Phương pháp giải các chủ đề căn bản giải tích 12 lê hoành phò

515.076
PH561P

NGirr.ThS. LE H O A N H

a ren luyen
kl nang
lam bai

H a No

NHA XUAT BAN DAI HOC QUOC GIA HA NOI

GIAI

PHO


NGIfT.ThS. LE HOANH PHO

GIAI

CAC
C H U
C A N

D E
B A N

(Boi dm!smmmi!mifi
cs^G NHA XUAT BAN flAI HOC QUOC GIA HA NOI

Ha Nol

.

.


NHA XUAT BAN DAI HOC QUOC GIA HA NQI
16 Hang Chuoi - Hai Ba Trang - Ha Npi
Dien thoai: Bien tap-Che ban: (04) 39714896:
Hanh chinh: (04) 39714899: Tonq bien tap: (04) 39715011
Fax: (04)39714899

*

Chiu

trdch

nhiem

xuat

* *

ban:

Gidm doc - Tong bien tap:

Bien tap:

TS.PHAM T H I TRAM

L A N HUONG

bdi:

N G U Y E N KHCJl M I N H

Che ban:

N H A SACH HONG A N

Saa

Trinh

bay bia:

VO THI T H I T A

Doi tdc lien ket xuat ban:
N h a sach H O N G A N

S A C H LIEN KET

CAC CHU DE CAN BAN GIAI TfCH 12
Ma so: 1L- 154DH2014
In 2.000 cuon, khd 17 x 24cm tai Cong ty Co phan VSn hoa Van Lang.
Giay phep xuat ban so: 463-2014/CXB/10-99 OHQGHN, ngay 14/03/2014

Quyet dinh xuat ban so: 154LK-TN/Q0-NXB OHQGHN, v ^ i

>

In xong va nop iuu chieu quy II nam 2014.


Nham muc dich giup cac ban hoc sinh Idp 10, Idp 1 1 , Idp 12 nam
vOng kien thifc can ban ve mon Toan ngay tCr luc vao THPT cho den
khi chuan bi thi Tot nghiep, tuyen sinh Cao dang, Dai hoc, tac gia da
bien scan bo sach P H l / O N G P H A P G I A I gom 6 cuon:
- C A C C H U D E C A N B A N DAI S O 10
- C A C C H U D E C A N B A N HINH H Q C 10
- C A C C H U D E C A N B A N DAI S O - GIAI T I C H

11

- C A C C H U D E C A N B A N H I N H H Q C 11
- C A C C H U D E C A N B A N GIAI T I C H 12
- C A C C H U D E C A N B A N HINH H O C 12

TL/ nen Toan can ban nay, cac ban c6 the nang cao dan dan, bo
sung va md rpng kien thufc va phUdng phap giai Toan, ren luyen ky
nang lam bai va tC/ng bade giai dung, giai gpn cac bai tap, cac bai toan
kiem tra, thi CLT.
Cuon C A C C H U D E C A N B A N G I A I T I C H 1 2
npi dung la phan dang Toan, tom tat kien thac
cac chu y; phan tiep theo la cac bai toan chpn
vdi nhieu dang loai va mac dp; phan cuoi la 8
hay dap so.

nay c6 16 chu de vdi
va phUdng phap giai,
Ipc can ban minh hpa
bai tap c6 hadng dan

DO da CO gang kiem tra trong qua trlnh bien soan song khong tranh
khoi nhQng sai sot ma tac gia chaa thay het, mong don nhan cac gop y
cua quy ban dpc, hpc sinh de Ian in sau hoan thien hdn.
Tac gia

LE HOANH

PHO

3


O CHUD E I

T i N H t>ON t>IEU
DANG TOAN
•

1.

TiM KHOANG DONG BIEN VA NGHjCH BIEN

Dinh nghia: Hdm so f xdc dinh tren K Id mot khodng, doan hodc nica khodng.
- f dong hien tren K neu vai moi xi, X2 e K: x/ < X2 =^f(x\) < f(x2)

-f nghich hien tren K neu vai moi x/, X2 e K: xi < X2 =>f(xi) > f(x2).
Dieu kien can de ham so dffn difu
Gid su hdm so c6 dao hdm tren khodng (a; b) khi do:
- Neu hdm so f dong bien tren (a: b) thif '(x) > 0 vai moi x e (a; b)
- Neu hdm so f nghich bien tren (a; b) thif'(x) ^0 vai moi x e (a; b).
Dieu kien dii de hdm so dffn dieu
Gid su hdm so f c6 dqo hdm tren khodng (a; b) khi do:
Neuf'(x) > 0 vai moi x e (a; b) thi hdm so f dong bien tren (a; b)
Neu f'(x) < 0 vai moi x e (a; b) thi hdm so f nghich bien tren (a; b)
Khi f '(x) = 0 chi tgi mot so hiru hgn diem cua (a; b) thi ket qud tren vdn dung.
Neu hdm so f dong bien tren (a; b) vd lien tuc tren nita khodng [a;b); (a;bj;
doan [a:b] thi dong bien tren nica khodng [a;b); (a;bj; doan [a;b] tuang icng.
Tuang tu cho nghich bien.
Phumtg phdp xet tinh dffn dieu:
- Tim tap xdc dinh
- Tinh dao hdm, xet ddu dqo hdm, lap bang bien thien
- Ket ludn
Chiiy:
1) Cong thirc vd quy tdc dao hdm
y^C

=>y' ^0:y=x

^y'

= l;y = x" =^y' = nx"-';

^y=J—(x>0);
y=
'4^=>y'=—

2Vx
n"-47
y = sinx =:>y' = cosx; y = cosx =>y' = -sinx;

y= ^

y = tanx =^y =

1r — ; y = cotx =>y , = — -r1— .
cos X
sin X

(u + v)' = u' + vV (u-v)'

= u'- v';

5


(u. v)' = u'.v + u.v';

u .V - u.v ;f'.-f'u.u'x..

2) Phuang trlnh luang gidc ca ban:
X = a + k27i
sinx = sina <=> X = 71 - a + k27i (keZ)
X = a + k27i
cosx = cos a <=> X = - a + k27r (keZ)
tanx ^ tan a c^x a + kTt (k e Z)
cotx = cot a <:=>x = a + kn(k e Z).

