Tải bản đầy đủ

10 trọng điểm bồi dưỡng học sinh giỏi môn toán lớp 12 phần 1

"'"1

510.76
M558T

Nha giao mi tu - Th.s LE H O A N H P H O

lOtrpngdiem

• Danh cho hoc sinh I6p 12 chtrong trinh chuan va nang cao
fi On tap va nang cao kT nang lam bal
••. Bien soan theo no! d u n g va cau true de thi cua Bo GD&.DT

XUAT BAN DAI HOC Q U 6 C GIA H A

NOI


to trong diem

MON TOAN

LdpU

• Danh cho hoc sinh Idp 12 churdng tnnh chuan va nang cao
• On tap va nang cao kT nang lam bai
• Bien soan theo noi d u n g va cau true de thi cua Bo GD&.DT

OEK]
Ha N » i

NHA XUAT BAN OAI HOC Q U 6 C GIA HA N6I


C/y TNHH MTV DWH Hhang Vi$t

Chuyen ad 1: TiNH D O N D I ^ U Vn CITC TBI
1. K I E N T H U C T R Q N G T A M

®Aw

^#

A .

Djnh li Lagrange: Cho f Id mOt hdm li^n tgc tr§n [a, b], c6 dgo ham tr§n
(a,b). Luc do ton tgi c e (a,b) d l :
^
f(b)-f(a)
= f '(c) hay m - /(a) = (b - a)f (c)
b-a

mhm

muc dich gii'ip cdc ban hoc sinh

them dcndng

l&p 10, lap 11, lap 12 co tu lieu doc

y


cao trinh do, cdc ban hoc sinh gidi tu hoc bo sung them kien thuc ky

M

nnng, cdc ban hoc sinh chuyen Todn tu nghien cim them cdc chuyen de, nhd sdch
KHANG

VIET hap tdc bien soqn bo sdch BOl DlXONG

DlTONG

CHUYEN

TRONG

DIEM

TOAN

LOP 10

-

TRONG

DIEM

TOAN

LOP 11

-

TRONG

DIEM

TOAN

LOP 12

DIEM

TOAN

O

LOP 12 nay co 21 chuyen devai

ngi dung la

tom tdt kien thuc twng tarn cua Todn pho thong vd Todn chuyen, phan cdc bdi
todn chgn Igc co khodng 900 bdi voi nhieu dang hai vd muc do tic co ban den phuc
tap, bdi tap tu luycn khodng 250 bdi, c6 huang dan hay ddp so.
Cudi sdch CO 3 chuyen
NGHIEM

NGUYEN

Dii dd cogdng

(C)y=f(x)

BOI

TOAN gom 3 cudn:

-

Cudn TRONG

HOC S/NH GlOl,

vd TOAN

de ndng

cao: DA

THU'C,

PHlTONG

TRINH

SUYLUAN.

kicm tra twng qud trinh bien tap song cung khong trdnh khoi

nhieng khiem khuye't sai sot, mong don nhan cdc gop y cua quy ban dgc de'lan in
sau hodn thien han.
Tdc gid

rMV'i'X

LEHOANHPHO

a c

O

Djnh ly Rolle: Cho fla mot ham li§n tuc tren [a,b], c6 (J?o ham tr§n (a, b) va
/•(a) = f(b). Luc do t6n tai c e (a,b) d4 f'(c) = 0.
Djnh ly Cauchy: Cho f va g Id hai ham lien tyc tren [a, b], c6 dgo ham tren
(a,b) va g'(x) 0 tai moi x € (a,b).

Lucd6t6ntaiC€(a,b)d|ii^^(^ = lM
g(b)-g(a)

g'(c)

Tinhdcndi^u
,j
Gia su' ham s6 f c6 dao ham tren khoang (a; b) khi do:
i :
N4U f d6ng bi§n tren (a; b) thi f '(x) > 0 voi mpi x e (a; b).
N^u f nghjch bi§n tren (a; b) thi f '(x) < 0 voi mpi x e (a; b).
N6U f '(x) > 0 voi mpi x € (a; b) vd f '(x) = 0 chi tgi mpt so hu-u hgn di6m cua
(a; b) thi ham s6 d6ng bi4n tren khoang (a; b).
N§u f '(x) < 0 voi moi x e (a; b) va f '(x) = 0 chi tai mpt so hOu hgn diem cua
(a; b) thi ham s6 nghich bi§n tren khoang (a; b).
N§u f d6ng bi4n tren khoang (a; b) va lien tyc tren [a,b) thi dong bi§n tren [a,b); va
lien tyc tren (a,b] thi dong bi§n tren (a,b]; lien tyc tren [a,b] thi d6ng bi§n tren [a,b].
Nlu f nghich bidn tren (a; b) va lien tyc tren [a,b) thi nghich bien tren [a,b); lien tuc
tren (a,b] thi nghich bi§n tren (a,b]; lien tyc tren [a,b] thi nghich bi^n tren [a,b].
N§u f '(x) = 0 voi mpi X e D thi ham so f khong d6i tren D.
Cyc trj cua ham s6

gficr,

:

Cho ham s6 f xac dinh tren tap hgp D vd XQ e D.

Xo du-oc gpi Id mpt di4m eye dai cua f neu ton tai mpt khoang (a; b) chii-a
dilm Xo sao cho (a; b) c D vd f(x) < f(Xo), V x e (a; b) \.


W tTQng diSm bSl dUCing h
. • .

vio

Xo dugc goi 1^ mpt d i i m cue tieu cua f n^u t6n tgi mpt khoang (a; b) chiJa
d i l m Xo sao cho (a; b) c : D
f(x) > f(Xo), V x e (a; b) \.

cf^ iNi-iMmrvL/wfinnang

(
-

-

B6 d § Fermat: Gia si> ham s6 c6 dao ham tren (a;b). N§u f dat ci/c trj tgi
d i l m Xo € (a;b) thi f '(XQ) = 0.
I
Cho y = f(x) lien tuc tren khoang (a;b) chii-a Xo, c6 dao ham tren c6c khoang
(a;;'o) va (xo;b):
N I U f '(x) doi diiu tu" am sang du-ong thi f dat eye ti§u tai XQ
N I U f (x) d6i d^u ti> d u K a n g sang am thi f dat CLfC dai tai Xo .
Cho y = f(x) CO dao ham eSp hai tren khoang (a;b) chua XQ;
Neu f '(xo) = 0 va f "(Xo) > 0 thi f dat cue t i l u tai XQ
N§u f '(xo) = 0 va f "(xo) < 0 thi f dat eye dai tai

XQ

IJng dung vac phiKcng trinh
-

-

-

N I U ham s6 f don dieu tren K thi phuang trinh f(x) = 0 eo t6i da 1 nghiem.
N I U f(a) = 0, a thuoc K thi x = a la nghiem duy nh^t cua phuang trinh
f(x)=0.
N§u f CO dao ham c^p 2 khong d6i dSu tren K thi f ' la ham dan di$u nen
phuang trinh f(x) =0 c6 t6i da 2 nghiem tren K. N§u f(a) = 0 va f(b) =0 vai a
b thi phuang trinh f(x)=0 chi c6 2 nghiem la x = a va x = b .
N l u f la mot ham lien tuc tren [a, b], c6 dao ham tren (a,b) thi phuang trinh
f ' ( x ) M j l [ i ? l c 6 it n h i t mot nghiem c e (a,b).
b a
Oac biet, neu /(a) = f(b) = 0 thi phuang trinh f'(x) = 0 eo it nhlit mot nghiem
c e (a, b) hay gi&a hai nghiem cua f thi c6 it nhSt mpt nghiem cua dao ham f'.

Chu y:
1) Tung dp eye trj y = f(x) tai x = X o :
Ham da thue: y = q(x). y' + r(x) => yo = r(Xo)
L j - u - »•
s u(x)
u ( X n ) u'(x.)
Ham huu ti: y = f(x) =
= > Vo = — ^ =
°
v(x)

v(x,)

v'(Xo)

Dae bi$t; Vai ham y = f(x) bae 3 c6 CO, CT va neu y = q(x). y' + r(x) thi
phuang trinh duang thing qua CD, CT la y = r(x).
2) S 6 nghiem cua phuang trinh bae 3: ax^ + bx^ + ex + d= 0, a 0.
N§u f '(X) > 0, Vx hay f '(x) < 0, Vx thi f(x) = 0 chi eo 1 nghiem.
N§u f "(x) = 0 c6 2 nghiem phan biet va:

.

, ^

'^m^-

2 . CAC BAI TOAW
o

vi^c

Vai yco ycr > 0 : phuang trinh f(x) = 0 chf c6 1 nghiem
Vai yco ycr = 0 : phuang trinh f(x) = 0 c6 2 nghiem (1 dan, 1 k6p)
Vai yco VCT < 0 : phuang trinh f(x) = 0 c6 3 nghiem phan biet

Bai toan 1 . 1 : Chung minh eSc h^m s6 sau Id hdm kh6ng d6i
a) f(x) = cos^x + cos^(x + - ) - cosxcos(x + - )
3
3
b) f(x) = 2 - sin^x - sin^(a + x) - 2cosa.cosx.cos(a + x).

<^'"'
>

Hu'd'ng din giai
a) f '(x) = -2cosxsinx - 2cos(x + - )sin(x + - )
3
3
+ sinxcos(x + - ) + cosx.sin(x + - )
3
3
= -sin2x - sin(2x + — ) + sin(2x + - )
3
3
= -sin2x - 2cos(2x + - ).sin - .
2
6
= -sin2x - cos(2x + ^ ) = 0, vai mpi x.
Do do f h i n g tren R nen f(x) = f(0) = "I

^ " ;^ = | -

b) Dao ham theo biSn x (a la hiing so).
i
f '(x) = -2sinxeosx - 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^a - 2cos^a = sin^a.
Bai toan 1 . 2: Cho 2 da thue P(x) va Q(x) thoa man: P'(x) = Q'(x) vdi mpi x vd
P(0) = Q(0). Chung minh: P(x) = Q(x).
Hu'O'ng din giai
f n.i
Xet ham s6 f(x) = P(x) - Q(x), D = R.
Ta C P f "(x) = P'(x) - Q'(x) = 0 theo gia thilt, do do f(x) la hdm h l n g nen
f(x) = f(0) = P(0) - Q(0) = 0 vai mpi x.
^ f(x) ^ 0 :^ P(x) = Q(x).
Bai toan 1 . 3: Chung minh ring:
a) arcsinx + areeosx = - , I x | < 1
2
2x
b) 2 aretan x + arcsin
= -n, x < - 1 .
l + x^

,


W tr
Cty TNHHMWOWH Hhang Vi$t

Hhi>

Hu^ng din gidi
a) Ham s6 y = f(x) = 2x^ + x - 4 lien tyc tren [-1,2] v^ c6 dgo ham f '(x) = 4x +1,
theo dinh ly Lagrange thi t6n tai s6 c e [-1;2] sao cho:

Himng din giai
a) N 4 U X = 1, X = -1 thi dung.
N^u - 1 < X < 1 thi x6t h^m so f(x) = arcsinx + arccosx
^

f'(x) = -

^

+-

^

=0 ^

b) Vai X < - 1, x6t f(x) = 2 arctan x + arcsin
2 • 2x2
Tac6f(x) = ^

2x

1+

+- | l i ^ = ^

1 + x2

1-x%2

l ( 2 M t l ) . f . ( c ) « ^ = 4c.1c>4c = 2 « c = I.
2-(-i)
' '
3
2

f(x) = C = f ( l ) = ^ .

b) Ham s6 y = f(x) = arcsinx lien tyc tren [ 0;1] va c6 dgo ham f'(x) =
N/I^'

X

theo dinh ly Lagrange thi ton tgi s6 c e [0;1] sao cho:

1+x

1-0

+^

V S + x^^
Suyraf(x) = C = f ( - 1 ) = - J + ^
2
4

1

?- =0 (vix<-1).

