tài liệu bồi dưỡng học sinh giỏi toán 12

ThS.
LE
HOANH
PHO
Nhd gido
Uu tu
C
c c
Mil
BOIDU8NG
7
HOC
SINH GIOI TOAN
DAI
SO-GIAI TICH
-
Ddnh
cho
HS
lop 12 on tap
&
nang cao
kinang
lam bai.
-
Chudn
bi cho cdc
ki
thi
quoc gia do
Bo

GD&DT
to choc
Mil
NHA XUAT BAN DAI
HQC QUOC
GIA
HA NQI
Bin
DUSNG
,
HQC
SINH
GO
TOAN
OAI
SO
-GIAI
TICH
Boi
duQng hoc sinh
gioi
Toan Dai so 10-1.
Boi
duQng hoc sinh
gioi
Toan Dai so 10-2.
-
Boi duQng hoc sinh
gioi
Toan

Hinh
hoc 10.
-
Boi duOng hoc sinh
gioi
Toan Dai so 11.
Boi
duQng hoc sinh
gioi
Toan
Hinh
hoc 11.
Bp
de thi tif luan Toan
hoc.
Phan
dang va pht/Ong
phap
giai
Toan So phtfc.
Phan
dang va phucing
phap
giai
Toan To hop va
Xac
suat.
1234 Bai tap tir luan
dien
hinh Dai so

giai
tich
1234 Bai tap ta iuan
dien
hinh
Hinh
hoc
li/ong
giac
ThS.
LE
HOANH
PHO
Nha gido iCu tu
BOIDUQNG
,
HOC SINH GIOI TOAN
DAI SO-GIAI TICH
12 *
-
Danh cho
HS
lap 12 on
rflp &
ndng cao kfndng lam bai.
-
Chudn bj cho cdc
ki thi
qudc gia do Bo
GD&DT

td choc.
Ha
Npi
NHA XUAT BAN DAI HQC
QUOC
GIA HA
NQI
NHA
XUAT
BAN DAI HOC
QUOC
GIA HA N0I
16
Hang
Chudi
- Hai Ba Trirng Ha Npi
Dien
thoai:
Bien
tap-Che
ban: (04)
39714896;
Hanh
chinh:
(04)
39714899;
Tdng
bien
tap: (04)
39714897

Fax: (04)
39714899
Chiu trach nhiem xuat bdn:
Giam ddc PHUNG QUOC BAO
Tong bien tap PHAM THI
TRAM
Bien tap noi dung
THUY
NGAN
Sda bdi
NGOC
HAN
Che bdn
CONG TI ANPHA
Trinh bay bia
SON KY
Ddi tdc lien ket xudt bdn
CONG TI ANPHA
SACH
LIEN
KET
BOI DUONG HQC SINH GIOI TOAN DAI SO GIAI TICH 12 -TAP 1
Ma so:
1L-177DH2010
In
2.000
cuon,
khd 16 x 24 cm tai
cong
ti TNHH In Bao bi

Hung
Phu
So'
xua't
ban:
89-2010/CXB/11-03/DHQGHN,
ngay
15/01/2010
j
Quyet
dinh
xua't
ban sd: 177LK-TN/XB
In
xong
va nop
ltiu
chieu
quy
II
nam 2010.
Ldi
N6I DAU
De giup cho hoc sinh
lap
12 co them tai lieu
tu
boi
duong, ndng
cao

va
ren luyen
ki
ndng gidi todn theo chuong trinh phdn ban mdi. Trung
tdm
sdch gido
due
ANPHA
xin
trdn trong giai thieu
quy
ban dong nghiep vd
cdc
em hoc sinh cuon: "Boi dudng hqc sinh
gioi todn Dai so' Gidi tich
12 "
nay.
Cuon
sdch
nay nam trong
bo sdch
6
cuon
gom:
-
Boi
ducmg
hoc
sinh gidi
todn

Hinh
hoc
10.
- Bdi ducmg
hoc
sinh gidi
todn
Dai so' 10.
-
Boi
dudng
hoc
sinh gidi
todn
Hinh
hoc
11.
- Boi dudng
hoc
sinh gidi
todn
Dai
so
-
Gidi
tich 11.
- Bdi dudng
hoc
sinh gidi
todn

Hinh
hoc
12.
-
Boi
dudng
hoc
sinh gidi
todn Gidi
tich 12.
do nhd gido
uu tu,
Thac
sTLe Hoanh Phd
to'chirc
bien
soan.
Ndi
dung
sdch
duoc bien
soan
theo
chuong trinh phdn
ban:
co
bdn
vd nang
cao
mdi

ciia
bd
GD &
DT, trong dd mot
so van de duoc
md
rdng
vdi
cdc
dang bdi tap hay vd kho dephuc
vu
cho
cdc
em
yeu
thich
mud'n
ndng
cao
todn
hoc, cd
dieu kien phdt trien tot nhat kha nang
ciia
minh. Cuon
sdch
la
su ke
thira
nhung hieu bii't 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 co'gang sap xep, chon loc cdc bai todn tieu bieu cho
tirng the
loai
khdc nhau
ung vdi
ndi dung
cua
SGK.
Mdt
sd'bdi tap cd the khd nhung
cdch
gidi duqc dua tren nen tdng
kien
thuc vd

ki
nang co
bdn.
Hqc sinh can
tu
minh
hoan
thien
cdc
ki
nang ciing
nhu
phdt trien
tu duy
qua viec gidi bdi tap cd trong sdch trudc khi ddi
chieu vdi
led
gidi
cd
trong
sdch
nay, cd the mdt soldi
gidi cd
trong
sdch
con
cd dong, hqc
sinh
cd thetu minh lam
rd

hon, chi
tie't
hon, ciing
nhu
tie
minh dua ra nhung
cdch
lap ludn mdi
hon.
Chung tdi
hy
vong bd sdch
nay se
la
mdt
tdi lieu
thie't
thuc, bo ich cho ngudi day vd
hqc,
dqc biet
cdc
em hqc sinh yeu thich mdn todn vd hqc sinh chuan
bi
cho cdc
ky thi
qudc
gia (tot nghiep THPT, tuyen sinh DH
-
CD) do
bq

GD &
DT
to chirc sap
tdi.
Trong qua trinh
bien soqn,
cudn
sdch
nay khdng the tranh khoi nhirng thieu sdt, chung
tdi
ra't
mong nhdn duqc gop
y
ciia ban
dqc
gdn
xa
debq
sdch
hoan
Men hon
trong lah
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:

Xin
chan thanh cam on!
MUC
LUC

Chuong
I:
tTng dung dao ham de khao sat va ve do thi cua ham so
§
1.
Tinh
don dieu cua ham so 5
Dang 1: Dong bien, nghich bien, ham hang 5
Dang 2: Ung dung
tinh
don dieu 17
§2.
Cue tri
ciia
ham so 37
Dang 1: Cue dai, cue tieu , 37
Dang 2: Ung dung
ciia
cue tri 48
§3.
Gia tri Ion nhat va gia tri nho nhat 58
Dang 1: Tim gia
tri
ldn nhat, nho nhat 58
Dang' 2: Bai toan lap ham so 69
Dang 3: Ung dung vao phuong
trinh
77
§4.
Duong tiem can cua do thi ham so 88