Bai toan 1: Tim khoang dong bien, nghich bien cua ham so:
a)y = x^-8x + 5
b) y = x^ - 2x^ + x + 1.
Gidi
a) Tap xac dinh D ^= R. Ta c6 y' = 2x - 8.
Cho y' = 0 « 2x - 8 = 0 <=> x = 4.
4
+00
Bang biSn thien (BBT) x -00
0 +
y'
y
Vay ham so nghich bien tren (-oo; 4), dong bien tren (4; +oo).
b) D = R. Ta CO y' = 3x^ - 4x + 1
Cho y' = 0 « 3x^ - 4x + 1 = 0 o x = - hoac x = 1.
+00
1/3
BBT x -00
1
+ 0
0 +
y'
^
^
y

^ ^--^

Vay ham so dong bien tren moi khoang (-co; ^) va (1; +co), nghich bi6n tren
khoang ( ^ ; 1).

Bai toan 2: Tim khoang dong bien, nghich bien cua ham so:
a)y-x''-2xl
b) y = x'"* + 9x^ - 3.
Gidi
a) Tap xac dinh D = R.
y' = 4x^ - 4x = 4x(x^ - 1), y' = 0 <=> x = 0 hoac x = ±1.
6


BBT

X

-co

-

y'
y

0

-1
0

+

0

+00

1
-

+

0

+00

+00

^ - 1 ^
Vay ham so dong bien tren moi khoang (-1;0) va ( 1 ; +oo), nghich bien tren
khoang (-oo;-1) va (0; 1).

b) D - R. Ta CO y' = 4x^ + 18x = 2x(2x^ + 9), y' = 0 <=> x = 0.
y' > 0 tren khoang (0; +co) ^ y dong bien tren khoang (0; +QO)
y' < 0 tren khoang (-oo; 0) => y nghich bien tren khoang (-co; 0).
Bai toan 3: Tim khoang dong bien, nghich bien cua ham so:
a)y =

3x-9

b)y = x + - .
x
Giai

a)D = R \.
Ta CO y' =

-6
(l-.r)

< 0 vai moi x ^ 1 nen ham so nghich bien trong cac khoang

(-oo; l ) v a ( l ; + o o ) .

b) TapxacdinhD = R \.
Taco
BBT:

1 - ^ = ^ ^ - ^ , y ' - 0 < » x = ±V3.
x^
X"
X

-30

y'

+

0

+00

0

-V3

-

-

0

+

y
Vay ham so dong bien tren khoang (-oo; - V3 ) va (V3 ; +oo), nghich bien tren
moi khoang (- V3 ; 0 ) v a ( 0 ; V 3 ) .
Bai toan 4: Tim cac khoang don dieu cua ham so:
K^

x+ \
jc +8

2x
X -9

Gidi
a) D = R. Ta c6: y' =

-X-

y' = 0

-2x + 8

{x'+%f
- X - 2x + 8 = 0

X = -4 hay X =2
7


BBT:

X

-00

-

y'

-4
0

4

2
0

+00

-

y

Vay ham so dong bien tren khoang ( -4; 2) va nghich bien tren cac khoang
-4), (2; + 0 0 ) .

(-co;

b ) D = R \6 y' = ~^^^

< 0, V x ^ ± 3 .
(x'-9)'

Do do y' < 0 tren cac khoang (-oc; -3), (-3; 3), (3; +oc) nen ham so da cho
nghich bien tren cac khoang do.
Bai toan 5: Xet su bien thien ciia ham so tren doan, nua khoang:
a) y = ^|9-x^

b)y= V x ' - 2 x + 7 .
Gidi

a) Dieu kien 9 - x" > 0 » -3 < x ^ 3 nen D - [-3; 3].
Voi -3
BBT:

< X <

3 thi y' -

X

- X

^

, y' = 0

^

y

0
0

y'

+

^

^

X =

0.

^

Vay ham so dong bien tren khoang (-3; 0) va nghich bien tren khoang (0; 3). Do
ham so f lien tuc tren doan [0; 2] nen ham so dong bien tren doan [-2; 0] va nghich
bien tren doan [0; 2].
b) V i A' = 1 - 3 < 0 nen x^ - 2x + 7 > 0, Vx => D = R.
„

. ,
2x-2
x-1
Fa CO y = —,
= ,
.
2Vx'-2x + 7 Vx--2x +7
y'^0<:^x> l,y'<Oc>x< 1
Va f lien tuc tren R nen ham so nghich bien tren nura khoang (-oo; 1 ] va dong
bien tren nua khoang [1; +oo).
Bai toan 6: Xet su bien thien cua ham so:
b)y =

a)y
Vl6^
a) DK: 16 - x^ > 0 «

X'

x+2

Gidi
< 16 <=> -4 < X < 4. D = (-4; 4).

8


Ta CO

16

. > 0 , Vx e (-4; 4).

y' =

(16-x-)Vl6-xVay ham so dong bien tren ichoang (-4; 4).
b) D = [0; +00). Vai x > 0, y' =
BBT:

+00

-

0

+

y'

, y' = 0 <» x = 2.

2Vx (x + 2)

2

0

X

2-x

y
Vay ham so dong bien tren (0; 2) va nghich bien tren (2; +oo).
Bai toan 7: Tim Ichoang don dieu cua ham so
a) y =

\fx

- X -

X

b) y =
Giai

a) D = R . V a i x 5 ^ 0 . t a c 6 : y ' = y' = 0

«

X' =

1

X =

y' > 0 <=> Vx^ > 1

^ - ^ = ^ ~ ^

±1.

x^ > 1 hoac x < -1 hoac x > 1.

y'<0<=>Vx^Vay ham so dong bien tren cac Ichoang
khoang (-1; 1).

(-cx);

-1) va (1; +oo), nghich bien tren

b) Tap xac dinh D = (-QO; - V6 ) U ( V6 ; +oo).
Taco: y'=
BBT:

(x'-6)Vx'-6

X

y'

y ' = 0 <z> x = ±3.

