
1-1

x

X

-

In 9
r(2)
f(x) == f(4)=i
0 co toi da 2 nghiem ma f(2) = f(4) = — nen S = {2; 4 } .
1/2 + 1
b) D K : 3x - 1 > 0, 31og (3x - 1) > 1 o x >
2

3
x = log (3y-l)
2

Dat y = log (3x - 1) thi co he:
2

y = log (3x-l)
2

Do do log (3x - 1) + x = log (3y - 1) + y
2

2

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

3

X

2

o2 -3x+l=0. Xet g(x) = 2 - 3x - 1, x > x

X

3
Ta co g'(x) = 2 .ln2 - 3, g"(x) = 2 .ln 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}.
x

18 -BDHSG DSGTW2-

x

2

c) D K : x > 0. Xet x = 2 thi PT thoa man.
VM ^^i u- + , 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.
2Q
d) Phuong trinh: log = (2x + 4x + 5 ) - ( x + x + 3)
2x + 4x + 5
<=> log (x + x + 3) + (x +x + 3) = log (2x +4x + 5) + (2x +4x + 5)
xx

2

+

x

+

x

+

d

2

2

3

2

2

2

3

2

3

Xet ham so f(t) = log 1 +1, (t>0) thi f'(t)= —+ l>0,Vt>0
3

t. In 3
Do do f(t) dong bien, nen phuong trinh f(x + x + 3) = f (2x + 4x + 5)
2

2

<=> x + x + 3 = 2x + 4x + 5 <=> x + 3x + 2 = 0
2

2

2

Vay phuong trinh co 2 nghiem x = -1 va x= -2
Vi du 20: Giai cac phuong trinh sau:
a) log (cotx + tan3x) - 1 = log (tan3x)
b) 21og3(cotx) = log (cosx)
2

2

2

Giai
s r^r^ [cotx + tan3x > 0
a) D K {
[tan3x>0

thi PT <» cotx + tan3x = 2tan3x

<=> 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

«k2n[cosx>0

cotx>0

2

Dat log (cosx) =2t=> cosx = 4 => cos x = 16 .
l

2

l

2

Do do 21og (cotx) = 2t => cotx = 3 => cot x = 9
l

2

l

3

nen 9t = o9'= 144' + 16*<=>
1-16'
16

t

Suy ra PT co nghiem duy nhat t = —
TC

Chon nghiem x = — + k2n, k e Z.
"
•
3

-BDHSG DSGT12/2-

f
T+f
I 9 J \
1 4 4

16\
I

<=> cosx = —
2'
2

V i du 2 1 : T i m 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 l ; 3
Giai
a) Dat t = 9

s,n2 x

, v i 0 < sin x < 1 nen 1 < t < 9.
2

PT <=> t + — = m. Xet f(t) = t + —. 1 < t < 9.
t
t
f ' ( t ) = - ^ ; f ' ( t ) = O k h i t = 0.
t
BBT: X 1
3
—

f
f

0

9

h

10

10
*• 6

Vay dieu kien f(t) = m co nghiem thoa l < t < 9 1 a 6 < m < 1 0 .
>/3 o l < t < 2
b) Dat t = ^/logg x + 1 , x
1; 3
}%

PT <=> t - 1 + t - 2m - 1 = 0 « • t + t = 2m + 2
Xet f(t) = t + 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<6o0V i du 22: T i m dieu kien de phuong trinh co nghiem duy nhat
2

2

2

a) ( S + l ) + 2m(V5 - l ) = 2x
x

b) log^(x

x

+ 3) = log (ax)
3

Giai
a)

PTo

S

+l

+ 2m

(Vs-i

=1

rp , V 5 + l V 5 - l
,
fV5+l
Taco:
=
1.
Dat
t
2
PT: t +

2m

A

,t>0.

l o t - t + 2m = 0.

Xet t = 0 => m = 0 thi PT: t - t = 0 <=> t = 0 hay t = 1 (thoa man).
Xet t * 0, didu kien co nghiem t > 0: ti < 0 < t hoac 0 < t, < t
2

2

2

O P < 0 hoac (A > 0, P > 0, S > 0) o m < 0 hoac m = Vay: m < 0 hoac m
20

1

-BDHSG DSGT12/2-

Cach khac: Xet ham so va lap bang bien thien.
b) PT: 21og (x + 3) = log (ax)'
fx + 3 > 0
,
„
<=>
<=> (x + 3) = ax, x + 3 > 0
|log (x + 3) =log (ax)
o x + 6x + 9 = ax, (x > - 3 )
x + 6x + 9
(x>-3)
Xet x = 0 (Loai). Xet x * 0 thi co: a
x
3

3

n

2

3

3

2

2

:

Datfl;x)=^^,(x>-3vax^) f (x) = ^

f

l

J

L

:

0

X
f

0

3

-

-

0

—a )

,

( x ) = Oix=3

+co
+

+00

f

!