Dang 1: Tim cac tiem can 88
Dang 2: Bai toan ve tiem can 96
§5.
Khao sat va ve ham da thuc 103
Dang 1: Ham bac ha 104
Dang 2: Ham trung phuong 113
§6.
Khao sat va ve ham hOu ti 126
Dang 1: Ham so v =
ax +
k (c * 0 va ad be * 0) 126
cx + d
2 i
Dang 2: Ham s6 v =
&X +
(a * 0. a' * 0) 135
a'x + b'
§7.
Bai toan thuong gap ve do thi 148
Dang 1: Tuong giao, khoang
each,
goc 149
Dang 2: Tiep tuyen. tiep xuc 159
Dang 3: Yeu to co dinh. doi xung - quy
tich
170
Chirong
II:
Ham so luy thua ham so mu va ham so logarit
§

1.
Quy tac bien doi va cac ham so 186
Dang 1: Bien doi luy thua - mu - logarit 188
Dang 2: Cac ham so mu. luy thua, logarit 200
Dang 3: Bat dang thuc va
GTLN,
GTNN
212
CHUONG
I: UNG
DUNG
DAO HAM OE
KHAO
SAT
VA VE DO THj CUA HAM SO
§1. TINH DON DIEU CUA HAM SO
A.
KIEN
THLTC
CO BAN
•
Dinh
nghTa: Ham so f xac dinh tren K la mot khoang,
doan
hoac
nira
khoang.
-
f
dong bien tren K neu

vdi
moi
Xi,
X2
6 K:
X]
<
X2
=>
f(xi)
<
f(x2)
-
fnghich
bien tren K neu
vdi
moi
xi.
xi
e K:
Xi<X2=>f(xi)>f(x2).
•
Dieu kien can de ham so don dieu
Gia
sir ham so co dao ham tren khoang (a; b) khi do:
-
Neu ham so f dong bien tren (a; b) thi f'(x) > 0 vdi moi x e (a; b)
-
Neu ham so f nghich bien tren (a; b) thi f'(x) < 0 vdi moi x e (a; b).
•

Dieu kien du de ham so don dieu
-
Gia sir ham so f co dao ham tren khoang (a; b)
Neu
f'(x)
> 0
voi
moi x e (a; b) thi ham so
f
dong bien tren (a; b)
Neu
f'(x)
< 0
voi
moi x e (a; b)
thi
ham so nghich bien tren (a; b)
Neu
f'(x)
= 0 vdi moi x e (a; b) thi ham so
f
khong doi tren (a; b).
-
Gia sir ham so f co dao ham tren khoang (a; b)
Neu f
'(x)
> 0
(hoac
f
'(x)

< 0) vdi moi x e (a; b) va f
'(x)
= 0 chi tai mot
so huu han diem cua (a; b) thi ham so dong bien
(hoac
nghich bien) tren
khoang (a; b).
B.
PHAN
DANG
TOAN
DANG
1: B6NG
BliN,
NGHICH
BIEN,
HAM
HANG
•
Phuong
phap
xet
tinh
don dieu:
-
Tim tap xac dinh
-
Tinh
dao ham, xet dau dao ham, lap
bang

bien thien
-
Ket luan
Chii
y: - Dau nhi
thuc
bac
nhat:
f(x) = ax + b, a ^ 0
x
-00 -b/a +co
f(x)
trai dau a 0 ciing dau a
- Dau tam thuc bac hai: f(x) = ax
2
+ bx + c, a * 0
Neu A < 0 thi
f(x)
luon ciing dau vdi a
Neu.A
= 0 thi f(x) luon cung dau vdi a,
trir
nghiem kep
Neu A > 0 thi dau "trong trai - ngoai
ciing"
X
-CO X] X
2
+00
f(x)

ciing
dau a 0 trai dau a 0 ciing diu a
-BDHSG DSGT12/1-
Vi
du 1: Xet chieu bien thien
ciia
ham sd:
a) y = x
2
- 6x + 5
c) y = x
3
- 2x
2
+ x + 1
b)y=
-x
3
3
2x
2
+ x - 3
d)
y = -x
3
+ 4x
2
- 7x + 5
Giai
a) Tap xac dinh D = R. Ta co y' = 2x - 6.

Cho y' = 0 » 2x - 6 = 0 » x = 3.
Bang bien thien
X
—oo 3 +00
y'
-
0 +
y
—
Vay
ham so nghich bien tren (-oo; 3), dong bien tren (3; +oo).
b)
D = R. Ta cd y' = 4x
2
- 4x + 1 = (2x - l)
2
> 0 vdi moi x
y'
= 0 o x = —. Vay ham so dong bien tren R.
2
c)
D = R . Ta co y' = 3x
2
- 4x + 1
Cho y' = 0 o 3x
2
- 4x + 1 = 0 <=> x = - hoac x = 1.
J
3
BBT

X
—00
1/3
1
+00
y'
+
0 0
+
y
^*
—^
Vay
ham so dong bien tren moi
khoang
(-co; —) va
(1;
+oo), nghich bien
3
tren
khoang
(—; 1).
3
d)
D = R Ta cd y' = -3x
2
+ 8x - 7
ViA'
=
16-21<0

nen y' < 0 vdi moi x do do ham so nghich bien tren R.
Vi
du 2: Xet chieu bidn thien cua cac ham so sau:
a) y = x
4
- 2x
2
- 5 b) y = x
4
+ 8x
2
+ 9
Giai
a) D = R. Ta co y' = 4x
3
- 4x = 4x(x
2
- 1)
Cho y' = 0
<=>
4x(x
2
- 1) = 0
<=>
x = 0
hoac
x = ±1
BBT
X
—00

-1
0
1
+00
y'
-
o •+
0 -
0
+
y
^
^* ^ ^
Vay
ham sd nghich bien tren moi
khoang
(-co; -1) va (0; 1), ddng bidn
tren moi
khoang
(-1;
0) va
(1;
+=»)•
6
-BDHSG DSGT12/1-
b)
D = R. Ta co y' = 4x
3
+ 16x = 4x(x
2

+ 4),y' = 0ox = 0.
y'
> 0 tren khoang (0; +co) => y dong bien tren khoang (0; +co)
y'
< 0 tren khoang (-co; 0)
=>
y nghich bien tren khoang (-co; 0).
Vi
du 3: Xet su bien thien cua ham so:
a) y = x +
—
x
b)y
c)y
a) Tap xac dinh D = R\ {0}
_3_
,.2
Ta co y' = 1
BBT:
Giai
, y' = 0 <» X
x
2
-3
3x-8
1-x
:V3"
d)y
1
(x-4)