-3

-00

+

0

-V6

-

3

V6

-

0

+00

+

y
Vay ham so dong bien tren cac khoang (-oo; -3), (3; +oo), nghich bien tren cac
khoang(-3;-V6),(V6;3).
Bai toan 8: Xet su bien thien cua ham so:
a) y = 4sinx - 3
b) y =

X

+ cos x.
9


Gidi

a) D = R . T a c6 y ' = 4cosx
Xet y ' >0 «

4cosx > 0 < » -

- + k 2 ; r < x < - + k 2 ; r , k e Z n e n
2
2

h a m so

d o n g bien tren cac k h o a n g (- ^ + k 2 ;T ; ^ + k 2 ;T ), k e Z .

X e t y ' < 0 <=> 4cosx < 0 < »
^

- +k2;r2

n g h i c h b i e n t r e n cac k h o a n g ( — + k 2 ; T ; —

—
2

+ k 2 ; r , k e Z n e n h a m s6

+ k 2 ; r ) , k e Z.

b) D = R . T a CO y ' = 1 - 2cosxsinx = 1 - sin2x
y' = 0 «

sin2x = 1 o

x

— + k7i, k e Z .
4

H a m so l i e n tuc tren m o i doan I — + kn; — + ( k +
4

l)7rj v a

4

y ' > 0 t r e n m o i k h o a n g (— + kn; — + ( k + l)7t) n e n d o n g b i e n t r e n m o i doan
4
4

[ - +k7r;
4

- +(k+
4

l)7i],k e Z.

V a y h a m so d o n g bien tren R .
Cach khac: lay a, b bat k y thuoc R v a a < b.
T r e n k h o a n g (a;b) t h i y ' > 0 v a y ' = 0 t a i h i j u han d i e m n e n h a m s6 f dong

b i e n , d o d o f ( a ) > f ( b ) . V a y theo d i n h n g h i a t h i h a m so f d o n g b i e n tren R .
B a i toan 9 : T i m k h o a n g d o n g b i e n , n g h i c h b i e n cua h a m so:
a) y = X - sinx t r e n [ 0 ; 2n]

b) y = x + 2cosx tren ( 0 ; n).
Gidi

a) y ' = 1 - cosx. T a c6 V x [ 0 ; 27t] =:> y ' > 0 v a y ' = 0 <=> x = 0 hoac x = 2n.
V i h a m so l i e n tuc t r e n doan [ 0 ; 2n] nen h a m so d o n g b i e n t r e n doan [ 0 ; 2::].
b) y ' === 1 - 2 sinx. T r e n k h o a n g ( 0 ; n).
y' > 0 <=> s i n x < - <r> — < x <
2
6

—
6

,
•
1
r>
7r , _ 571
7t
y' < 0 <=> s m x > - < » 0 < x < — hoac — < x < —.
2
6
6
6
V a y h a m so d o n g b i e n t r e n k h o a n g (—; — ) , n g h i c h biSn tren m o i k h o a n g
6

6

(0;5)va(^;H).
6

6

10


DANG TOAiy
2.

CHUfNG MII\iH DQfN DIEU

Neu f'(x) > 0 v&i moi x e (a; h) thi ham s6fdong hien tren (a; h)
Neu f '(x) > 0 vai moi x e (a; b) vd f '(x) = 0 chi tai mot so hitu hgn diem cua
(a; h) thi hdm so dong bien tren khodng (a; b).
Neuf'fx) < 0 vai moix e (a: H) thi hdm so nghich bien tren (a; b)
Neu f'(x) < 0 vai moi x e (a; b) vd f'(x) ^ 0 chi tai mot so hihi hgn diem cua
(a; b) thi hdm so nghich bien tren khodng (a: b).
Neu hidm sof dong bien tren (a; b) vd lien tuc tren nua khodng [a;b); (a;bj; doqn [a;bj
thi dong bien tren mm khodng [a;b); (a;b]; dogn [a;b] tuang icng.
Neu hdm so f nghich bien tren (a; b) vd lien tuc tren nua khodng [a;b); (a;bj; dogn
[a;b] thi nghich bien tren nua khodng [a;b); (a:hj: dogn [a;b] tuang ung.
Chiiy:

I) Ddu nhi thuc bgc nhdt: f(x) = ax -i- b, a ^0
- 00
-b/a

f(x)

trdi ddu a

2) Ddu tam thuc bgc hai: f(x)

0

+ 00

ciing ddu a

ax^ ^ bx + c, a ^0

Neu A< 0 thif(x) luon cung ddu vai a
Neu A

0 thif(x) luon cung ddu vai a, trie nghiem kep

Neu A> 0 thi ddu "trong trdi - ngodi cung "
-OO

f(x)

X/

cung ddu a 0 trdi ddu a

X2

+00

0 ciing ddu a

3) Gid su hdm so f xdc dinh tren khodng (a; h) vd Xo e (a; b). Hdm so f duac
goi Id lien tuc tgi diem Xg neu: lim f (x) = f(Xo). Hdm so khong lien tuc tai diem
Xo duac goi Id gidn dogn tgi diem XQ.
Hdm so f lien tuc tren mot khodng (a:b) neu no lien tuc tgi moi diem thuoc
khodng do.
Hdm so flien tuc tren nua khodng (a; bj neu no lien tuc tren khodng (a; b)
va Vim/(x)

=f(h).

Hdm sof lien tuc tren nua khodng fa; b) neu no lien tuc tren khodng (a; b)
vd lim f(x)

=f(a).
11


Ham so flien tuc tren dogn [a; b] neu no lien tuc tren khodng (a; b) vd
\\mf(x) =f(a), \xmf(x) =f(b).
x->a

X—>b

Bai toan 1: Chung minh rang cac ham so sau day dong bien tren R.
a) f(x) =

- 6x^ + 20x - 13

b) f(x) = 2x - cosx + Vs sinx.
Gidi

a) f'(x) = 3x^-12X + 20.
Vi A' = 36 - 20 < 0 nen f (x) > 0 voi moi x, do do ham so dong bien tren R.
b) y' = 2 + sinx - \ cosx = 2(1 + ^ sinx - ^

cosx).