4_
< » 1 - c 2 = 4 r « c 2 = 1 - ^.2, C h p n c = j l - — .

=-f
4

Bai toan 1. 4: Tinh gpn arctanx + a r c t a n - vai x

1

Bai toan 1. 6: Xet chilu bi§n thien cua ham s6:
o.

a) y = x^ - 2x' - 5

X

1

b)y =

(x-4)

Hifang din giai
HiPO'ng din giai
a) D = R. Ta c6 y' = 4x^ - 4x = 4x(x^ - 1)

Xet f(x) = arctanX + a r c t a n - . D = ( - « ; 0 ) u ( 0 ; + * )
X

Max X G (0;+ x ) thi f lien tyc va c6 dgo ham

,,

-1
f'(x) = — ^ + — ^ =—^
l + x^ l + x^
1 + x^

Cho y' = 0 » 4x(x2 - 1) = 0 o X = 0 hoSc x = ± 1 .
BBT
X
-X
-1
0
1
+x

^ - O n e n f h a n g t r e n ( 0 ; + * ).
Ux^

x^
Do<36f(x) = f(1)= ^ + ^ = f
Vai X G ( - X ;0 ) thi f lidn tyc va c6 dgo ham f'(x) = 0 nen f hing tren (-oc
Dod6f(x) = f(-1) = - ^

V$y arctanX + arctan—=

- - khi x < 0
2

y'

-

0 +

y

^

^

-

0

^

+

^

Vay ham s6 nghjch bi§n tren moi khoang (-oo; - 1 ) va (0; 1), d6ng bi4n tr§n
moi khoang ( - 1 ; 0) va (1; + x ) .
b) D = R \. Ta

CO

y' =

-2

• •..'•I

(x-4)^

4o'.

/ < 0 tren khoang (4; + x ) nen y nghich bien tren khoang (4; +x).
y' > 0 tren khoang (-x; 4) nen y dong bien tren khoang (-x; 4).
Bai toan 1. 7: Tim khoang dan dieu cua ham s6

X

-khi x>0
12
Bai toan 1. 5: Tim s6 c trong dinh ly Lagrange:
a) y = f(x) = 2x^ + X - 4 tren [ - 1 , 2 ]
b) y - f ( x ) = arcsinx tren [ 0;1] .

0

b)y = x + 1

a)y =

.1

5?.

'

N/X2-6

Hu-^ng din giai

aj Tap xac djnh D = ( - x ; - N/G ) u ( N/G ; + x ) .

V l ^ '
•••• tx)

i -K, :


W trqng

diSm bSi dUOng

Tac6:y^
BBT:

y ^ Q « x = ±3.

2x^(x^-9)
(x^ - 6 ) V x ' - 6
x

Cty TNHHMTVDWH

h
-V6

- 0 0 - 3

y'

+

0

76

-

3
_

0

+00

+

y
Vay ham so ddng bi§n tren cac khoang ( ^ ; - 3 ) , (3; + « ) , nghjch bi§n tren
cac khoang (-3; - Ve), ( N/6 ; 3).
>0, V x < 1 .

b ) D = (-oo; 1). T a c 6 y ' =
2V(1-x)3

V$y ham s6 d6ng b i l n tren khoang ( - « : 1).
Bai toan 1. 8: Xet si/ bi§n thien cua ham s6:
a) y = x + cos^x

b) y = x - sinx tren [0; 2n\.
Hirang din giai

a) D = R. Ta CO y' = 1 - 2cosxsinx = 1 - sin2x
y' = 0 o sin2x = 1 <=> x = - + k7t, k e Z.
4

_ sin(x + b)cos(x + a) - sin(x + a)cos(x + b) ^

sin(b-a)

sin''(x + b)

sin^(x + b)

Vi y' lien tuc tai moi d i l m x ?t - b + kn,
a - b ^ k;t nen y' giu- nguyen mpt
d i u trong moi khoang xac dinh => dpcm.
Bai toan 1.10: Tim cac gia trj cua tham s6 d l ham so:
a) y = (m - 3)x - (2m + 1)cosx nghjch bi§n tren R.
b) y = x^ + 3x^ + mx + m chi nghjch biln tren mpt dogn c6 dp dai bing 3.
Hifo-ng din giai
a) y' = m - 3 + (2m - 1)sinx
Ham s6 y khong la ham h^ng nen y nghich bi§n tren R:
y'< 0 , Vx « m - 3 + (2m - 1)sinx < 0, Vx
Dat t = sinx, - 1 < t < 1 thi m - 3 + (2m - 1)sinx = m - 3 + (2m - 1)t = f(t)
Di4u kien tu-cng du'ang: f(t) < 0, Vt e
f(-1)<0

[-

+ kTi; -

-m - 4 < 0

<=> sf(1) < 0

3m - 2 < 0

o

,

1]

2

- 4 < m < —.
3

b) D = R, y' = 3x^ + 6x + m, A' = 9 - 3m
Xet A' < 0 thi y' > 0, Vx : Ham luon d6ng bi§n (logi)
Xet A' > 0 « m < 0 thi y' = 0 c6 2 nghiem x,, X2 nen x, + X2 = - 2 , XiX2 = m
BBT:

X

H^m so lien tyc tren moi doan [ - + kn; - + (k + 1)7i]
4
4
y' > 0 tren moi khoang ( - + kn; - + (k + 1)7t) n6n d6ng bi^n tren moi dogn
4
4

X2

Xi

y'

0

+

-

0

+00

+

^

y

^

Theo d l bai: X2 - x, = 3 » (X2 - x,)^ = 9 <=> x^ + x^ - 2x^X3 = 9

+ (k + 1)7r], k e Z.

4
4
Vay ham so dong bi4n tren R.
b) y' = 1 - cosx. Ta c6 Vx [0; 27i] => y' > 0 va y' = 0 o x = 0 hoac x = 2n.
Vi ham s6 lien tyc tren doan [0; 2n] nen ham s6 d6ng bi§n tren doan [0; 27:].
Bai toan 1. 9: Chieng minh cac ham s6
a) y = cos2x - 2x + 5 nghich bi4n tren R.
b) y = ^ ' " ^ ^ ^ ^ ^ (a * b + kTt; k e Z) dan di?u tren moi khoang x^c dinh.
sin(x + b)
Hirang din giai
a) Vxi, X2 e R, Xi < X2 LSy hai s6 a, b sao cho a < Xi < X2 < b.
Ta c6: f '(x) = -2(sin2x + 1) < 0 vai mQi x e (a; b).
Vi f '(x) = 0 chi tai mOt s6 hO-u han d i l m cua khoang (a; b) nen hSm so f
nghich b i § n tren khoang (a; b) => dpcm.
b) Dieu kien x ^ - b + kn (k € Z).

Hhang Vi$,

ci.(x2 + Xi)^ -4x1X2 = 9 4 - - m = 9c:> m = - —(thoa)
3
4
Bai toan 1.11: Tim ci/c trj cua cac ham so sau:
a)y = (x + 2 ) ^ ( x - 3 ) ^

b) y = | x | ( x + 2).
Hu'O'ng din giai

a) y' = 2(x + 2)(x - 3)^ + 3(x + 2f

[x-Zf

= 5x(x + 2)(x - 3)^

Ta CO y' = 0 o X = - 2 hogc x = 0 ho0c x = 3
X -X
BBT
-2
0
3
y'

+

0

-

0

+

0 +

y

Vay d i l m eye dai (-2; 0) va eye tieu (0; -108).
b) Ham s6 y = f(x) lien tyc tren R.Ta c6:


W trgng diS'm bSi duOng h
s_

2)
2)

Vai X
Vai X
BBT

<
>

0, f
0, f

'(X) =
'(X) =

X

ic Hodnh

TNHHMTVDWHHhong Vi
Phd

Bai toan 1.13: Tim cue trj ciia h^m s6
a) y = x - sin2x + 2
b) y = 3 - 2cosx - cos2x.
Hirang din giai

khi x < 0
khi X >0

|-x(x +
|x(x +

r

-2x + 2; f ' ( x )
2x + 2 > 0.

=



-1

-X

+

y'

im;

-1.

x =

,„f

a) D = R, y' = 1 - 2cos2x

+

Ta CO y " ( - - ^ + k?:) = 4 s i n ( - - ) = -2 N/S < 0 nen ham s6 dgt ci^c dgi t^i didm
6
3

1
^

0/

x = - -

6

X +

1

ux

+k7t+—

2

+2.

t

= - + kn, k 6 Z; y c T = - + k7t - — + 2.
6
6
2
b) y' = 2sinx + 2sin2x = 2sinx(1 + 2cosx):
X

sinx = 0

, _ x^ + 8 - 2x(x + 1) _ -x^ - 2x + 8
a) D = R. Ta c6 y' =
(x^ +8)^
(x2+8)2
y' = 0 <=> X = -4 hoac x = 2
X
BBT
-X
-4
0

y
y

2
+

1/4

0

^

Ta c6y"(± — + 2k7r) = 2 cos — + 4 c o s — - 6 c o s — = - 3 < 0 nen ham s6
3
3
3
3

0

Ham s6 dat CD tai x = 2; yco = ^ , dgt CTtgi x = -4; y c T =
b) Tap x^c dinh D = ( — — N/6 ) u (



; +oc)
2x^(x^-9)

x^-6
X = 0 hoac x = ±3.

y'
y

-V6

-3

-X

+

0
-9V3

3

V6

-

-



+x

dat eye dai tai diem: X = ± — + 2k7i, k G Z,
3

yco = - •
2

Bai toan 1. 14: Chu-ng minh ham s6

3x^7x^-6-^
7x^-6 _3x^(x^-6)-x''

X

2

1 o X = k7i hoacx = ± — + 2kTt, k G Z.
'•'^^ '
cosx = -—
3
2
y" = 2cosx + 4cos2x
Ta CO y"(kn) = 2cosk7i + 4cos2k7t = 2cosk7i + 4 > 0, vai mpi k e Z, nen ham
s6 da cho dat eye ti§u tai cac dilm x = kn, ycT = 2 - 2cosk7t bSng 0 khi k
chin va bing 4 khi k le.