2
X
-co
QN/3
+00
y'
+
0 -
-
0 +
y
Vay
ham so dong bien tren khoang (-co; -^3 ) va (J3 ; +oo), nghich bien
tren m6i khoang (-S ; 0) va (0; V3 ).
b)
D = R\
{0}.
Tacdy' = 1
>
0 vdi moi x ^ 0 nen ham so dong bien
tren moi khoang (-oo; 0) va (0; +co).
-5
c)
D = R \
{1}.
Ta cd y' = - < 0 vdi moi x
-t-
1 nen ham so nghich
3
(1-x)

2
bien trong cac khoang (-co; 1) va
(1;
+oo).
d) D = R\ {4}.Tacdy'= ———-
(x-4)
3
y'
< 0 tren khoang
(4;
+co) nen y nghich bien tren khoang (4; +co).
y'
> 0 tren khoang (-co; 4) nen y ddng bien tren khoang (-co; 4).
Vi
du 4: Tim cac khoang don dieu
ciia
ham so:
,
x-2 2x
a) y = -5 b) y
X
+
X
+ 1
a) D = R. Ta cd: y'
x
2
-9
Giai
-x

2
+ 4x + 3
(x
2
+ x + l)
2
y'
= 0 e> x
2
- 4x - 3 = 0
<=>
x = 2 ± ^7
BBT:
X
-co 2-V7 2+V7 +°o
y'
-
0 + 0 -
y
—^—*
- *
-BDHSG DSGT12/1-
7
Vay
ham so dong bien tren khoang (2 - yfl ; 2 + V7 ) va nghich bien
tren cac khoang (-co; 2 - 77 ) va (2 + ; +oo).
b) D = R\{-3;3}.Tac6y'=^^ <
0
,Vx*±3.
(x

2
-9)
2
Do
do y' < 0 tren cac khoang (-co; -3), (-3; 3), (3; +oo) nen ham so da
cho nghich bien tren cac khoang do.
Vi
du 5: Xet su bien thien cua ham sd:
a) y =
V4-x
2
c)y
Vl6-:
b)y
= Vx
2
-2x + 3
d)y
x
+ 2
Giai
a)
Dieu
kien 4-x
2
>0<=>-2<x<2 nen D = [-2; 2]
Vdi
-2 < x < 2 thi y'
BBT:
V4~

,y'
= 0<=>x = 0.
X
-2 n 2
y'
+
0 -
y
^
^ — ^
Vay
ham so dong bien tren khoang (-2; 0) va nghich bien tren khoang
(0;
2). Do ham so f
lien
tuc tren
doan
[-2; 2] nen ham so dong bien tren
doan
[-2; 0] va nghich bien tren
doan
[0; 2].
b)
Vi A' = 1 - 3 < 0 nen x
2
- 2x + 3 > 0, Vx => D = R.
•p , 1 2x-2 x-l , . ,
Ta co y = — = = =. y = 0 o x = 1.
2vx
2

-2x
+ 3
Vx
2
-2x
+ 3
y'>0ox>l,y'<0ox<l
nen ham so nghich bien tren khoang (-co; 1)
va dong bien tren khoang
(1;
+00).
c)
DK: 16 - x
2
> 0 o x
2
< 16 o -4 < x < 4. D = (-4; 4).
Ta co v' =
16
> 0, Vx e (-4; 4).
(16-x
2
)Vl6-x
2
Vay
ham so dong bien tren khoang (-4; 4).
d)
D = [0; +00). Vdi x > 0, y' =
X
_, y'

2^y^(x
+ 2)
2
BBT:
X
0 2 +00
y
+
0
y
^
Vay
ham sd ddng bien tren (0; 2) va nghich bien tren (2; +00).
8
-BDHSG DSGTu/1-
Vj
du 6: Tim khoang don dieu cua ham so
a) y = Vx(x- 3)
c)y
b)y
= -x
7x
2
-6
d)y
=
x
+ 1
Vl-x
Giai

a) D = [0; +oo). Vdi x > 0, ta cd:
y
BBT:
1
2Vx
(x-3)
+
r
3NRX-1)
vx
= , y
2x
0<=>x=
1.
X
0 1 +GO
y'
0 +
y
———^
Vay
ham so nghich bien tren khoang (0; 1) va dong bien tren khoang
(i';+°o).
b)
D = R. Vdi x ^ 0, ta co: y' =—-
3 3vV 3vV
y'
= 0
<=>
x

2
= 1
<=>
x = ±1.
y'
> 0 o ^/x
2
"
>l<=>x
2
>lci>x<-l
hoac
x > 1.
y'
< 0
%/x
2
"
<le>x
2
<lo-Kx<l.
Vay
ham so dong bien tren cac khoang (-co; -1) va (1; +co), nghich bien
tren khoang
(-1;
1).
c)
Tap xac
dinh
D = (-co; -^6 ) U (x/6 ; +oo).

Tacd: y' = -^^
7
^£L,y
1
= 0»x = ±3.
(x
2
-6)v'x
2
-6
BBT:
X
—CO
-3
V6
V6
3 +CO
y'
+
0 -
•
_
0 +
y
•
Vay
ham so dong bien tren cac khoang (-co; -3) va (3; +oo). nghich bien
tren cac khoang (-3; -vo ) va (vo ; 3).
d)
D =

(-QO;
1). Tacdy'= ,
3
~
X
> 0,
Vx<l.
2V(l-x)
3
Vay
ham sd dong bien tren khoang (-co: 1).
-BDHSG DSGT12/1-
Vj
du 7: Xet su bien thien cua ham sd:
3
a)y x + smx
b
)
y = x
+
cog
2
x
Giai
3
a) D = R. Ta cd y' = +
cosx
< 0, Vx nen ham sd nghich bien tren R.
b) D = R. Ta cd y' = 1 - 2cosxsinx = 1 - sin2x
y' = 0o sin2x = 1 <=> x = - +kit,keZ.

4
Ham sd lien tuc tren moi doan [- + kn; — + (k +
4 4
va y' > 0 tren moi khoang (- + kn; - + (k + 1)TI) nen ddng bien tren
4 4
moi doan [- + kn; - + (k + l)7tl, keZ.
4 4
v
'
1
Vay ham so dong bien tren R.
Vi du 8: Tim khoang dong bien, nghich bien cua ham so:
a) y = x - sinx tren [0; 2TT] b) y = x + 2cosx tren (0; n).
Giai
a) y' = 1 - cosx. Ta cd Vx [0; 2n] => y
1
> 0 va
y' = 0 <=> x = 0 hoac x = 2n. Vi ham so lien tuc tren doan [0; 2n] nen ham
so ddng bien tren doan [0; 2n].
b) y' = 1 - 2 sinx. Tren khoang (0; 7t).
y'>0o sinx <-<=> - < x < —
2 6 6
y' < 0 » sinx > - <=>0<x< - hoac — < x < -
2 6 6 6
Vay ham so ddng bien tren khoang (-; —). nghich bien tren moi
6 6
khoang (0; — ) va (—; n).
6 6
Vi du 9: Chung minh cac ham sd sau nghich bien tren R:
a) f(x) = vx