+00
^

A

1 2 ^ "

Bieu k i f n co nghiem duy nhat: a < 0 hay a = 12.
DANG 2: BAT PHUONG TPJNH MO, LOGARIT
- Bat phuong trinh mu:
Neu a > 1
:a
< a
» f(x) < g(x)
N i u 0 < a < 1: a < a
<=> f(x) > g(x)
- Bat phuong trinh logarit:
'f(x) > 0
f(x)

g(x)

m

g(x)

Neu a > 1: log f(x) < logag(x) <=> \ g(x) >0

<=> 0 < f(x) < g(x).

a

f(x)f(x)>0
Neu 0 < a < 1: log f(x) < logag(x) <=> lg(x)>0

• » f(x) > g(x) > 0.

a

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 < m o x < log m (vod m > 0 va a > 1)
a < m o x > log m (voi m > 0 v a 0 < a < l )
log x < m <=> 0 < x < a (voi a > 1)
log x < m o x > a (voi 0 < a < 1).
V i du 1: Giai cac bat phuong trinh sau:
x

a

x

a

m

a

m

a

a) ( 0 , 5 ) i > 0,0625

b) 3 " ' > 9

c) I x I

d) (3 + 2 V 2 ) > ( 3 - 2 V 2 ) 2x-5

-BDHsn

<1

r>sn.Ti?/?-

| x

2

| x + 1 1

X

21

Giai
2
11 1 - 4x i
o - < 4 o
4
x
x
b)
- >. 3T2|X+1I o |
x-2|
o x - 4x + 4 > 4(x
| x

2 |

2 | x + 1 |

3

2

2

)

<=> UJ

16

^ 12


< 0 o x < 0 hoac x > x
4
x - 2 | > 2 | x + l | (vicas63>l)
+ 2x + 1) o 3x + 12x < 0 » - 4 < x < 0.
2

Ixl >1, x -x-2<0
c) BPT o
0
|x| >1, - l < x < 2

2

O

-x-2>0

2

0 < |x| < 1, x < -1 hay x > 2

I x | > 1 (vai - l < x < 2 ) o l < x < 2 .

b) BPT«(3 + 2V2) >(3 + 2V2) - (vi3 + 2V2 > 1)
x

5 2x

<=>x>5-2xox>-

3
Vi du 2: Giai cac bat phuong trinh:
a) <

b) 2 + 2 "
X

3 +5

3

X

X+1

-1

c) (2+ V3) + (2- V3) >4
x

x

x+1

- 3< 0

d) (0,4) -(2,5) >1,5
X

X+I

Giai
1
a) D K : x ^ - l , d a t t = 3 , t > 0 thi B P T o — ! — < — ! —
t + 5 3t - 1
f3t - 1 < t + 5
l
~l3t-l>0 -3x

b) Dat t = 2 , t > 0 thi BPT <=> t + ~ - 3 < 0 <=> t - 3t + 2 < 0 » 1< t < 2
X

2

ol<2"<2o0c) Ta co: (2 + V3 )(2 - V3 ) = 1 nen dat t = (2 + V3 ) , t > 0 thi BPT:
x

t+->4ot -4t+l>0
t
<=> t < 2 - V3 hoac t>2+V3 »x<-l hoac x > 1.
2

d) BPT: (0,4) - 2,5.(0,4)- - 1,5 > 0. Dat t = (0,4) , t > 0 thi BPT trd thanh
t - l,5t - 2,5 > 0
x

x

x

2

<=> t < -1 hoac t > 2,5. Chpn nghiem t > 2,5 nen (0,4) > 2,5
o(0,4) >(0,4) 'ox<-l.
Vi du 3: Giai cac bat phuong trinh
x

x

:

a) 3

2x+1

o
22

-2
4

2x+1

X

- 5.6 < 0 b) 2 " " - .2 ~
X

2x2

4x

2

2x

x2+1

- 2 <0

4

-3

5

<4

qX
d) ^^<3
3 -2
X

-BDHSG DSGT12/2-

Giai
a) Chia 2 ve cho 2
3

fsY
UJ

2 x

> 0, BPT

f3Y
Y3Y
~
-5 -2<0<=> [2
UJ
r

of

2

3

X

3

+ 1 <0

-

f'3Y
.3
— < 2 o x < l o g 2 ( v i c o s o — > 1)
3

b) Dat t = 2 x'-2x-l t > 0. Bat phuong trinh t - — - 2 < 0
ot -2t-4<0o(t-2)(t
3

+ 2t + 2 ) < 0 o t < 2 .

2

Dooo0< 2 - " < 2 < = > x - 2 x - 2 < 0 o l - V3 < x < 1 + V 3
x2

2x

1

2

c) DK: x * 0, xet x < 0 thi V T < 0 < 4 diing
Xet x > 0 thi 4 > 3 nen BPT: 4 < 4(4 - 3 )
o 4.3 < 3.4 o 3
<4
o x - l > 0 o x > l .
V a y S = (-oo;0) u ( l ; + o o ) .
d) DK: x * log 2, BPT:
X

X

X

X

X

X_1

X

X

X_1

3

3
3

3

X

-2

X

...