-

0

' ^'^'^

y' = 0 ct>

+x

-1/8 "

y' = 0
BBT

6

Ta CO y " ( - + k7r) = 4 s i n - = 2 v's > 0 nen ham so dat cue tilu tai cac diem:
6
3

Hipang din giai

y' =

+k7r,keZ,ycD = - -

S

Vaydi^m CD(-1; 1), CT(0; 0).
Bai toan 1.12: Tim cu'c trj cua ham s6
a)y =

.

y' = 0 o cos2x = - « X = ± - + kn, k e Z; y " = 4sin2x.

0

0

y

: j,

0

+x

+


Ham s6 dgt C D tgi x = -3; yco = -9 Vs . dgt C T tgi x = 3;ycT = 9 N/S

2x
khi X < 0
a) ^(x) = 1
X
khong c6 dao ham tai x = 0 nhung dgt cue tri tai
s i n - khix>0
.
2
dilm do.
~ '
b) y = f(x) =(x - a)(x - b)(x - c), a ;i e luon eo eu-c dai va eye tilu, ^ '
'
Hipang din giai
,
a) Ham s6 f xac djnh va lien tyc tren R. Ta CO
- ;i
-2x
khi x < 0
f '(X) = 1
X
nen lim f'(x) = -2;^ lim f'(x) = - , do do f
—C O S khix>0
x-,o~
2
[2
2
khong c6 dao ham tgi x = 0 va BBT tren khoang ( - K ; TI).


10 trgng diSm bSi dudng h
,•

Kinh

mHHMTVDWH

Phd

-sin^ X + 2asinx - 1

X
+

y'

y" =

-

acos^ X
Vdi sinx = a thi y" =

y

ham s6 dat cu-c dai tai X = 0
ycD = y(0) = 0.
b) D = R. y' = (x-b)(x-c) + (x-a)(x-c) + (x-a)(x-b)
= 3x^ - 2(a + b + c) + ab + be + ca.
A' = (a + b + c)^ - 3(ab + be + ea) = a^ + b^ + e^ - ab - be - ea
V$y

= ^ [(a-b)^ + (b-e)^ + (e-a)^] > 0 vai a

:

i

,

« sinx = a CO 3 nghiem thuoc khoang (0; ^ ) W ^ i ^ }<=> 0 < a < ^ .
Bai t o a n 1. 16: Tim m d4 ham s6:

e.

a) y =
b) y =

dat eye dgi tgi dilm (-2; -2).
x+1
971
a sin X - cos X - 1
dgt eye tri tgi 3 diem thupe (0; — ) .
b) f(x) =
aeosx
4
Hipang din giai

-i-Vq

y'

+

0

-1
-

Ta CO xi + X2 = 2, X1X2 =

^2

f(-2) = -2

[p-1

3
x

nen yco- yci < 0.

c> 4m^XiX2 + 2m(2 - 4m)(xi + X2) + (2 - 4m)^ < 0.
c:. -12m + 2m(2 - 4m) + (2 - 4m)^ < 0 c=> 4 - 20m < 0 c=> m

- i + Vq
0

+

Ham s6 dat eye dgi tgi di4m (2; -2) khi va ehi khi

J7q = i

^ ChLcng minh du-ong thSng

m
« (2mxi + 2 - 4m)(2mx2 + 2 - 4m) < 0

-Q c6 hai nghiem phSn

y

-i-Vq = -2

^

06 thi CO 2 eye tri ct- m 0,. V > 0, g(1) 0 0 m < -3 hoac m > 0.

va X2 = - 1 + x/q .

-X

"^'^"^ ^'^'^

. , mx 2mx - 3
,
2
Ta CO y =

, dat g(x) = mx - 2mx - 3.
(x-1)

(x +1)2

X

^'^

Hu'O'ng d i n giai

- 1 : loai.

NIU q > 0 thi phyang trinh: f '(x) =

T"^

^ . . ^ . .. ^ .
. •, .•
9'^ '^'''''^
^^i^-

a) Oi4u kien: x / I .

, vai mpi x ? t - 1 .

x+1
N§u q < 0 thi f '(x) > 0 vai mpi x

mx^ t (2 - 4m)x ^4m + 1
^
— ^ ' ^2

AB song song vai duang thing 2x - y - 10 = 0.

a) f(x) = X + p +

a) Ta c6 f '(x) = 1 -

^
* 0, do do ham s6 dgt eye trj t^i 3 dilm
sinx cos X

thuoc khoang (0; — )
4

^

Do do y' = 0 CO 2 nghiem phan biet va doi d^u 2 Ian khi qua 2 nghiem nen
luon luon c6 mpt ci/c dai va mpt eye tilu.
Bai toan 1.15: Tim tham s6 thye sao cho ham s6

bi^t xi = -1 - ^
BBT:

Hhang Vi$t

lq = i
IP = 1

b) Di§u ki^n x ;t - + kn. Ta c6 y' = - — , y ' = 0 o sinx = a.
2
a cos X

, , x^ - 2x + 2m - 2
b) OK: X / 1. Ta c6 y =
k \

> -.
<

T-

(X - \

•)1 iV

i


• ''

Di^u kien c6 2 eye tri A, B la A' > 0 va g(1) 0.
3
0 3 - 2m > 0 va 3 - 2m ^ 0 o m < - . Ta c6
2
-h P
A(1- 73 " 2 m ; 2-2m-2 73 - 2m )va B(1 + 73^ 2m ;2 -2m + 2 7 3 - 2 m ).
I

He s6 goc cua du'ang thing AB la: k = ^^^2)-yi^^)
X2 - X,

^

;

4^|3-2m ^ g. 273 - 2m

Va 2x - y - 10 = 0 <=> y = 2x - 10 nen h^ so goc bing nhau => dpcm.


Cf^y INMH

W trgng diSm b6i du&ng h
Bai toan 1. 17: Vi§t p h u a n g trinh du-ang thSng di qua d i l m c u e dai, c u e tieu
cua do thj:
a) y =
+ 3mx^ + 3(m^ - 1 )x +
- 3m
'
b)y =

x^ - 2mx + 5m - 4 - m^
, i!

a) y' = 3x^ + 6mx + 3(m^ - 1), A ' = 1 > 0, Vx nen d6 thj luon luon c6 CD va CT
v a i hoanh dp Xi, X2.
Ta c6: y(x) =

1
m
- X + — y'(x) - 2(x + m).
3
3

D o d 6 ; y i = y(xi) = [ ^ ^ i + y y ' ( x i ) - 2 ( x i + m) = -2(Xi + m)

y2

= y(x2) =

1
m
- X , + — y'(x2) - 2(X2 + m) = - 2 ( X 2 + m)
3 ^ 3 j

nen d u a n g t h i n g qua CD, CT la y = - 2 ( x + m).

m-m^

A = 2

12A + 4B + C = 0

8 =-9

y(1) = - 7

A + B + C + D = -7

C = 12

y(2)= -8

8A + 4B + 2C

D--12

^
. , 2x3-3x2-a
b) Ta CO y ' =
;
,x

y = 3x2-6x + 3c6djnh.
Bai toan 1.19: Giai cac p h u a n g trinh:
4

4 ^ v x ^ - 2x + 4 = 2(^/3 - 1)

b) 2x^ - X' + ^J2x^ - 3x + 1 = 3x + 1 + ^ x ^ + 2 .
Hu'O'ng d i n giai

{X-2Y

a) Xet ham s6 f(x) = v x ^ + 2x + 4 - V r

Di'eu kien c6 CO va CT la m - m^ > 0 o 0 < m < 1.

X+1
f'(X) = - ;
v'x^ ^ 2x + 4

Goi xi, X2 la hoanh do CD, CT thi x, < 2 < X2. Ta eo
y(xi) = x , - 2 ( m - 1 ) +

(X, - 2)

= Xi - 2(m - 1) + (xi - 2) = 2xi - 2m.

(X2-2)

= X 2 - 2 ( m - 1) + ( x 2 - 2 ) = 2x2-2m.

Vay p h u o n g trinh d u o n g t h i n g qua CD va CT la y = 2x -

2m

t

Xet ham s6 g(t) =

1)
X

+ 1 CO 3 cue trj va e h u n g minh 3

cue tri nay thupe mpt parabol c6 djnh.
Hipo-ng d i n giai
- -3^2
a) Dat A = 3a^ - 1, B = - ( b ' + 1), C =
Zc\ = 4d, thi ham s6 da cho la:

X +^

"

x f 1

>x-^

+

" '

3

+ 3

>0

(t^ + 3 ) V t 2 +

fj."

3

X -1

1)2 + 3

i(x -1)2

f'(x) >0

+ 3

nen ham s6 f(x) dong b i l n tren R, do do:
vx2 f 2x + 4

-

N/X2

- 2x + 4 - 2(^3 - 1) o

f(x) = f(2)

x = 2.

*,

Vay nghi^m duy n h i t x = 2.
b) PT o 2x^ - 3x + ^J2x^ - 3x + 1 = x2 +1 + ^x2 + 2
Xet ham s6: f(t) = t +

tren R,

> 0 nen ham so f(t)

f'(t) = 1 +
3^(t +1)

y = Ax~* + Bx^ + Cx + D.
Ta c6: y' = 3Ax^ + 2Bx + C

'

^[{x^f

1)2 + 3

. . _
tren R, g'(t) =
+3

x-1

nen ham s6 g(t) d6ng b i l n tren R, do do:

Bai toan 1. 18:

b) Tim a d l d6 thi ham s6 y = -^^

X f1

v/x^ - 2x + 4

N/(X

a) Cho d6 thi cua ham s6: y = (3a^ - 1)x^ - (b^ + 1)x^ + 3c^x + 4d c6 hai
didm c u e tri la (1; - 7 ) , (2; - 8 ) . Hay tinh t6ng M = a^ + b^ + c^ + d^.

- 2x + 4 tren R.

X-1

V't^

y(X2) = X 2 - 2 ( m - 1 ) +

0.

T u tpa dp cac d i l m c u e trj suy ra cac d i l m c u e trj n^y n i m tren (P):

(x-2)'^-~(m-m^)

m - m


'

y ' = 0 o 2x^ - 3x^ - a = 0 o a = 2x^ - 3x^, x * 0.
B i n g each xet ham s6 g(x) = 2x^ - 3x^ , x
0 va lap bang bi^n thien thi
di§u kien ham s6 cho eo 3 cue trj khi g(x) = 0 c6 3 nghi^m phan bi^t khac 0
la-1 < a < 0 .

2

(X - 2Y

= -8

Vl^t

yay M = a ' + b^ + e^ + d^ = 1^ + 2^ + 2^ + 3^ = 18.

a) v'x^ + 2x

m - m
b) DK: X ^ 2. Ta CO y = X - 2(m - 1) +
X -2
.
, ,
nen y = 1

3A + 2B + C = 0

y'(2) = 0

N e n d u p ' e a = ± 1 , b = 2, e = ±2, d = - 3 .