2
+1 - x b) f(x) = cos2x - 2x + 5.
Giai
a) Tacdf'(x) =
T
^=-l.
Vx
+1
Vi
Vx
2
+1 > Vx
2
=
I
x | > x, Vx nen f
'(x)
< 0, Vx do dd ham sd f nghich
bien tren R.
b) f'(x) = -2(sin2x+ 1)<0 vdi moi x.
10 -BDHSG DSGT12/1-
f'(x)
= 0osin2x = -lc^>2x = +2knox =
+kn,k&
Z.
2 4
Ham f(x)
lien
tuc tren moi doan [ + kn; ~ + (k + \)n] va
f'(x)

< 0 tren
moi khoang (-— +kn; — + (k+l)n) nen ham so nghich bien tren moi doan
4 4
[
+k;r;
+(k + l)n], k e Z.
4 4
Vay
ham sd nghich bien tren R.
Cach khac: Ta chung
minh
ham sd f nghich bien tren R:
VXj,
x
2
e R, x
x
< x
2
=>
f(Xj)
>
f(x
2
).
That vay, lay hai sd a, b sao cho a <
X|
<
X2
< b.

Ta cd: f'(x) = -2(sin2x + 1) < 0 vdi moi x e (a; b).
Vi
f
'(x)
= 0 chi tai mot sd huu han diem cua khoang (a; b) nen ham sd f
nghich bien tren khoang (a; b) => dpcm.
Vi
du 10: Chung
minh
rang cac ham so sau day dong bien tren R.
a) f(x) = x
3
- 6x
2
+ 17x + 4 b) f(x) = 2x -
cosx
+ S sinx.
Giai
a) f'(x) = 3x
2
- 12x + 17. Vi A' = 36 - 51< 0 nen f'(x) > 0 vdi moi x, do dd
ham so dong bien tren R.
V3
b)
y' = 2 + sinx - v3
cosx
= 2(1 +
—
sinx cosx).
2 2

= 2[1 + sin(x - —)] > 0, vdi moi x.
3
Vay
ham sd ddng bien tren R.
Vi
dull:
Chung
minh
ham so:
x-2
a) y = ddng bien tren moi khoang xac dinh cua nd.
x
+ 2
-x
2
—
2x + 3
b)
y = nghich bien tren moi khoang xac dinh cua nd.
x
+ 1
a) D = R \
{-2}.
Ta cd y' = — > 0 vdi moi x * -2
Giai
4_
(x
+ 2)
2
Vay

ham so dong bien tren moi khoang (-oo; -2) va (-2; +oo).
x
2
-2x-5
b)
D =
R\{-l}.Tacdy'=
~ < 0 vdi moi x
*-l
(vi
A'
= 1 - 5 < 0).
(x
+
1)
2
'
v
'
Vay
ham so nghich bien tren mdi khoang (-oo; -1) va
(-1;
+oo).
Vi
du 12: Chung
minh
ham so:
-BDHSG DSGT12/1-
1 1
a) y -

i + x2
dong bien trong khoang (-1; 1) va nghich bidn trong cac
khoang (-co; -1) va (1; +oo).
,
. sin(x + a) , , _ , ,
b
)
y ~ (
a
^ b +
krt;
k e Z) don dieu Uong mdi khoang xac dinh.
sin(x + b) • °
Giai
,
,
l(l
+ x
2
)-2x.x 1-x
2
a)
y
= (i
+
x
2
)
2
=

(T^'
y
=0ox
= ±1
-
Ta cd y' > 0
<=>
1 - x
2
> 0
<=>
-1 < x < 1.
y'<0<=>l-x
2
<0<=>x<-l
hoac
x > 1.
Tir
do suy ra dpcm.
b)
Ham sd gian doan tai cac diem x = -b + kn (k e Z).
,
_ sin(x + b) cos(x + a) - sin(x + a) cos(x + b) _ sin(b-a)
sin
2
(x + b) sm
2
(x + b)
(do a - b * kn)
Vi

y' ?t 0 va y'
lien
tuc tai moi diem x * -b + kn, nen y' giu nguyen mot
dau trong moi khoang xac dinh, do do ham so don dieu trong moi khoang
do.
Vi
du 13: Chung minh:
a) sin
2
x + cos
2
x = 1, Vx. b) cosx + sinx. tan— = 1, Vx e (-— ; —).
2 4 4
Giai
a) Xet
f(x)
= sin
2
x +
cos
2
x,
D - R.
f
'(x) = 2sinxcosx - 2cosxsinx = 0, Vx.
Do
dd f(x) la ham
hang
tren R nen
f(x)

= f(0) = 1.
b) Xet f(x) = cosx + sinx tan-, D = (-—; — ).
2 4 4
r-,/ x x sinx . x x
1
(x) = -smx + cosxtan— + = -sinx + cosx.tan— + tan—
•
2
2cos
2
-
2 2
2
X
X i
X
=
-sinx + tan
—
(1
+ cosx) = -sinx + tan— .cos —
2 2 2
—sinx + sinx = 0 voi moi x e (— ; —)
4 4
,
, TT 71
Suy ra rang f la mdt ham
hang
tren khoang (-— ;
—

).
Do dd f(x) = f(0) = 1 vdi moi x e (-—; - )•
4 4
Vi
du 14: Chung minh cac ham so sau la ham khong ddi
-BDHSG DSGTU/1-
a)
f(x)
= cos
2
x + cos
2
(x +
—)
- cosxcos(x + ^)
3 3
b) f(x)
- 2- sin
2
x - sm
2
(a + x) -
2cosa.cosx.cos(a
+ x).
Giai
a)
f'(x)
= -2cosxsinx - 2cos(x +-
)sin(x
+-

)
+ sinxcos(x +^) + cosx.sin(x +^ )
3 3 o o
o 71 7T
=
-sin2x - sin(2x + —) + sin(2x + - ) = -sm2x - 2cos(2x +
-).sin-
3 3 2 b
= -sin2x - cos(2x + — ) = 0, vai moi x.
2
113
Do
do f hang tren R nen
f(x)
=
f(0)
= 1 + = -
6 w w 4 2 4
b)
Dao ham theo bien x (a la hang so).
f
'(x)
= -2sinxcosx - 2cos(a + x)sin(a + x)
+
2cosa[sinxcos(a + x) + cosx.sin(a +
x)].
=
-2sin2x - sin(2x + 2a) + 2cosa.sin(2x + a) = 0.
Do
do f hang tren R nen

f(x)
=
f(0)
= 2 - sin
2
a -
2cos
2
a
= sin
2
a.
Vi
du 15: Cho 2 da thuc P(x) va Q(x) thoa man: P'(x) = Q'(x) vdi moi x va
P(0)
= Q(0).
Chung
minh:
P(x) = Q(x).
Giai
Xet
ham so
f(x)
= P(x) -
Q(x),
D = R.
Ta
cd f
'(x)
= P'(x) - Q'(x) = 0 theo gia thiet, do do

f(x)
la ham hang nen
f(x)
=
f(0)
= P(0) - Q(0) = 0 vdi moi x.
f(x)
= 0
=>
P(x) ^
Q(x).
Vi
du 16: Xac
dinh
ham so
f(x)
thoa man:
f(0)
= 8;
f(x).f
'(x)
= 1 - 2x (*).
Giai
Ta cd (*) -(f (x))* = l-2xo (f
3
(x))' = 3 - 6x.
3
Xet
ham sd
g(x)