= 2[1 + sin(x - y)] ^ 0. voi moi x.
Vay ham so dong bien tren R.
Bai toan 2: Chung minh cac ham so sau nghich bien tren R:
a) f(x) =

Vx'+l -

X

b) f(x) = cos2x - 2x + 5.
Gidi

a) T a c 6 f ' ( x ) = - , —

1.

Vx-+1
Vi V x ^ ^
> Vx^ = I X I ^ X , Vx nen f '(x) < 0, Vx do do ham so f nghich bien

tren R.
b) f (x) = -2(sin2x + 1) < 0 voi moi x.
r(x) = 0 « s i n 2 x - - l « 2 x = - - + 2 k K « x = - - + k 7 i , k € Z .
2
4
Ham f(x) lien tuc tren moi doan
4

+ krc; -— + (k + 1)TC] va f (X) < 0 tren moi
4

khoang (-— + k7i; — + (k+l)7i) nen ham so nghich bien tren moi doan
4
4
[ - - + k 7 T ; - ~ + ( k + l ) 7 r ] , k e Z.
4
4
Vay ham so nghich bien tren R.
Cach khac: Ta chung minh ham so f nghich bien tren R:
V x i , X2 e R, X i < X 2 => f ( X | ) > f ( X 2 ) .

That vay. lay hai so a, b sao cho a < xi < X2 < b.
Ta c6: f'(x) = -2(sin2x + 1) < 0 voi moi x e (a; b).
Vi f '(x) = 0 chi tai mot so huu han diem cua khoang (a; b) nen ham so f nghich
bien tren khoang (a; b) =^ dpcm.
12


Baai toan 3: Chung minh cac ham so sau don dieu tren R:
a ) y = -x^-2x2 + x - 3

b) y = - - x '

+2x'--x'

Gidi
a) D = R. Ta c6 y' = 4x^ - 4x + 1 = (2x - 1)^ ^ 0 vai moi x
y' = 0 <=> X = ^ . Vay ham s6 dong bi^n tren R.
b) D = R. Ta

CO

y' = -3x^ +8x^ -7x^ = (-3x^ + 8x - 7 ) x l

V i A' = 16 - 21 < 0 nen -3x^ + 8x - 7 < 0 v o i mgi x.
Do do y' < 0 vai moi x. y' = 0

x = 0.

Vay ham so nghjch bien tren R.
Bai toan 4: Chung minh ham so"
dong bien tren moi khoang xac dinh cua no.
x+2 ,
— x~ — 2x + 3
'
b)y =
nghich bien tren moi khoang xac dinh cua no.
X + 1
•
a)y =

x-2

Gidi
a ) D = R \6 y ' =

—

^

4
(x + 2)^ > 0 vai moi

xit-2.

Vay ham so dong bien tren moi khoang (-QO; -2) va (-2; +oo)
b ) D = R \.
-x'
Ta

CO

y' =

-2x-5
1 — < 0 vai moi x ^ -1 (vi A' = 1 - 5 < 0).
(x + 1)-

Vay ham so nghich bien tren moi khoang (-oo; -1) va (-1; +oo).
Bai toan 5: Chung minh ham so: y = f(x) = ^ x^ + 2x^ + 3x - 1

a) nghich bien tren doan [-3;-!].
b) dong bien tren cac nira khoang ( - x ; -3] va [-1; +oo).
Gidi
D = R. Ta c 6 f ' ( x ) = x^ + 4x + 3
f ' { x ) = 0 « x^ + 4x + 3 = 0 < » x BBT

X

-00

-f-

y'

-3

-1

0

0

-00

+
+00

^ -1

y

+00

-7/3'

•3 hoac X = - 1 .


a) Ta CO f "(x) < 0 tren khoang (-3;-l) nen f nghich bien tren khoang (-3;-l) va f
lien tuc tren doan [ - 3 ; - l ] nen f nghich bien tren doan [ - 3 ; - l ] .
b) Ta CO f '(x) > 0 tren cac khoang (-co; -3) va ( - 1 ; + 0 0 ) nen f dong bien tren
khoang ( - 0 0 ; -3) va ( - 1 ; + 0 0 ) va f lien tuc tren cac nira khoang ( - 0 0 ; -3J va [ - 1 ; +GO)
nen f dong bien tren cac nua khoang ( - 0 0 ; -3] va [ - 1 ; +co).
Bai toan 6: Chung minh ham so: y =
dong bien trong khoang ( - 1 ; 1) va
1 + x^
nghich bien trong cac khoang (-QO; -1) va ( 1 ; + 0 0 ) .
Giai
Tap xac dinh D = R.
l(l + x ' ) - 2 x . x _ 1 - x '

y' = 0<:i>x = ± l .

Ta

CO

y' < 0 «

1 - x^ > 0 <=> - 1 < X < 1.

y' < 0 < ^ 1 - x^ < 0
1 u do suy ra dpcm.

X < -1 hoac X > 1.

Bai toan 7: Chung minh ham so: y =

sin(x + b)

(a ^ h + kn; k e Z) don dieu

trong moi khoang xac dinh.
Gidi
Ham so gian doan tai cac diem x = -b + k7i (k 6 Z).
, _ sin(x + b)cos(x + a) -sin(x + a)cos(x + b)
sin"(x + b)
sin(b - a)

?t 0 (do a - b ?t kji)

sin"(x + b)
Vi y ^ 0 va y' lien tuc tai moi diem x -b + kTi, nen y' giiJ nguyen mot dau
trong moi khoang xac djnh, do do ham so don dieu trong moi khoang do.

DANG TOAN

BAi TOAl\ CO THAM SO

•

3.

Gia sif ham so cd duo ham tren khoang (a; b):
Dung dieu kien can
- Neu ham so f dong bien tren (a: b) thif'(x) ^0 vai moi x e (a; b)
- Neu ham so f nghich bien tren (a; b) thi f'(x) <0 vai moi x e (a; b).
14


Dung dieu kien dii
Neuf'(x) > 0 vai moi x e (a; b) thi ham sof dong hien tren (a; b)
Neuf '(x) > 0 vai moi x e (a; h) va f '(x) = 0 chi tgi mot sii huu hctn diem cua
(a: h) thi ham so dong bien tren khodng (a; b).
Neuf'fx) < 0 vai moi x e (a; h) thi ham so nghich biin tren (a; b)
Neuf '(x) <0 vai moi x e (a: b) va f '(x) = 0 chi tgi mot s6 him han diem cua
(a; b) thi ham so nghich bien tren khodng (a; b).