Hu'O'ng d i n giai

va

T a c6:

y'(1) = 0

mang

MTVUVVM

d6ng b i l n tren R, do do:


lO trpng diSm b6i duOng hQC sinh gidi mon lodn If
PT: f(2x^ - 3x) = f(x^ + 1) »

' = ' o •/? FHO

Ctj/ TNHHMTVDWH Hhong Vi$t

2x^ - 3x = x^ + 1

Do d6 5x^+ 7x* = 5x^+ 7x* o f(x) = f(y) » x = y
N6n (8x' + 1)=* + 27 = 162y <=> (8x^ + 1)^ = 162x - 27

<::> 2x^ - x^ - 3x - 1 = 0 c:> (X + ^ )(2x^ - 2x - 2) = 0
1 ,

0$t u = 2x, phu-ang trinh: (u^ + l f = 27(3u - 1) <=>
Lgi dat V = \/3u-1 o

1±V5

Ta c6 h$:

Bai toan 1. 20: Giai cac phuang trinh :
a) 9x' - 54x + 72 =

1

1

2x - 5

X -

2

2x

J vai t > 0. Ta

+1 = 3v

o

-

= 3(v - u)

1

5

- 3(x - 1)2

+1 = 3v

(u - v)(u^ + vu +

I

a) DK :x / 1, - , PT 3(2x - 5)

+1 = 3u

+ 1 = 3u

+1 = 3v

1

b) 4 2x - 1 ! (x^ - X + 1) = x^ - 6x^ + 15x - 14.
Hu'O'ng dan giai

Xet f(t) = 3 t^ -

,

+ 3) = 0

u-v = 0

Do d6
+ 1 = 3u hay 8x^ - 6x + 1 = 0
X6tx e [ - 1 ; 1 ] n § n d $ t x = cost
x-1

;0r • .

PT: 2 ( 4 c o s ^ t - 3 c o s t ) = -1c:>cost = - - < = > t = ± — + — , (!<€ Z)
Ti> d6 c6 3 gid trj cua x
27t
8TI
x = cosx = cos9
9

c6:

f (t) = 6t + — > 0 nen f dong bien tren (0; +-r)

2
9
3
cOng chinh \k 3 nghi0m cua phu-ong trinh b$c 3:
1471

x = cos-

V|y nghi$m h? x = y = c o s — ; c o s — ,cos

9
9
9
b) ( 2 ) o ( y - 1 ) [ ( x + y ) 2 - 1 ] = |x + y| [ ( y - l ) 2 - i ]

Phuang trinh: f ( i 2 x - 5 l ) = f ( ! x - l ! ) < » i 2 x - 5 l = ! x - 1
o 4x" - 20x + 25 = x^ - 2x + 1 o 3x^ - 18x + 24 = 0.
c:> x' - 6x + 8 = 0 cr.-- x = 2 hoac x = 4 (chon)

V6i y = 1: (3)

Vay nghiem x = 2 hoac x = 4

V6i x + y = 0 (3) <=> y = 1 => x = - 1 ; I
I
1 P + 3 I 2x - 1 I

b) PT: [ 2x - 1 .[(2x - 1 ) ^ + 3] = (x - 2)^ + 3x - 6

<:> I 2x -

= (X

+ 1 = 3^3u - 1

x = - 1 : I
V6ix-.y^0.y>1:(3)c>(^"y)'-^Jy-^)'-^
y-1
x+y

- 2)^ + 3(x - 2)

X6t ham s6 f(t) = t' + 3t, D = R.
o

Ta CO f '(t) = 3t^ + 2 > 0 nen f d6ng bi§n tren R.

I

PT: f( 2x - 1
X

I)

= f(x - 2) «

! 2x - 1
x>2

-2> 0

3x2 ^ 3

(2x --1)2 . (X - 2)

I

=x- 2

(VN). Vgy S = 0.

- X

Bai toan 1. 2 1 : Giai cac he phuang trinh:
a)

= y-1-

y-1

X6f hdm s6 f(t) = t - ;J. D = (0 ; + 00)

P T o f ( | x + y|) = f ( y - 1 ) c > |x + y| = y - 1

(8x^ +1)^ t 27= 162y

(y-1)r(x + y f - 1

x+y

f'(t)= 1 + ^ > 0 , V t € D ^ h d m s 6 d 6 n g bi4ntr§nD

5x' + yx'^ - 5y^ f 7y^

[y>1

x2 +y2 ^ 5 ; y > 1
b)

ix + y

(1)
X

f y (y2 - 2y) (2)

Hu'O'ng d i n giai
a) Xet f(t) = 5t'+ 7t^ t G R thi f'(t) = 35t^ + 35t' " 0, V t nen f d6ng bi§n tren R,

[x = - 1

hay

' 'ill •'. •

x=:1-2y
1-2N/24


Ctj/ TNHHMTVDWH Hhong Vi$t

10 trQng diSm bSi dUOng h
Bai toan 1. 22: Giai cac he phucng trinh
- 2x + 1 = 2y

3 6 x ^ y - 6 0 x ^ + 25y = 0

- 2y + 1 = 2z .

a)

36y^z-60y^ +25z = 0

b)

3 6 z ^ x - 6 0 z 2 +25x = 0

- 2z + 1 = 2x
'

Hu'ang d i n giai

>

f(x) = g(y)
tren (0; +«). Ta c6 h$ f(y) = g(z)
f(z) = g(x)

0 < f(x) < 1.
a)DK

0 < f(y) < 1 => 0 < g(z) < 1 => 0 < z < 1.
g(y) > g(z) > g(x)

- 2 x + 3 - Vx^ - 6 X + 11

7 3 ^ - >/x^.
HifO'ng d i n giai

-2

f(x) =

,

Vgy h$ CO 2 nghiem x = y = z = 2 + yfs , x = y = z = 2- \l3 .

36x^+25

4

Do do BPT: f(x) > f(1)

«

eoz^

361^ + 25

x > 1. Vay: S = (1; 4)

^~^<;=>1[3-x>0

.t>0.

Jt::"

7(x - 1)2 + 2 + ^Jx^^ > V(x - 3)2 + 2 + 7 3 ^

Xet ham s6 y = f(t) =

Tu" h0 suy ra x, y, z khong am. Neu x = O t h i y = z = 0 suy ra (0; 0; 0) Id
nghiem cua he phu'ong trinh.

f '(t) > 0, Vt > 0 nen f d6ng bi^n tren (0; +oo).

> 0
274-X

•If

36f + 25

60t'

272x3 +3x2 +6X + 1

BPT: 7x2 _ 2x + 3 + 7 x ^ >7x2 - 6x + 11 + 7 3 ^

60y2

0 thi y > 0, z > 0. Xet ham s6 f(t) =

: If,

Suy ra f(x) la hdm s6 d6ng bi§n

Di§u ki$n:

60x2

36z^ + 25

X >

< X<

X + 1)+ 16 _
Xet: f(x) = 72x3 6(x2
+ 3x2+ +6X

Xet X < z < y thi cung nhan du-gc ^kx qua tren.

z=•

>

2x^+3x2+6x + 1 6 > 0

Ta CO PT t^ - 4t + 1 = 0 chpn nghiem: x = y = z = 2 - N/2.

N§u

5 5 5^
6'6'6

4-X>0

^ .^,

=> y > z > X nen x = y = z.

X =

,x

du-gc x(36x2 _ gQj^ ^ 25) = 0. Chpn x = 0; - .
6

b)

nen 0 < g(y) < 1 => 0 < y < 1 =^ f(0) > f(y) > f(1)

b) H0 phu-ang trinh tu-ong du-ong

-rm

60z2

36z2 + 25
Ti> tinh d6ng bien cua f(x) suy ra x = y = z. Thay vdo he phu'ong trinh ta

Ta CO PT: t^ - 4t + 1 = 0 chpn nghiem: X = y = z = 2 + N/3 .

y=

36y2 + 25

a) V2x^ +3x2 +6X + 16 >2V3 + N / 4 - X

< y < z => f(x) < f(y) < f(z)

f(x) > f(y) > f(z)

60y2

Bai toan 1. 23: Giai cac b i t phu'ong trinh

=> g(y) < g(z) < g(x) = > y < z < x n e n x = y = z.

Do do X < y < z

36x2 ^ 25

Vay tap nghiem cua he phuong trinh la \;

Gia su' X = min{x; y; z}. Xet x < y < z.

N§u 0 < x < 1 thi f(0) > f(x) > f(1)

z=

60x^

X =

Oat f(t) = t^ - 2t + 1, t > 0 thi f '(t) = 2(t - 1) nen f dong bien tr§n (1; +oo)
nghich biln tren (0; 1). Dgt g(t) = 2t, t > 0 thi g'(t) = 2 > 0 nen g d6ng bien

X

He phuong trinh du-gc v i l t l^i

'

a)Tac62y = x ^ - 2 x + 1 = ( x - 1 ) ^ > 0 = i > y > 0 . T i i ' a n g t y z , x > 0 .

N§u x > 1 thi 1 <

y=

£>9o ham: f'(x) =

J^Ti + 7t,

7 t ^

27t

D = (0, +«)

> 0 nen f d6ng bi§n tren (1, 3].

Do do BPT f ( x - 1) > f(3 - X) <=> X - 1 > 3 V^y nghiem cua b i t phu'ong trinh S = (2, 3].
Bai toan 1. 24: Giai cac b i t phuong trinh
a) 7 3 - X

+

x2 - 7 2 + x - x 2 < 1

X <=> X

> 2.
i•


10 trpng diS'm bSi duang HQC sinh gioi m6n Toon 12 - LS Hodnh Phd

+ 2V3 - 4x < 7 .

b) 4(7x)''
'

X6t h^m s6 f(t) =

x2 = y 3 +y2 + y + a

Hu'6'ng din giai

a) D|t t = x^ - X, BPT: N/sTt -

t,

Nen f(x) = 0 c6 nghiem duy nhit x < 0.
V|y phuang trinh cho c6 nghi^m duy nhlt.
Bai toan 1. 26: Chung minh he phuang trinh c6 nghi^m duy nhit:

Ta + t -

z2 = x^ + x2 + X + a

42^\ -3 < t < 2.