=
f
3
(x)
- 3x + 3x
2
thi g'(x)
=
(f
(x))'
- 3 + 6x = 0.
nen g(x) = C: hang so tren D, do do:
f(x)
- 3x + 3x
2
= C
^> f
3
(x)
= -3x
2
+ 4x + C.
nen f(x) =
N/-3X
2
+ 3x + C Vi f(0) = 8 => C = 64.
Vay f(x)
=
yj-3x
2

+ 3x + 64 , thu lai dung.
Vi
du 17: Tim cac gia
trj
cua tham so a de ham so dong bien tren R.
a)
f(x)
= - x
3
+ ax
2
+ 4x + 3 b)
f(x)
= ax
3
- 3x
2
+ 3x + 2
•
3
Giai
a) f
'(x)
= x
2
+ 2ax + 4,
A'
= a
2
- 4

-
N6u a
2
- 4 < 0 hay -2 < a < 2 thi f
'(x)
> 0 vdi moi x e R nen ham so
ddng
bien tren R.
-BDHSG DSGT12/1-
13
-
Neu a = 2 thi f
'(x)
= (
x
+ 2)
2
> 0 vai moi x * -2 nen ham sd dong bien
tren R.
-
Neu a = -2 thi ham sd f '(x) = (x - 2)
2
> 0 vdi moi x * 2 nen ham so
dong bien tren R.
-
Neu a < -2
hoac
a > 2 thi f '(x) = 0 cd hai nghiem phan biet nen f
1
co

doi
dau:
loai.
Vay
ham sd ddng bien tren R khi va chi khi -2 < a < 2.
b)
V
(x)
= 3ax
2
- 6x + 3.
Xet
a = 0 thi f
'(x)
= -6x + 3 cd doi dau:
loai
Xet
a * 0, vi f khong phai la ham
hang
(y' = 0 tdi da 2 diem) tren dieu
kien
ham so dong bien tren R la f
'(x)
> 0, Vx
(a>0 fa>0 f
a
>0
<=>
<^=> <=> «a> 1.
[A'<0

[9-9a<0 |a>l
Yl
du 18: Tim cac gia tri cua m de ham sd nghich bidn tren R:
a) f(x)
mx
- x
b)
f(x) = sinx - mx + C.
Giai
a) y' = m - 3x
2
-
Neu m < 0 thi y' < 0 vdi moi xeR nen f nghich bien tren R
-
Neu m = 0 thi y' = -3x
2
< 0 vdi moi x e R,
dang
thuc chi xay ra vdi
x
= 0, nen ham so nghich bien tren R.
m
-
Neu m > 0 thi y' = 0 o x = ±^
BBT
X
—00
Xl
x
2

+00
y'
0 H 0
r
y
—»•
Do
do ham so dong bien tren khoang
(xi,
x
2
):
loai
Vay
ham so nghich bien tren R khi va chi khi m < 0.
b)
Vi f(x) khong la ham
hang
vdi moi m va C nen
f(x)
= sinx - m + C nghich
bien tren R
<=>
f
'(x)
=
cosx
- m < 0, Vx
a>
m >

cosx,
Vx o m > 1.
Vi
du 19: Tim m dd ham sd dong bien tren moi khoang xac dinh:
a) y
(3m
-l)x- m
2
+ m
by
= x + 2 +
x
+ m
m
x-l
Giai
a) D = R\
{-m}.
Ta co:
,
_ (x + m)(3m -1) - [(3m - l)x - m
2
+
Y
=
' (x + m)
2
14
m
4m

2
-2m
(x
+ m)
2
-BDHSG DSGTU/l-
Ham so dong bien tren moi khoang xac dinh
<=>
4m
2
- 2m > 0
<=>
m < 0
hoac
m > —.
2
b)
Ta cd y' = 1
m
vdi
moi x * 1.
(x-l)
2
-
Neu m < 0 thi y' > 0 vdi moi x * 1. Do do ham sd dong bien tren moi
khoang (-oo; 1) va
(1;
+oo).
XT'
, x

2
-2x + l-m
-
Neu m > 0 thi y = -„
(x-l)
2
y'
= 0ox
2
-2x+l-m =
0<=>x=l
+ Vm
BBT
X
—oo
1 —Vm
1
1
+ Vm
+oo
y'
+
0 -
0 +
y
Ham sd nghich bien tren moi khoang (1 - Vm ; 1) va
(1;
1 + Vm ):
loai.
Vay

ham sd ddng bien tren moi khoang xac. dinh cua nd khi va chi khi
m<0.
Vi
du 20: Tim a de ham sd:
a) f(x) = x
3
- ax
2
+ x + 7 nghich bien tren khoang
(1;
2)
b) f(x) = — x
3
- — (1 + 2cosa)x
2
+ 2xcosa + 1, a e (0; 2rr) dong bien tren
3 2
khoang
(1;
+oo).
Giai
a) f'(x) = 3x
2
-2ax+ 1
Ham so nghich bien tren khoang (1; 2) khi va chi khi y < 0 vdi moi
xe (1;2)
ff(l)
< 0 f
4-2a<0
13

<=>
<
<=><
<=>a>—
[f(2)<0
[l3-4a<0 4
b)
y' = x
2
- (1 +
2cosa)x
+
2cosa.
Ta cd 0 < a <
2TT.
y'
= 0 o x = 1
hoac
x =
2cosa.
Vi
y' > 0 d ngoai khoang nghiem nen ham so dong bien vdi moi x > 1 khi
va chi khi 2cosa < 1 cosa < — o — < a < —
2 3 3
Vi
du 21: Tim m de ham sd y = (m - 3)x - (2m + l)cosx nghich bien tren R.
Giai:
y'
= m - 3 + (2m - l)smx
Ham so y khdng la ham

hang
nen y nghich bien tren R:
y'
^ 0, Vx « m - 3 + (2m - l)smx < 0, Vx
D5t
t = sinx, -1 < t < 1 thi m - 3 + (2m - l)smx = m - 3 + (2m - l)t =
f(t)
-BDHSG DSGT12/1-
15
Dieu
kien tuang duong: f(t) < 0, Vt e [-;1 1]
[f(-l)<0
f-m-4<0 '9
°lf(l)^0 °l3m-2^0 ^-
4
^
m
^
V»
du 22: Tim m de ham sd y = x
3
+ 3x
2
+ mx + m chi nghich bidn tren mpt
doan cd dp dai
bang
3.
Giai:
D
= R, y' = 3x