Bai toan 1: Tim cac gia tri cua thiam so a de ham so f(x)

+ 4x + 3

dong bien tren R.
Gidi
Tap xac dinh D = R.
f'(x)-=x^ + 2ax + 4, A' = a^-4
- Neu a' - 4 < 0 hay -2 < a < 2 thi f '(x) > 0 vai moi x e R nen ham so d6ng
bien tren R.
- Neu a = 2 thi f '(x) = (x + 2)^ > 0 vai moi x

-2 nen ham so dong bien tren R.

- Neu a = -2 thi ham so f '(x) = (x - 2)^ > 0 vai moi x
tren R.

2 nen ham so dong bien

- Neu a < -2 hoac a > 2 thi f'(x) = 0 c6 hai nghiem phan biet nen f' c6 doi dau: loai.
Vay ham so dong bien tren R khi va chi khi -2 < a < 2.
Bai toan 2: Tim cac gia tri cua tham so a de ham so f(x) = ax^ - 3x^ + 3x + 2 dong
bien tren R.
Gidi
Tap xac dinh D = R. Ta c6 f '(x) = 3ax^ - 6x + 3.
Xet a = 0 thi f (x) = -6x + 3 c6 d6i dSu: loai
Xet a 9i 0, vi f khong phai la ham hang (y' = 0 toi da 2 diem) nen dieu kien ham
so dong bien tren R la f '(x) > 0, Vx
a>0
o •

a >0
9 - 9a < 0

fa > 0
a>l

a ^ 1.

Bai toan 3: Tim cac gia tri cua m de ham so f(x) = mx - x nghich bien tren R:
Gidi
Tap xac djnh D = R.

Ta CO y' == m - 3x^
- Neu m < 0 thi y' < 0 vai moi x G R nen f nghich bien tren R
- Neu m = 0 thi y' = -3x^ < 0 vai moi x e R, dang thuc chi xay ra vai x = 0, nen
ham so nghich bien tren R.
15


- Neu m > 0 thi y' = 0 «
BBT

X

X=

±iy

-00

-

y'

X2

0

+

0

+00

-

y
Do do ham so dong bien tren khoang (xi, X2): loai
Vay ham so nghjch bien tren R khi va chi khi m < 0.
Bai toan 4: Tim cac gia tri cua m de ham so f(x) = sinx - mx + 4 nghich bien tren R.

Gidi

Tap xac dinh D = R.
Vi f(x) khong la ham hang voi moi m nen f(x) = sinx - m + 4 nghich bien tren R
<=> f'(x) = cosx - m < 0, Vx
m > cosx, Vx <::> m ^ 1.
Bai toan 5: Tim m de ham so y = x + 2 +

x-1

dong bien tren moi khoang xac dinh.

Gidi
Tap xac dinh D = R \ 1}.
Ta

CO

y' = 1 -

m

(x-1)-

, vai moi x

?t

1.

x^\

- NcLi m < 0 thi y' > 0 vai moi

Do do ham so dong bien tren m6i khoang (-00; 1) va (1; +oc).
- Neu m > 0 thi y'= x" - 2x + l - m
(x-1)^
y' = 0 « > x ^ - 2 x + l - m = 0<=>x = l ± V m
l-Vm

X

BBT

+

y'

1+

1

0

v'm

0

+00

+

y
Ham so nghich bien tren moi khoang (1 - v m ; 1) va (1; 1 + v m ): loai.
Vay ham so dong bien tren moi khoang xac dinh cua no khi va chi khi m < 0.
Bai toan 6: Tim a de ham so: f(x) = x - ax" + x + 7 nghich bien tren khoang (1; 2).

Gidi

Ta CO f'(x)

3x^ - 2ax + 1 la tam thuc bac hai.

16


Ham so nghich bien tren khoang ( 1 ; 2) khi va chi
khi y' < 0 voi moi x e ( 1 ; 2)
[/'(1)^0
/'(2)<0

<=>

4-2a<0

13
<=> a> —
[13-4a<0
4

<

Bai toan 7: Tim m de ham so y =
doan CO do dai bang 3.

+ 3x^ + mx + m chi nghich biSn tren mot

Gidi:
D = R, y' = 3x^ + 6x + m . A' = 9 - 3m
Xet A' < 0 thi y' > 0, V x : Ham luon d6ng bien (loai)
= 0 CO 2 nghiem X|, X2 nen x i + X2 = -2, X|X2 =

Xet A' > 0 <=> m < 3 thi
X

-00

X7

+

y'

y

0

-

+00

0

+

^

Theo de bai: X2 - x i = 3 »

m

^
(x2 - x i ) ' = 9 <=> x^ + x^ -2X|X2 = 9

<=> (X2 + xi)^ - 4xiX2 = 9 <=> 4 - ^ m = 9<=> m = - ~ (thoa)
Bai toan 8: Tuy theo tham so m , xet sir bien thien cua ham so:
y = - x^ - 2mx^ + 9x - m.
3

Gidi
D = R. Ta CO y' = x^ - 4mx + 9; A' = 4m^ - 9
— /i™2

- NIU A' < 0 <=> 4m^ < 9 <» | m | < ^ thi y' > 0, Vx nen ham s6 dong bien tren R.
- Neu A' > 0 « 4m^ ^ 9 <::> m

> - thi y' = 0 CO 2 nghiem phan biet
xi,2 = 2 m ± V 4 m ^ - 9 .

Lap bang bien thien thi ham dong bien tren khoang

(2m - V4m' - 9 ; 2m +V4m^ - 9 )
Va nghich bien tren moi khoang (-oo; 2m - V4m' - 9 ), (2m + V4m" - 9 ; +oo).
Bai toan 9: Xet sir bien thien cua ham so: y = ^ ^ ^ " ^ theo tham so m .
x-1

Gidi
D = R \.
iHij

viEN imn BiNH THUA;-.;!