Hipang din giai
X6t h^m f(t) = t^ + t2 + t + a CO f '(t) = 3t2 + 2t + 1 > 0 do d6 f(t) 1^ h^m d6ng

V 6 i - 3 < t < 2 thi f (t) =

/
+
/
>0
2V3 + t
2V2-t
ngn f d6ng bi§n tren (-3; 2).
Ta c6 f(1) = 2 - 1 = 1 n§n b^t phu-ang trinh:
f(t) < f(1)

t < 1 » x^ - X - 1 < 0

b) DK: 0 < X <

- . PT o
4

x2 = f(y)
biln. He PT:

Khong giam tong qu^t gia su x Ib-n nhit trong 3 s6.
=> z2 > x2 > y2. N^u
x2 > y2 > z2

V6i X =0 thi BPT I4

thi g '(X) = 8x - 8 x ( - - 2x2) -

^

3
ndn g(x) nghjch bi^n trSn ( 0 ; - ) ,
4

g(l) o x >

+ 273 - 4x

= 4x(4x2 _ 3) -

4
V3-4X

V3-4X

2

2

Bai toan 1. 25: Chufng minh phu-ang trinh:
x^^ - x^ + 3x'' - 3x2 + 1 = 0
nghj^m duy nhlt.
HiTO'ng din giai
O0t f(x) = x^^ - x^ + Sx" - 3x2 + 1, D = R
X6t X > 1 thi f(x) = x^(x^ - 1) + 3x2(x2 - 1) + 1 > 0: v6 nghi^m
X6t 0 < X < 1 thi f(x) = x^^* + (1 - x2f > 0 : v6 nghi^m
X6t X < 0 thi: f (X) = 13x^2 _ g^s + -,2x3 _ 6x
= 13x^2 _ 6x(x - 1 )2 > 0 n6n f d6ng bi^n
Bang bi6n thidn:

-00

z>Othix>y>z>0

x2 = y2 = z2 ^ f(x) = f(y) = f(z) ^ x = y = z

N§u X < 0 = > 0 > x > y > z = i > x 2 = y2 = z2=>x = y = z
Neu X > 0 > z. Khi do y2 = f(z) < f(0) = a =^ a > 0
Lgi CO z2 = f(x) > f(0) = a ^ z = - N / S
=^y2 = f ( z ) < f ( - 7 a ) = - V a ( N / a - 1 ) 2 < 0 : v 6 l i .

<0

1
g ( - ) = 7 n6n b i t phu'cng trinh g(x)

^ . V§yt$pnghiemS = ( I ; | ] .

-00

-

_ Xet X > y > z =^ f(x) > f(y) > f(z)

+ 2V3 - 4x < 7

V6i 0 < X < - . Xet ham s6 g(x) = 4x2 + ^ - 2 x 2
4
2

y2 = f(z)
z2=f(x)

< x<

4x2 + ^ - 2 x 2
2

+ z2 + z + a

y2 =

< 1, -3 < t < 2.

~ X6t X > z > y

''''

z2 > y2 > x2

{-^ \

Tuang tu-nhu tren n l u y > 0 hay X < 0 ta suy ra X = y = z
N4U x>0>y=:i>x2 = f(y) < f(0) = a
z2 = f(x) > f(0) = a. N§u z > v/a
thi X > z > v'a => x2 > z2 => z2 = y2 = z2
=> X = y = z trai v^i x > 0 > y

'

'•

N l u z < - Va li luan nhu tren ta dan d§n mSu thuan.
V^y he CO nghiem duy nhit x = y = z = to6'd6toia nghiem duy nhit cua
phuang trinh: t^ + t2 + t + a = 0.
Bai toan 1. 27: Chung minh he i

~ c6 dung 3 nghiem phSn bi$t.
y2 + x^ = 1

Hu'O'ng din giai
Tru 2 phuang trinh v6 theo v§ va thay the ta dugc:
x'(1-x)-y2(1-y) = 0 ^ ( 1 - y ' ) ( 1 - x ) - ( 1 - x ' ) ( 1 - y ) = 0
=^ (1 - x)(1 - y)[1 + y + y 2 - ( 1 + x + x2)] = 0.
=^(1 - x ) ( 1 - y ) ( y - x ) ( 1 + x + y) = 0.

,


Xet X = 1 thi h$ c6 nghiem (1; 0).Xet y = 1 thi h$ c6 nghiem (0; 1)
Xet X = y thi x^ + y 3 = 1
x^ + x^ - 1 = 0.
Dat f(x) = x ' + x-^ - 1, D = R. Ta C O f(1) = 1 ^ 0.

Xetfx =

X

-co

+

y'

-2/3

0

0

0

X . 1
2

-{

f

+ 00

1

, 3x^+1
, x>--,x^Othif'(x)=

2
x"^
+ 00
0
0

+

-co

^

Do do f(x) = 0 CO 1 nghiem duy nhit XQ > 0, XQ ^ 1 nen he c6 nghiem (X(
Xet 1 + X + y = 0
y = - X - 1 nen y^ + x^ = 1 o x^ + x^ + 2x = 0
o x(x^ + X + 2) = 0 » x = 0. Do do he CO nghiem (0; 1).
Vay he c6 dung 3 nghiem phan biet.
Bai toan 1. 28: Tim tham s6 de phu-ang trinh

1
<=:> f(x) = m CO 2 nghiem phan biet x>--,x=^0<^

a) ( N / X + V x - 2 ) m^Jx+-.

^

3 ^ x ( x - 2 ) = 2 CO nghiem ftfim^i Ja>

b) 37tanx + 1 .(sinx + 2cosx) = m(sinx + 3cosx) c6 nghiem duy nhit thuoc
khoang (0, - ) .
Hyang din giai

Hu-o-ng din giai

a) Oi§u kien: x > 2

a) Xet f(x) = ^To< + ^/l7x , D = R
3^(1-x)2

3^(1+ x)2

9
^ - 2

Vx-2

b) vx^ + mx + 2 = 2x +1 CO 2 nghiem phan bi^t.

1

^-7

Bai toan 1. 29: Tim m d l phuang trinh

a) ^1 + X + \ / l - x = a CO nghiem

1

+ 00

+ CO

Oi^u kien phu-ong"trinh da cho CO 2 nghiem phan biet

-co

f '(x) = -

+

+

f

-23/27

y

+4X-1

X
X

BBT:

f (X) = 3x^ + 2x, f '(X) = 0 <=> X = - - ho$c x = 0.
BBT

3x^

(x^±1)

= i ^ J ^ , f ( x ) = o«

2

PT« 2

r

x= 0

<=> V x ^ +

= lim (^1 + x - ^1 - x )

= lim

^

Vx-2

=0

\

m .vx + V x - 2

N/X-2

3^(1-x)^ . ^(1 + x)2

lim f(x)= lim (^/Toc + ?/l + x )

2

Vx-2

-3.Vx(x-2)

=2(^fx-4y^

-3.t/x(x-2) = V x - 7 x ^
- 3.t/x(x-2) = (1 - m2)7x

-3.fM=1-m^

D a t t = f P — ^ , 0 < t < 1.PT: — - 3 t = 1-m^ , 0 < t < 1 .
V

Tu-ang t u lim f(x) = O.Lap BBT thi PT c6 nghiem » 0 < a < 2.

X

ii • J • '

X->-oo

b) PT<^

2x + 1 > 0
<=> 3x^ + 4x - 1 = mx, X
2
x'^ +mx + 2 = (2x + 1)^

Vi X = 0 khong thoa man nen:

Xetf(t) = 4 - 3 t , t € (0; 1)=> f ' ( t ) = - - | - 3 < 0 .
e (0; 1). ,
(2
t3
T
t
0
1
Bang bi§n thien
f'(t)
-

> - -

^"^^ ^ = m, x > - |

+x

4
I*

99

f(t)




V^y phyang trinh cho c6 nghi^m khi
1> - 2 <=>-N/3 < m < N/3
0
tanx > - 1 .

b) Dieu ki^n : cosx

b) X6t x = 0 => 1 = 0: loai.
X6t x ;t 0. Chia 2
cho x^ phuang trinh:
1 3
1
x ' + 3x' + (6 - a)x + (7 - 2a) + (6 - a ) . 3 + ^ + — = 0

_+

D$t t = 7tanx + 1 > 0 , phu-o-ng trinh:
sinx + 3cosx
= mcosx
cosx
3 7 t a n x + 1 (tanx + 2) = m(tanx + 3)
Syjxanx

o

+ 1

sinx + 2cosx

<=> 3 7tanx+1 (tanx + 1 + 1) = m(tanx + 1 + 2 )
« 3t(t2 + 1) = m(t2 + 2) o
X6t h^m so y =
3t-

m = ^^J-^ .
t^ + 2

^
v6i t e (1 , + « ) ,
t^ + 2
+15t^ + 6
(t^ + 2f

V$y phu-ang trinh c6 nghi^m duy nhit khi m > y(1) <=> m > 2 .
Bai toan 1. 30: Tim tham s6 d l phu-ang trinh
a) (4m - 3) 7x + 3 + (3m - 4) Vl-x + m - 1 = 0 c6 nghi^m
b) x^ + 3x^ + (6 - a)x'' + (7 - 2a)x^ + (6 - a)x^ + 3x + 1 = 0 v6 nghi$m.
Hu>6ng din gial
a) Di§u ki$n: - 3 < x < 1 khi d6:
3x/x73 + 4>/irx +1
PT <=> m = — 7 = =

4Vx + 3 + 3V1 -

X

+1

Ta c6: (VxTi )2 + (VT^ )^ = 2 nSn d$t:

(x3+

X6tf(t)=^Zi;ii^
-5t2+16t + 7
,,

. te[0.1]

-5y-8t-60
(-61'+ l e t + 7)^

X
7
2
Vay dieu ki^n phu-ang trinh c6 nghi^m la f(0) < m < f(1) <=>- < m < - .
9
7

x^

x^

X

0 $ t t = x + - , Itl > 2 ^ t ^ = x ^ + 4 - + 2
X
x^
va t^ = x^ + — + 3(x + - ) n§n x^ + ^ = t^ - 3t.
x^
X
x^

Do do phu-ang trinh: t^ - 3t + 3(t^ - 2) + (6 - a)t + 7 - 2a = 0
(t + 2)a = t ' + 3t^ + 3t + 1
Khi t = - 2 thi phuang trinh khong thoa.
^,
.- u
t3+3t2+3t + 1 (t + 1)3
Khi t ^ - 2 thi phuang trinh: a =
=
—.
t+2
t+2

oat f(t) = lilJ)! , t < -2 hay t . 2 thi f '(t) = ^2^13^.
t+2

2(t + 2f

Lap BBT thi f(t) > — Vt e D n6n PT v6 nghi$m khi a < — .
4
4
Bai toan 1. 31: Tim tham so d§ bSt phuang trinh c6 nghi^m
a) sin^x + cos^x > m
b) cos^2x + 2 (sinx + cosx)^ - 3sin 2x + m > 0.
Hyo-ng din giai
a) Xet f(x) = sin^x + cos^x = (sinx + cosx) (1 - sinx.cosx)
D^t t = sinx + cosx ; |t| < N/2

7x + 3 = 2sin(p = 2 2t
; V/l—
- x- = 2cos(p = „2 l - t ^
1 + t'
l + t^
V a i t = taniP , 0 < cp < I ^ . 0 < t < 1 nen: m = - 7 t ' + 1 2 t + 9
2
4
-5t2+16t + 7

JL) + 3(x'+4

+_

x^ x
)+ (6-a)(x+ -)+7-2a = 0
X

t^ = 1 + 2sinx cosx => sinx cosx =
Ta

CO

h(t) = t 1 .