2
+ 6x + m, A' = 9 - 3m
Xet
A' < 0 thi y' > 0, Vx: Ham luon dong bien (loai)
Xet A' > 0 <=> m < 0 thi y' = 0 cd 2 nghiem x,, x
2
nen x, + x
2
= -2, X]X
2
= —
3
BBT:
x
—CO
*1
x
2
+00
y'
+
0
-
0
+
y
^
^
Theo de bai: x
2

-
X]
= 3 o (x
2
-x,)
2
= 9o x
2
+ x
2
-2x
t
x
2
= 9
4 15
<=>
(x
2
+ x
t
)
2
- 4x
t
x
2
= 9ci>4 — m = 9 o m= (thoa)
3 4
Vi

du 23: Tuy
theo
tham
&6
m, xet su bien thien cua ham sd:
\
1 3 ? , rx ix 2x + m
a) y = - x
3
- 2mx
2
+ 9x - m b) y =
3 x-l
Giai
a) D = R. Ta cd y' = x
2
- 4mx + 9; A' = 4m
2
- 9
-
Neu A' < 0 <o 4m
2
< 9 <=>
bien tren R.
I
m
[
< — thi y' > 0, Vx nen ham so dong
-
Neu A' > 0 co 4m

2
> 9 co
I
m | > - thi y' = 0 cd 2 nghiem phan biet
xi,
2
= 2m +V4m
2
-9 Lap bang bien thien thi ham ddng bien tren
khoang (2m - V^m
2
-9 ; 2m +V4m
2
-9) va nghich bien tren m6i
khoang (-00; 2m - \/4m
2
-9 ) va (2m +
V4m
2
-9 ; +00).
b)D = R\ {l}.Tacd y'= ~
2
~"!
(x-l)
2
-
Neu m = -2 thi y = 1, Vx * 1 la ham sd khong doi.
-
Neu m > -2 thi y' < 0, Vx *1 nen ham so nghich bien tren moi khoang
(-00; 1) va

(1;
+00).
-
Neu m < -2 thi y' > 0, Vx * 1 nen ham sd ddng bien tren moi khoang
(-co; 1) va
(1;
+00).
16
-BDHSG DSGT12/1-
DANG
2: UNG
DUNG
TINH
BON
DI$U
- Giai
phirong
trinh,
he phirong
trinh,
bat phuong
trinh:
Xet f(x)
la ham so v6
trai,
neu can thi bien doi, chpn xet ham, dat an phu,
Tinh
dao ham rdi xet
tinh
don dieu.

Neu ham sd f don dieu tren K thi phuong
trinh
f(x) = 0 cd toi da 1
nghiem. Neu f(a) = 0, a thupc K thi x = a la nghiem duy
nhat
cua phuong
trinh f(x)
= 0.
Neu f cd dao ham cap 2 khdng ddi dau thi f la ham don dieu nen phuong
trinh
f(x) = 0 cd tdi da 2 nghiem. Neu f(a) = 0 va f(b) = 0 vdi a * b thi
phuong
trinh f(x)=0
chi cd 2 nghiem la x = a va x = b .
- Chiing minh
bat
dang
thiic:
Neu y =
f(x)
cd y' > 0 thi
f(x)
dong bien: x > a =>
f(x)
> f(a); x < b
=>f(x)<f(b)
Doi
vdi y' < 0 thi ta cd bat
dang
thuc

nguoc
lai.
Viec
xet dau y' doi khi phai can den y",
y"\
hoac
xet dau bp phan,
chang
han tir so
ciia
mpt phan so cd mau duong
Neu y" > 0 thi y' dong bien tir do ta cd danh gia f '(x) rdi
f(x),
Vi
du 1:
Giai
phuong
trinh:
vo - x + x
2
- 72 + x - x
2
= 1
Giai
Dat t = x
2
- x thi phuong
trinh
trd thanh: 73+ t - 72-t =1,-3 < t < 2.
Xet

ham sd
f(t)
= 73 + t - 72-t, -3 < t < 2.
Vdi -3 < t < 2 thi f'(t) =
1
+ . > 0 nen f dong bien tren (-3; 2).
273 + t 272 -t
Ta cd
f(l)
= 2 - 1 = 1 nen phuong
trinh:
f(t) = f( 1) <=> t
=
1 o x
2
— x — 1 = 0 <=> x
=
l^H.
Vi
du 2:
Giai
phuong
trinh
72x
3
+3x
2
+6x + 16 = 273 + 74-x
Giai:
Dieu

kien xac djnh:
f2x
3
+3x
2
+6x
+ 16>0 f(x + 2)(2x
2
-x + 8)>0
<=>C
cs> -2<x<4
4-x>0 4-x>0
Phuong
trinh
tuong duong 72x
3
+ 3x
2
+ 5x +16 - 74 - x = 273
Xet
ham s6
f(x)
=
N/2X
3
+3x
2
+ 6x
+
16 - 74 - x 2 < x < 4

™,
cu
. 3(x
2
+ x + l) . x
Thi
f (x) =
—.
= + — > 0 nen i dong bien ma
72x
3
+3x
2
+6x + 16 274-x
f(l)
= 273 , do do phuong
trinh
trd thanh
f(x)
=
f(l)
o x = 1
Vdy
phuong
trinh
co nghiem duy
nhat
x=l
-BDHSG DSGT12/1-
17

Vi
du 3: Giai
phucmg
trinh yjx
2
-1 = Vx
3
-2 - x.
Giai
Dieu kien: x >%/2
Ta cd: Vx
3
-2 = x + Vx
2
-1 >x>l=>x
3
>3=i>x>v
/
3
Chia
2 ve cho vo? thi
phuong
trinh:
~
1
a.
x
2
.vx x
4

vx Vx V xVx
Xet f(x) la ham sd ve trai, x > tfi thi
,,
x
9 5x X 3
f'(x)=
B r
—
r
, <0.
2x
5
.Vx 2xVx
n 2
2
2x
Vx
Do do ham so f
nghich
bien
tren
khoang
(y3 ; +oo) ma f(3) = 0 nen
phuong
trinh cd
nghiem
duy
nhat
x = 3.
Vi du 4: Giai phuong trinh: 3x

2
- 18x + 24
1 1
2x-5 x-l
Giai
5
Dieu kien x * 1; —,
phuong
trinh trd
thanh:
2
(2x-5)
2
J_
=
(x-l)
2
|2x-5| |x-l|
Xet f(t) = t
2
- i vdi t > 0.
Ta co: f '(t) = 2t > 0 nen f dong bien tren (0; +oo)
Phuongtrinh:f(|2x-5|) = f(|x - l|)o 12x- 51 = |x-l|
<=>
4x
2
- 20x + 25 = x
2
- 2x + 1
<=>