17


„
, , -2-ni
Ta CO y =
(x-1)
- N6u m = -2 thi y = 2, Vx 7t 1 la ham so khong doi.
- Neu m > -2 thi y' < 0, Vx 9^1 nen ham so nghich bien tren moi khoang
(-o);!), ( ! ; + « ) ) .

- NIU m < -2 thi y' > 0, Vx ^ 1 nen ham so dong bien tren moi khoang

DANG TOAN
•

4.

UfniG DUNG VAO GIAI PHlTOfNG TRINH

Nku ham sSf dan dieu tren K va c6 M, N thuoc K thi phuang trinh
f(M)=f(N)
<^ M=N.
Neu ham so f dong bien tren K va cd M, N thuoc K thi bat phuang trinh
f(M) >f(N) ^ M>N.
Neu ham so f nghich bien tren K va cd M, N thuoc K thi bat phuang trinh
f(M)>f(N)
o MC/iiiy:
1) Ta CO the xet f(x) la ham so ve trdi, neu can thi bien doi, chon xet ham thuan
lai, dat anphu, .... Tinh dgo ham roixet tinh dan dieu.
Neu ham so f dan dieu tren K thi phmmg trinh f(x) = 0 co toi da 1 nghiem.
Neuf(a) = 0, a thuoc K thi x = a la nghiem duy nhdt ciia phuang trinh f(x)=0.
2) Neu f CO dgo ham cap 2 khong doi ddu thif ' la ham dan dieu nen phuang
trinh f '(x) = 0 cd toi da 1 nghiem do do phuang trinh f(x) = 0 cd toi da 2 nghiem.
Neu f(a) = 0 va f(b) =0 vai a ^ b thi phuang trinh f(x)=0 chi cd 2 nghiem la
x = a va X = b.
Bai toan 1: Giai phuong trinh: V3 - x + x^ - V2 + x - x^ = 1.
Gidi
Dat t = x^ - X thi phuang trinh tra thanh: V3 + t - V2-t = 1, -3 < t < 2.
Xet ham

s6 f(t) = V 3 +1 - V2 - 1 ,

Vai -3 < t < 2 thi f'(t) = —i=

+

-3 < t <
}

2.
> 0 nen f d6ng biSn tren (-3; 2).

2V3 +1 2V2 - 1
Ta CO f ( l ) = 2 - 1 = 1 nen phuang trinh:
f(t) = f ( l ) « t = 1 <» x^ - X - 1 = 0 « X =
18


Bai toan 2: Giai pliuang trinh ^j2x^ + 3x^ +6x + \6 = 2V3 + V 4 - x .
Gidi:
Dieu Icien xac dinh:

Ix"+3r+Gx+\6^0

f(x+2)(2r-x+8)>0

4-x>0

4-x^O

Phuong trinh tuong ducng^J2x^ + 3x^ +6x + \6 --^A-x

0-2
= 2Vs

Xet ham s6 / ( x ) = V2x' +3x- +6x + 16 - V4 - x -2 < x < 4
3(x' + X + 1)

Thi/'(x) =

> 0 nen f dong bien

+ -

V2x' + 3 x ' +6X + 16

2V4^

ma f(l)=2 V3 , do do phuong trinh tra thanh f(x) == f ( l )

x=l

Vay phuong trinh c6 nghiem day nhdt x = l .
Bai toan 3: Giai phuong trinh Vx - V l - x = 5 - 4x .
Giai
D i k kien: x ^ 0. PT

4x + V x - V T ^ = 5

Xet f(x) = 4x + Vx - VToc (x ^ 0) ^ f'(x) = 4 +

1
2V^

3^(1-x)^

Ma f '(x) > 0, Vx > 0 va f(x) lien tuc tren [0; +00)
Nen ham so f(x) dong bifin tren nua khoang [0; +00)
Khi X = 1 ^ f ( l ) = 5 nen X = 1 la nghiem PT
K h i x > 1 ^ f ( x ) > f ( l ) = 5:loai.
Khi 0 < X < 1 ^ f(x) < f ( l ) = 5: loai. Vay nghiem la x = 1.
Bai toan 4: Giai phuong trinh: 3x" -18x + 24 = —^2x-5

^—
x-1

Gidi
Dieu kien x^\;
(2x-5)'-

~, phuong trinh tra thanh:
'
=(x-l)^'
2x-5
x-1

1
Xet f(t) = t^ - j vai t > 0. Ta c6: f (t) = 2t + ^ > 0 nen f d6ng biSn tren (0; +00)

Phuong trinh: f( 12x - 5 | ) = f( | x - 1 | ) « 12x - 5 | = | x - 1
« 4x^ - 20x + 25 = x^ - 2x + 1 « 3x^ - 18x + 24 = 0.
<=>x'-6x + 8 = 0 < ^ x = 2 hoac x = 4 (chon).
19


Bai toan 5: Giai bk phuong trinh: 4 | 2x - 1 | (x^ - x + 1) > - 6x^ + 15x - 14.
Gidi
BPT: I 2x - 1 I .[(2x - 1)^ + 3] > (x - 2)^ + 3x - 6
«
I 2x - 1 P + 3 I 2x - 1 I > (x - 2)^ + 3(x - 2)
Xet ham s6 f(t) = t^ + 3t, D = R.
Ta CO f '(t) = 3t^ + 2 > 0 nen f ddng biSn tren R.
BPT: f( I 2x - 1 I ) > f(x - 2) » I 2x - 1 I > X - 2.
Xet x - 2 < 0 thi BPT nghiem dung.
Xet X - 2 > 0 thi 2x -1 > 0 nen BPT « 2x -1 > X - 2 <:i> X > -1: Dung
Vay tap nghiem la S = R.
Bai toan 6: Giai bk phuong trinh: V x + 1 + 2V'x + 6 < 20 - 3Vx + 13 .
Gidi
Diku Icien: x ^ - 1 . BPT v'lk lai: V x + T + 2Vx + 6 + 3Vx + 13 < 20
Xet f(x) la ham so ve trai, x ^ - 1 .
1
1
Ta c6: f (x) =

2Vxn

+ •V x + 6

2Vx + 13

> 0 nen f dong bien tren [-1; +oo).