2

- 1) = - ^ t ^ + | t v a i |t|< V2

3,2 . 3
h'(t) = - - t ^ + - = 0 « t = ± 1
^ ' 2 2
Lap BBT thi b i t phuang trinh CO nghi^m khi m < 1.

v •

b) D|t t = sinx + cosx, |t| < 72 va t^ = 1 + 2 sinx cosx ^ sin2x = t^ - 1
cos^ 2x = 1 - sin'2x = - t " + 2t^
BPT:

- t " + 2t^ -1' + m + 3 < 0 ; (|t| < V2)


-j^c^rnwn-t /VIIV uvvH hnang

Hirang din giai
Xet ham so g(x)= f(x) +x - 1 , khi do thi g(x) lien tyc v^ c6 dgo ham tren
[0;1].
Ta c6: g(0)= - 1 < 0 va g(1)= 1 >0 nen t6n tai so c thupc (0;1) sao cho g(c) =0.
Do do f(c) + c - 1 =0 hay f(c) = 1 - c
Ap dung dinh ly Lagrange cho f tren cac doan [0;c] va [c;1] thi :
"

Xet f(t) = -t^ + 2f' - t-^ + m + 3
f '(t) = - 2 t {2f - 3t + 1) ;

f '(t) = 0 =^ t = 0 ; - ; 1
2
Lap BBT suy ra dieu kien c6 nghiem la: m + 3 > 0 <=> m > - 3
Bai toan 1. 32: Tim 6\hu lx^-3x-4 < 0
(1)
X • '
- 3x Ixl - m^ - 15m > 0

(2)

,

.
Htpcng dan giai
Xet(1) : x ^ - 3 x - 4 < 0 o - 1 < x < 4
Ta tim dilu lf(x) = x^ - 3x |x| - m^ - 15m < 0 ; Vxe [ - 1 ; 4]
Vi f(x) = •

t6n tai ae(0;c) sao cho:

t
I
•t

' X.---

nin:C(a).f(b) =

->f'(x) =

3x^ - 6x ; 0 <

< 4

X

0 2<

X

Hii'O'ng din giai
X6t 2 h^m s6: g(x) = arctanx;h(x) =

=>f'(x) = 3x(x + 2 ) < 0
f '(X)

= 3x

(X

=> f '(x) = 3x

(X

dao ham tren (0;1),

- 2) > 0

Do do - m ^ - 1 5 m + 16<0<=>m< -16 v m > 1

Taco: g'(x) = — ^ ; h ' ( x ) =
1 + x^

Vay di4u kien c6 nghiem la - 1 6 < m < 1

?)<
(1 + x^)^

1 j,
1 + x^

Theo dinh ly Cauchy thi ton tai c G(0;1) sao cho:

Bai toan 1. 33: Cho 3 s6 a,b,c thoa man abc ?^Ova - + - + - = 0.
7 5 3
ChLKng minh phu'ong trinh : ax"* + bx^ + c = 0 c6 nghiem.
Hii'O'ng din giai

=
g(i)-g(0)

Xet hdm s6 F(x) = - x ^ + - x ^ + - x ^ , khi do F(x) lien tyc, c6 dao hSm
' '
7
5
3
F '(x) = x^. (ax"* + bx^ + c) = x^ .f(x) nen theo dung dinh II Lagrange tren [0,1]
t h i t d n t g i c e (0,1): ^ ^ ^ ^ ^ ^ = F ' ( c ) .
1-0

tren [0,1], khi do thi g(x),h(x) c6
1 +x

- 2) < 0

<4

"V

ChCfng minh b i t phuang trinh: f'(x) - f(x) < -(f(1) - 2f(0)) c6 nghiem.

<0

Khi-1
M!j!(£).(lz£)£
=,
C
1-C
c(1-c)

Vay t6n tai 2 s6 phan biet a; b thupc (0;1) sao cho f (a).f'(b) = 1.
' '
Bai toan 1. 35: Cho ham s6 f(x) c6 dao ham tren [0,1] va nhan gia tri du'ong.

x^ - 3x^ - m^ - 15m ; 0 < x < 4
X

'-S; v..
,

va ton tai be(c;1) sao cho: ^^^^~^^"^^ = f'(b)
1-c

x^ + 3x^ - m^ - 15m ; -1 < x < 0

3x^ + 6x ; -1 <

= f'(a)
c- 0

t

vi^t

hay
g'(c)

l:!^

=f ( c ) - ^ f ( c )
^ ^

4

ir.

^^

«

-(f(1)-2f(0)) = f ' ( c ) - ^ f ( c ) .
^
1 + c^
V i O < c < 1 nen 1+c^ >2c va vi f(c) > 0 nen f ' ( c ) - - ^ f ( c ) > f ' ( c ) - f ( c )
l + c^
=> dpcm.
Bai toan 1. 36: Gia si> f\a mpt ham xac dinh tren [a, b], c6 dao ham d i n d p
n + 1 tren (a,b) va XQ e (a,b). ChCrng minh t6n tai c n i m giOa x va XQ d l c6:

(J.''.ids

Ma F(0) = 0, F(1) = - + - + - = OnenF '(c) = 0 hay c^f(c) = 0.
7
5
3
Vi c G (0,1) nen
0 do do f(c) = 0 => dpcm.
Bai toan 1. 34: Cho ham s6 f c6 dao ham tren tren [0;1] va thoa m§n:
f(0) = 0; f(1) = 1. Chu-ng minh ton tgi 2 s6 phan bi^t a;b thupc (0; 1) sao
cho f'(a).f'(b) = 1.

fW-f(>^o)^(x-x,).!:|il(x-x„)^....

!!X)(x-x)".i!:::!(^(x-x)ft

n!

"""^

(n + 1)!^

°'


/c

Lr\fny

atom

tXJr

ainJtty

tiv^

srrm

rrnpn r c ^ n V 2 r — le

grar

rtOOnn

mo

HiTO'ng d i n giSi
Ta tim mOt da thCfc Pn(x) c6 bgc I
trong do c IS mOt di^m n l m giu-a x vS XQ.

/(Xo) = Pn(Xo) , f'(Xo) = P'n(Xo)

Cong thirc tren du'O'c gpi la c6ng thtpc khai trien Taylor cua hSm f tgi diem x = Xo.

/'"'(X)

= P<"' (Xo)

, M(3\: Pn(x) = Ao + A i ( X - X o ) + A 2 ( X - X o ) ^ + . . . + A n ( X - X o ) "
Luc d6:
.,
P'n(x) = Ai + 2A2(x - Xo) + ... + nAn(x - Xo)""'
'
P"n(x) = 2A2 + 3.2.A3(x - Xo) + ... + n(n - 1)An(x - XQ)""^

Pr(x)

,
'"
v

= n!An.

Do d6 thay x = Xo vdo c^c d i n g thii'c tr§n ta du'O'c:
: Pn(xo) = Ac , P'n(xo) = A i , P"n(Xo) = 2A2

P<"' (XQ) = n!An.

Nhu vgy: f(xo) = Ao, A , = r(Xo), 2A2 = r(xo), ... , /<"'(xo) = n!An n§n:
PnW = n X o ) + ^ ( X - X , ) . ^
Dat Rn(x) =

(

X - X „ ) ^ + ... + ! l ^ ( X - ^

- Pn(x) ta suy ra R|,"'(x) = f^"'(x) - P^"'(x)

nen: Rn(xo) = R'n(xo) = .... = Rj,"'(Xo)=0.
m

F(x) = ( X - xo)"*^ thi: F(xo) = FXxo) = ... = F<"'(Xo) = 0.

Vol X e (a,b) ta vi6t du'yc ^
^ ""^^^"
.
F(x)
F(x)-F(Xo)
R (x)
Theo djnh ly Cauchy ta c6
"Ta

lai c6

=

R'(£ )
" ^ vai \y n l m giOa x X Q .

Mll^MlLM^

theo dinh ly Cauchy ta du-^c:

F'(^i)-F'(Xo)

3 . B A I LUYfiN T A P
Bai t?p 1 . 1 : Tim cac khoang dan di$u cua hSm s6:
2x
us
X + 1
a) y = x ' - 9
b) y = Vx^ - x , +
Hu>ang d i n
'{.w/v^

r

-2(x^ + 9
X
a) K§t qua y' = — \ 0 nen hSm s6 d§ cho nghjch bien tren cSc khoang
(x^ - 9)^
(-«;-3), (-3; 3), ( 3 ; + = 0 ) .
^.^^ ,
b) K i t qua d6ng bien tr§n (-00; 1), nghjch bi§n (1,+00).
Bai t$p 1. 2: Tim m de hdm s6:
a) y = -—+ ("^ + 2)x - m + 3
^^^^ ^^^^
khoang x^c djnh
x + 1
b) y = - x^ - — x^ - 2x + 9 dong bi§n tren ( 1 ; + x ) .
3
2
Hirang d i n
a) Tap xac djnh D =
- 1 ) u ( - 1 , +00).
Tinh dao ham y ' vS l|p lu$n y ' > 0 tren D. K§t qua m > 1.
b) K§t qua m < - 1 .
Bai t$p 1. 3: Tim ciic trj cua hSm s6:
a) y =

y • •

,.

b)y=^(x-5).
7x^-6

Hifang d i n
a) Ham s6 le. Tinh dgo ham va l$p BBT.
K§t qua CO tai x = - 3 ; yco = - 9 \/3 , CT tgi x = 3 ; y c r = 9 \ / 3 .
R (x)

R'"^^'(C)
Sau n + 1 l^n ap dyng dinh ly Cauchy ta du-gc
:= — - - v^i c nSm
F(x)
F<"^'>(c)
giOa 4n va XQ, va do do c n i m giOa x va XQ.
Nhung R<"*^>{x) = / " ' ' ' ( x ) va F'"*^'(x) = (n + 1)! nen

= 1^^^
.
F(x)
(n + 1)!

Vay: f(x) = f(x„) / ^ ( x - x„) + l | | ° l ( x - x^)^ +...

nl

""'^

b) K§t qua CD tgi x = 0, yco = 0 va CT tgi x = 2, y c r = - 3 \ / 4 .
Bai t?p 1.4: Tim ci^c trj ham s6:
a) y = x - sin2x + 2
b) y = sin2x + cos2x.
Hu-angdln
a) T$p xSc djnh D = R, y ' = 1 - 2cos2x, y " = 4sin2x.
Dung d i u dgo hSm d p 2.
K§t qua: CD tgi x = - - + kn, k e Z, y c o = - - +
6
6

(n + 1)!^

X = - + kn, k e Z; ycT = - + krt - — + 2.
6
6
2

R
+—
2

*

: f
+ 2 ; dgt CT tai


W trqng diS'm bSi dUOng h
b) K§t qua diem ci^c dai x = — + kn , diem ci^c t i l u x = —

8

8

b) Vx^ - 2 > / 3 x - 2 + Vx^ - s T s x + 4 = 3 .

+ kn .

Hirang din

Bai tap 1. 5:
a) Tim m d§ h^m so y = 2x^ - 3 (3m + 1 )x^ + 12 (m^ + m) x + 1 c6 ci/c dai va
cu-c tilu. Vilt piiu-ang trinli du-ang thing di qua CD, CT.
X
x^+2mx + 1-3m^
,
CO hai d i l m eye th n l m v l hai
b) Tim m de ham so y = —
x -m
phia cua true Oy.

Hipang din
.