3x
2
- 18x + 24 = 0.
c^x
2
-6x
+ 8 =
0cox
= 2
hoac
x = 4
(chon)
Vi
du 5: Giai bat
phuong
trinh:
4 | 2x - 11 (x
2
- x + 1) > x
3
- 6x
2
+ 15x - 14
Giai
BPT: | 2x - 11 .[(2x - l)
2
+ 3] > (x - 2)
3
+ 3x - 6
<eo

| 2x - 11
3
+ 3 | 2x - 11 > (x - 2)
3
+ 3(x - 2)
Xet ham sd f(t) = t
3
+ 3t, D = R.
Ta cd f '(t) = 3t
2
+ 2 > 0 nen f
dong
bien
tren
R.
BPT:
f(
| 2x - 11) > f(x - 2) o
I
2x - 11 > x - 2.
Xet x - 2 < 0 thi BPT
nghiem
dung.
Xet x - 2 > 0 thi 2x - 1 > 0 nen BPT o2x -1>X-2<=>X>-1
:
£)
U
ng
Vay tap
nghiem

la S = R.
18 -BDHSG DSGT12/1-
Vi
du 6:
Giai
bat phuong
trinh: Vx
+
1
+ 2Vx + 6 < 20 -
3VX
+ 13.
Giai
Dieu kien:
x >
-1.
BPT
viet lai: Vx
+
1
+ 2%/x + 6 + 3Vx + 13 > 20
Xet f(x)
la ham so ve
trai,
x >
-1.
Ta co:
113 A
f'(
x

)
= —— +
—
+— > 0 nen f dong bien tren [-1; +oo).
2Vx
+ l
Vx
+ 6
2Vx
+ 13
Ta
cd f(3) = 20 nen
BPT:f(x) <f(3)ox<3.
Vay tap nghiem cua BPT la
S =
[-l;3].
Vi
du 7:
Giai
bat phuong
trinh:
3Vtanx + 1.
smx +
2c
°
SX
< 2
1
"^
sin

x +
3
cos x
Giai
Dieu kien
tanx > 0. Dat t = tanx, t > 0 thi
VT
=
3v^TT.^
=
f(t),
t > 0
t
+ 3
3 t + 2 1 '
Ta
cd f
'(t)
= —, . + 3vt +
1.
-> 0 nen ham so f dong bien,
2Vt+T
t + 3 (t + 3)
2
ma t > 0 =>
f(t)
>
f(0)
= 2.
Mat

khac
VP = 2
l
~
4
^ < 2 nen dau "='' ddng
thoi
xay ra
<=>
t = tanx = 0
<=>
x = krc, k e Z.
x
+ 3 = y
+Vy
2
+ 1
Vi
du 8:
Giai
he phuong
trinh
y+
3
= z +
%/z
2
+ 1
Z + 3 = X + N/X
2

+ 1
Giai
Xet
ham sd
f(t)
= t +
Vt
2
+1 -3,teR
fU'f.m
1
t
Vt
2
+1 + t
Vt7
+ t
tmf'(t)
= l+ . —; > , >0, Vt
vv+i
Vt
2
+ i vV + i
nen
f(t)
ddng bien tren R. Ta cd he phuong
trinh
x
=
f(y)

y
=
f(z)
z =
f(x)
Gia
su x > y thi
f(x)
>
f(y)
nen y > z do dd
f(y)
>
f(z)
tuc la z > x: vo li
Gia
su x < y thi
f(x)
<
f(y)
nen y < z do do
f(y)
<
f(z)
tuc la z < x: vo
11
Gia
su x = y thi
f(x)
=

f(y)
nen y = z do do x = y = z. The vao he:
x
+ 3 = x + v
/
x
2
+lco3 =
v
/
x
2
+lci>x
2
=2
<S-X
=
±V2
Thu
lai x = y = z =
+%/2
thi he nghiem dung.
Vay
he phuong
trinh
cd 2 nghiem x = y = z =
+
V2
-BDHSG DSGT12/1-
19

Vi
du 9: Giai he phuong
trinh:
Ta cd x
y
-y + i
1
x
-l
=
y(y-l)
<
y
3
- l = z(z-i)
z
3
- l =
x(x-l)
Giai
lY
3 3 1
-
+—>—>—=
2)
4 4 8
x
>
Tuong tu y, z > - Dat
f(t)

= V
1
l,t>
- thi
2
f
'(t)
= 2t - 1 > 0 nen f dong bien tren (—; +oo)
2
Ta cd he
<
3 2
y
= z
-y
+ l
z +
1
co
<
z
3
= x
2
-
X
+ 1
z
3
>

X
f(y)
f(z)
f(x)
3
Z>
X.
y
3
> z
3
=> y > z.
Gia
su x > y thi
f(x)
>
f(y):
nen
f(z)
> f(x)
Do
do x > y > z > x: vd
li.
Tuong tu x < y: vo li nen x = y => x = y = z. Ta cd t
3
= f(t)
ot
3
-t
2

+ t-l = 0o(t- l)(t
2
+ 1) = 0 co t = 1.
Vay
he co nghiem duy
nhat
x = y = z = 1.
x
3
-2x + l = 2y
Vi
du 10: Giai he phuong
trinh
Ta cd 2y = x
2
- 2x +
1
= (x -
l)
2
> 0
y
-2y + l = 2z
z
2
-2z + l = 2x
Giai
y
> 0. Tuong tu z, x > 0.
Dat

f(t)
= t
z
- 2t + 1, t > 0 thi f
'(t)
= 2(t - 1) nen f dong bien tren
(1;
+oo)
va nghich bien tren (0; 1). Dat g(t) = 2t, t > 0 thi g'(t) = 2 > 0
f(x)
= g(y)
nen g ddng bien tren (0; +oo). Ta cd he I
f
(y)
= g(z)
f(z)
= g(x)
Gia
su x =
min{x;
y; z}. Xet x < y < z.
-
Neu x >
1
thi 1< x < y < z ^>
f(x)
<
f(y)
< f(z)
=>

g(y) < g(z) < g(x) =o y < z < x nen x = y = z.
Ta co phuong
trinh:
t
2
- 4t + 1 = 0 nen chon nghiem: x = y =
z
= 2-
V3.
-Neu 0 < x <
1
thi f(0) >
f(x)
>
f(l)
=o 0 <
f(x)
< 1.
nen 0 < g(y) < 1 => 0 < y < 1 =o
f(0)
>
f(y)
>
f(l)
=o 0 < f(y) < 1 ^> 0 < g(z) <1=>0<Z<1.
20
-BDHSG DSGTn/i
Do
do x < y < z =>
f(x)

>
f(y)
>
f(z)
=> g(y) > g(z) > g(x)
=>
y > z > x nen x = y
= z
-
Ta cd phuong
trinh
t
2
- 4t + 1 = 0 nen chon nghiem: x = y = z = 2- 42
Xet
x < z < y thi cung nhan
duoc
ket qua tren.
Vay
he cd 2 nghiem x = y = z = 2+ v
/
3,x = y = z = 2- -Js
36x
2
y-60x
2
+25y = 0
36y
2
z-60y

2
+25z = 0
Vi
dull:
Giai he phuong
trinh:
36z
2
x-60z
2
+25x = 0
Giai
60x
2
36x
2
+25
He phuong
trinh
tuong duong vdi
<
z
60y
2
36y
2
+25
60z
2
36z