Ta CO f(3) = 20 nen BPT:f(x) < f(3) « x < 3.
Vay tap nghiem cua BPT la S - [-1; 3].
[ x ' ( l + y-) + y ' ( l + x ' ) = 4 V ^
Bai toan 7: Giai he phuang trinh: <
x"yvl + y " - " v l + x^ = x ^ y - x
Gidi
Dieu kien: xy > 0
Phuang trinh thu hai cua he tuang duang vai x - V l + x^ = x^y(l - -^1 + y^ )
Ta CO X = 0 khong thoa man phuang trinh
Vi

X-

Vl

+ x ' < 0 va 1 - ^|l + y^ < 0 nen suy ra y > 0. Do do x > 0.

Phuang trinh tuang duang: — - —^il + - ^ = y - yJl + y^
X
X V
X
Xet ham f(t) = t -1 V l + t^ tren (0; +oo)
Xac6f'(t)= 1 - /

- V l + t^ < 0 , v a i m o i t e (0; +oo)

Vi + t '

Suy ra ham f nghich bien tren (0; +oo).
Phuang trinh f( —) = f(y)
X

— = y <» xy = 1.

X

20


Thay vao phuong trinh thu nhat ciia he, ta c6:
1+

1

+ ^ ( l + x ' ) = 4 « x ' +-V = 2

1)^ = 0 «
Ket hop dieu kien, ta c6 nghiem x = y = 1.
»

Bai toan 8: Giai he phuong trinh

(x^ -

x^ =

1 <»

X= +

1

V x ^ - ^ = 8-x'
(x-ir=y
Gidi

Dieu kien x > 1, y ^ 0.
He phuong trinh tuong duong vdi:

V7^-(x-l)'+x'-8 =0

(1)

y = (x-l)^

(2)

Xet ham s6 f(t) = V t ^ - (t - 1)^ + t^ - 8, voi t ^ 1.
Ta c6f'(t) = -2(t-l) + 3t^ +

J

= 3t^ - 2t + 2 +
2^/^^

> 0 voi moi t > 1 nen

2Vt^

f(t) dong bien tren (1; +GC).
Phuong trinh (1) c6 dang f(x) = f(2) nen (1) <=> x = 2, thay vao (2) ta dugc y = 1.
Vay nghiem cua phuong trinh la (x; y) = (2; 1).
- 3x- - 9x + 22 = / + 3y- - 9y
I

Bai toan 9: Giai he phuong trinh:

1
X

+y - x +y =
Gidi

x' - 3x' - 9x + 22 = y' + 3y- - 9y
He \
1
x'+y-x+>'=-

x' - 3x' - 9x + 22 = / + 3y- - 9y
•!

ix-^y-+{y

+

^f=l

1
1
Dat u = x — ; v = y + —
2
2
He da cho thanh

2

_^u
4

= {v + Yf --{v
2

+ \f - — ( v + 1)
4

Xethamf(t)=

r'-|/'-^/

rac6f'(t)=

- 3 / - — < 0 v 6 i m o i t t h 6 a | t k 1 nenf(u) = f(v + 1)
4

u=v+ 1
21

Do do

(v +

1)

+ V =

1

<::> V =

Vay he da cho c6 nghiem la

(3
2

0 hay

V

1

v=0
hay
w=1

= - 1 <=>-^

1

v= -l
w=0

l l

2' 2

2

x--12x+35<0
Bai toan 10: Giai he bat phuang trinh:

(1)

x' - 3 x ' + 9 x + - >0 (2)
3
Gidi

Tac6(l)<=>x-- 12x + 3 5 < 0 « 5 < x < 7
Xet (2): Dat f(x) - x^ - 3x^ + 9x + ^ , D = R
f (x) = 3x^ - 6x + 9 > 0, Vx G R nen f(x) d6ng biln: x > 5 =^ f(x) > 286/3.
Do do f(x) > 0, VxG (5; 7)
'
Vay tap nghiem ciia he bat phuang trinh la S =(5; 7).

DANG TOAN
5.
•

tfniG DUNG VAO SO NGHIEM PHlTtfNG TRINH

Neu hum so f dan dieu tren K thi phuang trinh f(x) = 0 cd toi da 1 nghiem.
Neu f CO dgo ham cap 2 khong doi ddu thif ' la ham dan dieu nen phuang trinh
f '(x) ^~ 0 CO toi da I nghiem do do phuang trinh f(x) = 0 c6 toi da 2 nghiem.
Tit BBT cho ta cdc gid tri ciia y, neu y nhdn gid tri tie dm sang duang hay
nguac lai tren mot mien thi y 0 c6 dimg I nghiem tren mien do.
Bai toan 1: Chung minh rang phuang trinh 3x^ + 15x - 8 = 0 c6 mot nghiem duy rMt.
Gidi
Ham f(x) = 3x^ + 15x - 8 la ham so lien tuc va c6 dao ham tren R .
Vi f(0) = -8 < 0, f ( l ) = 10 > 0 nen ton tai mot s6 Xo e (0; 1) sao cho
tuc la phuang trinh f(x) = 0 c6 nghiem.

f(Xo)

= 0,

Mat khac, ta c6 y' = ISx"* + 15 > 0, Vx e R nen ham so da cho luon luon dong
bien. Vay phuang trinh do chi c6 mot nghiem duy nhat.
Bai toan 2: Chung minh phuang trinh: x'^ - x^ + 3x'' - 3x^ + 1 = 0 c6 nghiem duy nhSt.
Gidi:
Dat f(x) = x'^ - x*" + 3x^ - 3x^ + 1. D = R
Xet x > 1 thi f(x) = x''(x' ~ 1) + 3x^(x^ - 1) + 1 > 0: v6 nghiem
2x3
Xet 0 < X < 1 thi f(x) = x'^ + (1 - x^"
> 0: v6 nghiem

22

Xet

< 0 thi: f'(x) = 13x'^ - 6x^ + 12x^ - 6x
= 13x'^ - 6x(x - 1)^ > 0 nen f ddng bien
Bang bien thien:
0
X
—00
+
y'
X

y
-00

Nen f(x) = 0 CO nghiem duy nhat x < 0.
Vay phuong trinh da cho c6 nghiem duy nhat.
Bai toan 3: Chung minh rSng phuong trinh 2x^ Vx - 2 = 11 c6 mot nghiem duy nhat.