> ,3

b) K i t qua - 1 < m < 1.
Bai tap 1. 6: Chipng minh ham s6
a) y = x^ + ax^ - (1 + b^)x + a + 4b - ab luon luon c6 eye dai va eye tieu vd-i mpi
tham s6 a, b.
b) y =

1
3x - - ba diem eu-e tri phan biet A, B, C . Tinh di$n tieh tam

2
giae ABC.

X

b) K i t qua S =

Bai tap 1. 7: Giai eae phuang trinh :
a) 3x^ - 1 8 x + 24 =

^

I 2x - 5 1^

-

2x-5

a) N/X2-1 = V X ^ - 2 - X



a)

7x-i-Vy
(X -

=i-x3

b)

ir = y

f

(4x^ + 1)x + (y - 3)75 - 2y = 0
4x2 ^ y2 ^ 2^3 _ 4x ^ 7

Hifang din
a) D i l u kien x > 1, y > 0. He phuang trinh tu'ang du-ang vai:
(X

- 1)2 + x^ - 8 = 0

(1)
(2)


' 1 •!

- (t - 1)^ + t^ - 8, vai t > 1.

-6X

+ 11 < N / 3 ^ - V x ^ .

Hu'O'ng din
a) D i l u ki0n: x > - 1 . BPT vilt l^i: ^/xTl + 27x + 6 + sVx +13 > 20
Xet f(x) la ham so v l trai, x > - 1 thi:
1

1

3

f '(x) = — = = +

x-1

K i t qua X = 2 hoae x = 4.

Bai tap 1. 8: Giai eae phu-ang trinh;

= 0.

K i t qua nghiem duy nhit x = 3.
b) Ham dan di^u . K i t qua x =3.
Bai tap 1. 9: Giai eae h0 phu-cng trinh :

b) Vx^ - 2 x + 3 - Vx^

^
x-1

- I x li-^

b)Kltquax= 1 ^ ^ .

xVx

a) Vx7^ + 27x + 6 >20-3N/X + 13

Hirang din
^
2x-5

V

Bai tap 1.10: Giai bat phyang trinh:

x-1

b) V 3 - x + x2 - V2 + x-x2 =1

a) PT: (2x - 5)^ - (x - 1)^ =

• ....

b) K i t qua x=^;y = 2.

1

2x-5

VX

Xetham s6f(t)= TTl
Kit qua X =3, y =0.

27

1

X^'N/X

y = (x-1)'

a) y' e6 A' = a^ + 3(a + b") > 0, Va, Vb

J •;f!,X f

Chia 2 ve eho Vx"" thi du'p'e phu-ang trinh:

>/x^ -

Hu'6ng din

\/2.Tac6:

Vx^ - 2 = X + yjx^ - 1 > x > 1 = > x ' > 3 = > x >

\x^.\[x.

a) Tap xac dinh D = R. L i y y chia y'.
K § t q u a m # 1 v a y = - ( m - 1 ) ^ x + 2{m^ + m)(3m + 1 ) + 1 .

x^

a) D i l u kien: X >

'% .ist '30':ft

+ .
> 0 .Ket qua x > 3.
2Vx + 1 Vx + 6 2Vx + 13
b) K i t qua 1 < X < 2.
Bai t^p 1.11: Chiang minh phu-ang trinh c6 nghi$m duy nhit:
X

-x2 + 2 x - 5 = 0
Hiring din

- 5 x ' + 15x

Chung minh hdm VT ding biln tren khoang (0, +«), e6n khi x < 0 thi v6 nghiem.


W

trgng

diem

Churen

bSi dUOng

HQC sinh

aS 2:

gioi

mdn Toon

12 - L S Hodnh

Cty

Phd

MTVDWH

Hhang

Bu'ccc 3: Ve d6 thi
_ Tinh dgo ham d p hai, xet d i u d l chi ra dilm u6n cua ham da thu-c.
_ Cho vai gia tri d$c biet, giao d i l m vai hai true tog dp.
_ Ve dung d6 thi.

KHHO SIIT VA V€ D b THf HflM SO

1. K I ^ N T H U C T R O N G

TNHH

TAM

Vi$t

r,'

B i n dang dd thj ham bac 3: y = ax^ + bx^ + cx + d, a ^ 0 c6 tam d6i xij-ng la
d i l m u6n.

Tinh I6i lorn cua d6 thj:
Ham s6 f x^c djnh tr6n K Id mOt khoang, dogn ho$c nu-a khoang.
f gpi Id 15m tren K neu Va,p,a + p = 1: f(ax + py) < af(x) + pf(y), Vx,y s 0
f gpi Id I6i tr§n K n l u Va.p.a + p = 1: f{ax + py) > af{x) + pf(y),Vx,y > 0

Bon dang do thi ham trung phu-ang: y = ax + bx^ + c, a ?t 0

& i^fiiir -

(C)>{

O

Du'O'ng tiem c a n

Cho hdm so y = f(x) li§n tyc va c6 dgo ham cap 2 tr§n K

Duong thdng x = XQ dt^pc gpi la tiem can du-ng cua d l thi ham s6 y = f(x)
n l u it nhdt mpt trong cac dilu kien sau dupe thoa man:

-

f 16m tren K <=> f ' (x) > 0, Vx e K
f l6i tren K »

lim f(x) = + x ; lim f(x) = + x ; lim f(x) = - r \m f(x) = - «

f "(x) < 0, Vx e K.

Dilm udn cua dd thj:
D i l m U(xo;f(xo)) dLcp-c gpi Id diem u6n cua duang cong (C): y = f(x) n4u ton
tai mpt khoang (a;b) chupa d i l m Xo sao cho mpt trong 2 khoang (a,Xo), (xo,b)
thi tiSp tuy^n tgi difem U ndm phia tr6n d6 thj c6n 6- khoang kia thi ti^p tiiyln
ndm phia du'b'i d6 thj.
Cho hdm s6 y = f(x) c6 dgo hdm cdp 2 mOt khoang (a;b) chua d i l m XQ. N I U
f "(xo) = 0 vd f " ( X ) doi ddu khi x qua diem XQ thi U(xo;f(xo)) Id d i l m u6n cua
du'dyng cong (C): y = f ( x ) .

-

Duong

y = y ^ du-p-c gpi la tiem can ngang cua do thi

n l u lim f(x) = yo hode lim f(x) = yo.
-

Duong thing y = ax + b, a ^ 0 dup-c gpi la tipm cgn xien cua d6 thi y = f(x)
n l u lim [f(x) - (ax + b)] = 0 hoac lim [f(x) - (ax+ b)] = 0.

Chu y:
1) Neu chia tach du-gc y = f(x) = ax + b + r(x) va lim r(x) = 0 thi tiem can xien:
y

1) Neu y =p(x).y" + r(x) thi tung dO dilm u6n t?i XQ Id yo = r(xo)
2) N l u f l6i trSn dogn [a,b] thi GTLN = max{f(a); f(b)}
3) Neu f 16m tr6n dogn [a,b] thi GTNN = min{f(a); f(b)}
Khao sat va ve d6 thj ham da thirc: g6m 3 bu-d'c:
Bu'O'cl: Tdp xdc dinh
- Tdp xdc djnh D = R
- Xet tinh c h i n , le neu CO.
'
Bu-d'c 2: Si^ biln thi6n
- Tinh cac giai hgn.
- Tinh dgo hdm d p mOt, x6t d i u
- L$p bang b i l n thien r6i chi ra khoang d6ng biln, nghjch b i l n vd eye dgi,
eye tieu.

thing

= ax + b

b
Vx^TbxTc * x + —
2
Khao sat va ve d6 thj ham hOu t i : g i m 3 buac:
B u a c i : Tap xac dinh
~ Tim tap xac dinh
- Xet tinh c h i n , le n l u c6,
Buac 2: Chilu biln thien
2) Bilu thtpc tipm cgn khi x -> ± x :

Tinh cac giai han , tim cac tiem cdn
Tinh dao ham cdp mpt, xet ddu
Lap bang biln thien roi chi ra khoang dong biln, nghjeh b i l n va eye dgi,
cyctilu •


1 0 trQng

-

diem

b6i duang

hQC sinh

Hoanh Phi)

IP

g:oi mSn J,

Bii-dc 3: Ve
thj
Cho vdi gid trj dSc bi^t, giao diem v6'i hai tryc tog dOVe dung d6 thi, iu-u y tam doi xii-ng la giao diem 2 tiem c#n.
Hai dang d6 thj ham hu-u tf b^c 1/1: y =

cx + d

v6i c ^ 0, ad - be ^ 0

7

B6n dang do thj hdm hu-u tf :y =


a'x + b'

(a ^ 0, a V 0)

Chuy:
1) Tu- d6 thi (C): y = f(x) suy ra cac d6 thi:
y = - f(x) b i n g cdch Idy d6i xung qua true hodnh.
y = f(-x) bdng eaeh l l y doi xi>ng qua trye tung.
y = - f(-x) b i n g eaeh l l y d6i xung qua g6c.
y = I f(x)| b i n g eaeh l l y phin d6 thj a phia tr§n trge hoanh, eon phin
phia du'6'i trgc hodnh thi d6i xu-ng qua trye hoanh.
y = f(|x|) Id ham s6 ehdn, bIng edch l l y phIn d6 thj a phia ben phai true
tung, r6i l l y d6i xu-ng phIn do qua true tung.
2) Bai todn ve bien ludn s6 nghiem phu-ang trinh dgng g{x,m) =0
Ou-a phuang trinh ve dang f(x) = h(m) trong d6 v6 trai Id hdm so dang xet,
d § ve d6 thj (C): y = f(x). S6 nghiem Id so giao diem eua do thj (C) v6'i
du'6'ng thing y = h(m).
3) O i l m d$e biet cua hp d6 thj: (Cm): y = f(x,m)
-

Diem c6 dinh eua hp Id d i § m md mpi d6 thi d^u di qua:
Mo(xo, yo) e (Cm), Vm o yo = f(xo, m), Vm

-

Oi^m md hp khdng di qua Id d i l m md khpng c6 d6 thj ndp cua hp di qua vdi
mpi tham s6: Mo(xo, yo) « (Cm), Vm o yo

f(xo, m) Vm

Nhdm theo tham s6 vd dp dyng cdc m^nh d4 sau;
Am + B = 0, Vm o A = 0, B = 0
Am^ + Bm + C = 0, Vm «> A = 0, B = 0, C = 0
Am + B

5t

0, Vm <=> A = 0, B ^ 0

Am^ + Bm + C

?t

0, Vm <=> A = 0, B = 0. C ; t 0
ho0c A ^ 0, A = B^ - 4AC < 0

2.CACBAIT0AN

Bai toan 2.1: Tim d i l m u6n vd cdc khoang l6i fSm cua 66 thj:
a) y = x^ - 2x^ + x + 1
b) y = x" + 8x^ + 9.
Hiwng din giai
a) D = R • Ta c6 y' = 3x^ - 4x + 1, y "= 6x - 4.