2
+25
Tir
he suy ra x, y, z khong am. Neu x = 0thiy = z = 0 suy ra (0; 0; 0) la
nghiem cua he phuong
trinh.
Neu x > 0 thi y > 0, z > 0. Xet ham sd f(t)
3000t
60f
36t
2
+25
t>0.
Ta cd: f'(t)
(36t
2
+25)
2
>0,
Vt>0.
Do
do
f(t)
ddng bien tren khoang (0; +co).
y
=
f(x)
He phuong
trinh
duoc

viet lai I z =
f
(y)
x
=
f(z)
Tir
tinh
ddng bien cua f(x) suy ra x = y = z. Thay vao he phuong
trinh
ta
duoc
x(36x
2
- 60x + 25) = 0. Chon x =
6
'5 5 5
,6'6'6
Vay
tap nghiem cua he phuong
trinh
la
<j
(0;0;0);
:8-X
3
Vi
du L2: Giai he phuong
trinh
Vx-1

-Ty
(x-l)
4
=y
Giai
Dieu
kien x > 1, y > 0. He phuong
trinh
tuong duong vdi:
IVx-l
-
(x-l)
2
+ x
3
- 8 = 0 (1)
[y
= (x-D
4
(2)
-BDHSG DSGTU/1-
Xet
ham so f(t) =
VtTT
- (
t
- l)
2
+
t

3
- 8, vdi t > 1.
1
Ta co f
'(t)
=
-2(t-l)
+ 3t
2
+ =
3t
2
- 2t + 2
2vW 2v/Tl
>
0 vdi moi
2Vt-l
t
> 1 nen
f(t)
ddng bien tren
(1;
+oo).
Phuong
trinh
(1) cd
dang
f(x) = f(2) nen (1) co x = 2, thay vao (2) ta
duoc
y =1.

Vay
nghiem cua phuong
trinh
la (x; y) = (2; 1).
x
2
- 12x + 35 < 0 (1)
Vi
du 13:
Giai
he bat phuong
trinh:
x
u
- 3x
2
+ 9x + - > 0 (2)
3
Giai:
Ta cd (1) o x
2
- 12x + 35 < 0 co 5 < x < 7
Xet
(2): Dat
f(x)
= x
3
- 3x
2
+ 9x + -, D = R

3
f
(x)
= 3x
2
- 6x + 9 > 0, Vx eR nen
f(x)
ddng bien: x > 5 =o
f(x)
>
286/3
Do
do f(x) > 0, VXG (5 ; 7)
Vay
tap nghiem cua he bat phuong
trinh
la S = (5; 7).
YJ
du 14: Chung
minh
rang phuong
trinh
3x
5
+ 15x - 8 = 0 cd mot nghiem
duy
nhat.
Giai
Ham
f(x)

= 3x
5
+ 15x - 8 la ham sd
lien
tuc va cd dao ham tren R.
Vi
f(0)
= -8 < 0,
f(l)
= 10 > 0 nen ton tai mot sd x„ e (0; 1) sao cho
f(xo)
= 0,
tuc la phuong
trinh f(x)
= 0 cd nghiem.
Mat
khac, ta cd y' = 15x
4
+ 15 > 0, Vx e R nen ham sd da cho
luon luon
ddng bien. Vay phuong
trinh
dd chi cd mot nghiem duy
nhat.
Vi
du 15: Chung
minh
phuong
trinh:
x

13
- x
6
+ 3x
4
- 3x
2
+ 1 = 0 cd nghiem
duy
nhat.
Giai:
Dat
f(x)
= x
13
- x
6
+ 3x
4
- 3x
2
+ 1, D = R
Xet
x > 1 thi
f(x)
=
x
6
(x
7

- 1) +
3x
2
(x
2
- 1) + 1 > 0: vd nghiem
Xet
0 < x < 1 thi
f(x)
= x
13
+ (l - x
2
)
3
> 0: vd nghiem
Xet
x < 0
thi:
f
'(x)
= 13x
12
- 6x
5
+ 12x
3
- 6x = 13x
12
- 6x(x - l)

2
> 0 nen
f
dong bien
Bang bien thien:
x
—CO
0
y'
+
y
1
—OO
Nen
f(x)
= 0 cd nghiem duy
nhat
x < 0.
Vay
phuong
trinh
cho cd nghiem duy
nhat.
22
-BDHSG DSGT12/1-
Vi
du 16: Chung
minh
rang phucmg
trinh 2x

2
Vx-2
= 11 cd mot nghiem
duy
nhat.
Giai
Xet
ham so f(x) = 2x
2
Vx
- 2 thi ham sd xac
dinh
va
lien
tuc tren nua
khoang [2; +oo).
f'(x)=
2
f
n
I ^ x
2
) x(5x-8)
2xVx-2+—,
-
2Vx-2
>
0, vdi moi x e (2; +co)
Vx-2
Do

do ham so dong bien tren nua khoang [2; +oo).
Ham sd
lien
tuc tren doan [2; 3], f(2) = 0, f(3) = 18. Vi 0 < 11< 18 nen
theo
dinh
li ve gia tri trung gian cua ham sd
lien
tuc, tdn tai sd thuc c e (2; 3)
sao cho f(c) = 11 tuc c la mot nghiem cua phuong
trinh
f. Vi ham sd
dong bien tren [2; +co) nen c la nghiem duy
nhat
cua phuong
trinh.
Vi
du 17: Chung
minh
rang vdi moi x e (-1; 1), phuong
trinh:
sin
2
x +
cosx
= m
cd
mot nghiem duy
nhat
thudc doan [0;

TT].
Giai
Xet
ham sd f(x) = sin
2
x +
cosx
thi ham sd
lien
tuc tren doan [0; n],
Ta cd f'(
x
) = 2sinxcosx - sinx = sinx(2cosx - 1), x e (0;
TI)
Vi
sinx > 0 nen f
'(x)
= 0 o
cosx
=
—
o x =
—
2 3
BBT:
X
TC
0 — Tt
u
3

f(x)
+
0
f(x)
5
71 ' 71
Ham f dong bien tren doan [0;
—
] va nghich bien tren doan [— ; 7i].
3 3
'
71 7t 5
Ham so f
lien
tuc tren doan \—;
Til,
f(—)
= — va
fire)
=
-1.
Theo
dinh
li
3 3 4
5
ve gia tri trung gian cua ham so
lien
tuc, vdi moi m e (-1; 1) cz (-1; —).
4

ton tai mot sd thuc c e (—; 7i) sao cho f(c) = 0 tuc c la nghiem cua
3
phuong trinh. Vi ham so f nghich bien tren [—; 71] nen tren doan nay,
3
phuong
trinh
cd mot nghiem duy
nhat.
TC
5
Con vdi moi x e [0;
—
],
ta cd 1 < f(x) < — nen phuong
trinh
khong co
3 4
nghi?m suy ra dpcm.
Vi
du 18: Tim sd nghiem cua phuong
trinh
x
3
- 3x
2
- 9x - 4 = 0.
-BDHSG DSGT12/1-
23