Gidi
Xet ham so f(x) = 2x^ Vx - 2 thi ham so xac dinh va lien tuc tren nua khoang
[2;+«)).
1

\

x(5x-8)

f'(x)= 2 2 x V x ^ + ^ = —

V
2Vx-2

> 0, voi moi x

x-2

e

(2; +oo)

Do do ham so dong bien tren nua khoang [2; +oo).
Ham s6 lien tuc tren doan [2; 3], f(2) = 0, f(3) = 18. V i 0 < 11 < 18 nen theo
dinh l i ve gia tri trung gian cua ham so lien tuc, ton tai so thuc c e (2; 3) sao cho
f(c) = 11 tuc c la mot nghiem cua phuong trinh f.
Vi ham s6 d6ng bidn tren [2; +co) nen c la nghiem duy nhat cua phucmg trinh.
Bai toan 4: Tim so nghiem cua phuong trinh x^ - 3x^ - 9x - 4 = 0.

Gidi

Xet ham s6 y = x"* - 3x^ - 9x - 4, D = R.
y' = 3x^ - 6x - 9, y' = 0 « x = -\c
BBT

x

3

-1

-00

+

y'

0

+00

+

0

+00

1

y
-00

'

~^-31 "

Dura vao BBT thi phuong trinh y = 0 c6 diing 3 nghiem.
Bai toan 5: Chung minh he

x~ +

~1
CO dung 3 nghiem phan biet.

_y- + x' = 1

Gidi
Trir 2 phuong trinh ve theo ve va thay the ta dugc:
23


x2(l - x ) - y ' ( l - y ) = 0 = > ( l -y^)(l - x ) - ( l -x^)(l - y ) = 0
=^ (1 - x)(l - y ) [ l + y + y^ - (1

+

X

+ x^)J = 0.

^ ( 1 - x ) ( l - y ) ( y - x ) ( l + x + y) = 0.
Xet X = 1 thi he c6 nghiem ( 1 ; 0)
Xet y = 1 thi he c6 nghiem (0; 1)
Xetx ^- y thi
Dat f( x) = X -

+

-h y^ = 1
X

+ X^ -

1 = 0.

^- 1,D = R . Ta c 6 f ( l ) - 1 ^ 0 .

f ( x ) = 3x^ + 2x, f'(x) = 0 « x = - | hoac X =
BBT

X

-00

-2/3
+

y'

+00

0
0

-

0

+
+00

-23/27

y
-00-^

Do do f(x) = 0 CO 1 nghiem duy nhat XQ > 0, Xo?^ 1 nen he c6 nghiem (XQ; y o ) .
Xet 1

+ X

+ y = 0 =^ y =

-X -

1 nen y^ + x-^ = 1 o x^ + x^ + 2x = 0

<=> x(x^ + X + 2) = 0 <::> X = 0. Do do he CO nghiem (0; 1)
Vay he c6 dung 3 nghiem phan biet.
Tim m de phuong trinh sau c6 2 nghiem phan biet:

B a i toan 6:

Vx^ + m x + 2 = 2x + 1 .
Giai
2x + l > 0

PT<=>

,
<:^3x + 4x - 1 = mx, X

x ' + m x + 2 = (2x + l ) '

1
> - -

2

3x" + 4 x - l
1
= m, x ^ - —
2

V i X = 0 khong thoa man nen:

X

3X-+4X-1

^

1

^«u
3x^+1

Xet f(x) =
, x ^ — , X 7t 0 thi f (x) =
5—

X
2
x"
Lap BBT thi dieu kien phuong trinh cho c6 2 nghiem phan biet la
1
9
f(x) = m CO 2 nghiem phan biet x > - - . x ? ^ 0 < » m > B a i toan 7: Tim m de phuong trinh c6 nghiem
m(Vl + x '

- Vl-x-

+2)=

2 V 1 - X ' + V l + x-

- Vl-x'

.

Giai
DiSu kien -1 < x < 1 .Dat x = V l + x" - V l - x '

thi t ^ 0

24


va t- = 2 - 2 VT^x^ < 2, d4u = khi x^ = 1. Do do 0 < t < V2
PT: m(t + 2) = 2 - + t « m = ~^ "^^"^^
t+2

^ ^ ^ ^ , 0 < t < V2 , f '(I) = ——^ < 0 nen f nghich b i k tren
t+2
(t + 2)^ •

Xet f(t) =

[0;V2].
Dieu kien c6 nghiem:
min f(t) < m < max f(t)

DANG TOAN

f( V2 ) < m < f(0) « V2 - 1 < m < 1.

UfNG DUNG VAO BAT DANG THUfC

6.

Nei4y =f(x) xdc dinh tren Kc6y'>
bat dang thicc:

x>a

=>f(x) >f(a); x
Neuy f(x) xdc dinh tren K cdy'<
cd hat dang thitc:
x>a

0, Vx e K thif(x) dong hiin tren K nen cd

0, Vx e K thi f(x) nghich hien tren K nen

^f(x)
Chuy:
1) Co the f '(x) = 0 chi tgi mot so hint han diem ciia K.
2) Vice xet ddi4 y' doi khi phdi can den y
y
hogc xet ddu bo phdn ciia
ham so, chdng han tic so ciia mot phdn so cd man duang,....
Neu y" > 0 thi y 'dong hien tic do ta cd ddnh gid f '(x)

r6if(x),...

3) Ham so f xdc dinh tren K la mot khodng, dogn hogc nica khodng.
- f dong hien tren K neu vdi moi X/, x: e K: x/ < x? =^f(x\) < f(x2)
- f nghich hien tren K neu vdi moi x/, x^ e K: x/ < X2 =^f(xi) >

/(xo).

Bai toan 1: Chung minh cac bat dang thuc sau:
a) sinx < x vai moi x > 0, sinx > x vai moi x < 0.
-I

b) cosx > 1

vai moi x^Q.

2

Gidi
a) Vai X > — thi x > 1 nen sinx ^ 1 < x.

2

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