'

y"= 0 <=> X = - ; y"> 0 <=> x > - ; y"< 0 <=> x < "• * •'
3
3
3
J; ;
2
29
2
Vdy 6\km uon l ( - ; — ) , hdm s6 l6i tren khoang ( - < » ; - ) vd 16m tr6n
3
27
3
2
khoang ( - ; + « ) .
3
*''•'•»•': \ •
b) D = R. Ta c6 y ' = 4 x ' + 16x , y" = 12x^ + 16 > 0 V x
Vdy do thj khong e6 d i l m u6n vd hdm s616m trSn R.
j
Bai toan 2. 2: Tim d i l m u6n vd cdc khoang l6i 16m cua d l thj:
x^ - 2x + 3
2x +1

''"-^-T;!-

"'""ITT
Himng din giai

a)D = R \ { - 1 } . T a c 6 y = ^ ^ - l ? ^ = x - 3 + ^
x+1
x+1
N6ny'=1

^ . y " =
(x + 1)2

01

^ 0, V x * - 1
(x + 1)='

-

y"> 0 <=> X >-1 ; y"< 0 <=> x < - 1
Vdy do thi khong c6 d i l m u6n, hdm s6 loi tren khoang (-oo;-1) vd 16m tren
khoang ( - 1 ; +oo).
b) D = R \. Ta c6 y' =

,y " = —
{x-5f

^ ^ 0,Vx ^ 5
(x-5)3

y"> 0 o X > 5 ; y"< 0 o x < 5
Vdy do thj khdng c6 d i l m uon, hdm s6 l6i tr6n khoang ( ^ ; 5 ) vd 16m tr§n
lSai toan 2. 3: Chu-ng minh ring v6'i mpi a, d6 thj hdm s6 y =

^ " ^ ^ Iu6n
x^ + X +1

c6 ba d i l m u6n thing hdng.
Hu^ngdlngidi
Ta c6- y- - (x^ + X +1) - (X + a)(2x +1) _
(x2+x + 1)2

x^ + 2ax + a - 1
(x2 + x + 1)2
35


Ctj/

„ ^ 2(x^ + 3ax^ + 3(a - 1)x - 1)
^

. 0 6 thi:
y" = 6 x - 1 2 ,
y"=0 « x = 2

(x^ + X + ^f

y" = 0 <=> x^ + 3ax^ + 3(a - 1 )x - 1 = 0
Dal f(x) = x^ + 3ax' + 3(a - 1 )x - 1, x e R
lim f(x) = -co, Mm f(x) = +oo va d6ng thai ham s6 nay lien tgc ten t$p s6
thyc nen phuang trinh f(x) = 0 c6 ba nghi^m phan bi^t thupc cac khoang
(-x;-1),(-1;0),(0; + « )
Gia s(y hoanh dp cua m6t trong cdc diem uon Id Xo n § n

Suy ra yo =

x^ + Xg + 1

_

(XQ +

Chox = 0 t h i y = - 1 .
b) Ta CO y' = 3 x ' - 1 2 x + 3m
06 thj ham s6 c6 d i l m eye dai, eye ti§u

o m<4

-mi d(.r

Gpi cac dilm eye tri Id A(xi; y,). B(x2; y2).

< » ( x o + 3 a - 1 ) ( x ^ + X o + 1) = 3(xo + a)
+a

%

nghi?m phan biet » A ' = 36 - 9m > 0

+ Sax^ + Sax^ + 3a - 1 = Sx^ + 3a

XQ

Vi$t

khi va ehf khi phacng trinh y' = 0 c6 hai

x^ + 3ax^ + 3(8 - 1)x, - 1 = 0
XQ

Hhang

n6n tarn d6i xi>ng la diem uon 1(2; 1). **

Taco: f(0) = - 1 < 0 , f(-1) = 1 > 0

Ta c6:

TNHHMTVDWH

3a - 1)(x^ + x^, + 1) _ x^ + 3a - 1
3 ( x ^ x „ + 1)

Theo dinh li Viet

X, + Xj =
x^Xj

4

=m

Ta e6 yi= (2m - 8)Xi + m + 2, y2 = (2m - 8)X2 + m + 2

Vay cdc diem u6n cua d6 thi thupc du'ang thing y = ^ "'^

— - nen chung

AB=

J(x,-X2)'+(2m-8)2(X2-x,)2

=

( 1 + { 2 m - 8 ) ' (x^ + X2r-4x,x,

3
thing hang
Bai toan 2. 4: Chp ham so: y = x^ - 6x^ + 3mx - m + 2, m la tham so.
a) Khao sat sy bi§n thien va ve d6 thi cua ham s6 khi m = 3
b) Tim m sao cho d6 thi cua ham s6 da cho c6 cac diem eye dai, eye ti§u A
va B ma khoang each AB = 4v'65 .
Hu-o-ng din giai
a) Khi m = 3 ham s6 tra thanh y = x^ - 6x^ + 9x - 1
• Tap xac dinh D = R

= V(4m2 - 3 2 m + 65)(16-4m)
nen AB = 4^65

o (4m^ - 32m + 65)(16 - 4m) = 1040

o 4m^ - 48m^ +193m = 0 « m(4m^ - 48m +193) = 0
<=> m = 0 (thoa m § n ) .Vay m = 0.
Bai toan 2. 5: Cho ham so: y = - - x ^ + (m - 1)x^ + (3m - 2)x - - c6 d6 thj
3
3
(Cm) vai m la tham so.
i>

• Sy bi^n thien: y' = 3x^ -12x + 9
y' = 0 < = > x = 1 v x = 3
Bang bien thien:

a) Khao sat sy bi§n thien va ve do thi cua ham s6 khi m = 2.
b) Tim m d l tren d6 thi ( C J eo hai dilm phan biet e6 hoanh dp ciing d i u va ti§p
tuydn cua (Cm) tai m6i di^m do vuong goc vai du-ang thing d: x - 3y + 1 =0.

+ 00

Hu'O'ng din giai
a) Khi m = 2 ham s6 tra thdnh y = x ^

Ham s6 d6ng bi§n tren mpi khoang ( - « ; 1 ) va (3;+ « ) , nghich bien tren (1;3).

Ham

s6 dat eye dai khi x = 1, yco =

x = 3ycT=-1.
36

3 va dgt eye ti4u tgi

• T$p xac dinh D = R
• Sy bi6n thien: y' = -2x^ + 2x + 4;
y' = 0c:>x = - 1 v x = 2.

^

+ x^ + 4x - - .
%^
A » - I


Ctv TNHH MTViyWH
10 trQng diSm

bSl dt/Ong

HQC sInh gl6l

mdn To6n

J2 - L S Hodnh

Hhang

V/^f^

PM

Hoanh do giao d i l m cua d vd do thj (C) id nghi^m cua phuang trinh

Bang i>iln thien
-1

-00

Ix3+Ix2_^x +2 = m

+00

6
2
2
,
i ;
Phu'ang trinh c : > x ' - 3 x ^ - 9 x + 1 2 - 6 m = 0
(1)
Ou-ang thing d c i t (C) tgi A, B thoa m§n tam gide OAB can tgi O khi

-+
+00

2 ~

-00

6 vd phu'ang trinh (1) c6 nghi^m x i , - X i , X2 (trong 66 x^, -x^ Id

m^ 0
Hdm s6 d6ng biln tren khoang ( - 1 ; 2) vd nghjch biln tr§n m6i khoang (-^; -1),
(2; + x ) .
Ham s6 dgt eye t i l u tgi x = - 1 vd ycr = - 4 , dgt eye dgi tgi x = 2 vd yco = 5.
• 06 thj:

" j

(2)

xf = 9
x^Xg = 1 2 - 6 m

Suy ra 1 2 - 6 m = 27 « m = — .

^^m d6i xi/ng.

b) y ' = - 2x^ + 2(m - 1)x + 3m - 2
H? s6 g6e cua d: x - 3y + 1 = 0 1^ k = 3
Tiep tuyen cua (Cm) t^i moi dilm vu6ng goc vb-i dub-ng thing d: x - 3y + 1 = 0
khi y' = - 3
« - 2 x ^ +2(m-1)x + 3 m - 2 = -3
« 2 x ^ - 2 ( m - 1 ) x - 3 m - 1 =0
Phuang trinh c6 hai nghi^m X i , X j , thoa m§n X i . X2 > 0
A' = ( m - 1 ) 2 +2(3m + 1)>0

m^ + 4m + 3 > 0

m< -3

-3m-1

1
m< —
3

1
-1
>0

Phuang trinh <=> x^ - X2X^ - x^ x + x^ X2 =0

Oong nhit cac he s6 cua (1) vd (2):

n§n d6 thj nh$n

d i l m uon ' ( ^ • ^ )

Khi d6 X i , X2 la nghi^m cua phu'ang trinh (x^ - x^ )(x - X2) = 0

x,=3

0 6 thj cat Oy tgi ( 0 ; - - ),
3
y" = - 4 x + 2 ;
y" = 0 = > X = ^

hodnh dp cua A, B)

V | y m < - 3 hay - 1 < m < — .
3
Bai toan 2. 6: Cho ham so: y = - x ^ - - x ^ - - x + 2 . Tim m d l hai d i l m A, B
6
2
2
thu6c do thj (C) CO tung dO m vd g l e O tgo thdnh tam giac OAB cSn tgi O.
HuHyng din giai
Hai dilm A, B thupc d6 thj (C) e6 tung dp m nen thuOc dyang thing d : y = m.

Bai toan 2. 7: Khao sdt sy biln thien va ve 66 thj cua hdm s6
a)y = -x^ + 3 x ^ - 4 x + 2
b) y = x^ - 3x^ + 3x + 1. ^
Hu'O'ng din giai
a) y = -x^ + 3x^ - 4x + 2
• Tgp xdc dinh D = R
^ r
• Sy b i l n thien lim y = - c c vd lim y = + x
Ta CO y' = -3x^ + 6x - 4 < 0, Vx nen hdm so nghjch b i l n tr§n R. Hdm s6
khdng c6 eye trj.
Bang b i l n thi^n
x

—00

+x

-

y
y

0

• 0 6 thi: y" = -6x + 6, y" = 0
X = 1 nen d6 thj c6 d i l m u6n 1(1; 0).
Cho

X

= 0 => y = 2. Cho y = 0

o -x^ + 3x^ - 4x + 2 = 0
o ( x - 1 ) ( x ^ - 2 x + 2) = 0 o x = 1 .

-2

\2


W trgng diSm bSi dUOng hqc sinh gioi m6n To6n 12 - LS Ho6nh Phd

b) y =
- 3x^ + 3x + 1.
• Tap xac djnh D = R.
• Sy biln thien: Mm y = - ^ c

f

Cti^ TNHHMTVDWH Hhong Vi$t

• Tap xac dinh D = R, ham s6 chin.
. Sy bi§n thien: y' = 4 x ^ - 1 2 x = 4x( x^ - 3)

Mm y = +co

y' = 0 c:> X =0 hoac x = ± s/'i

Ta CO y' = 3x^ - 6x + 3 = 3(x - 1)^ > 0, Vx nen ham so d6ng bi§n tren R,
ham so khong c6 eye trj.
Bang bi§n thien;

Bang bi§n thien
X

^

-V3
- 0 +

-00

y'
X

1

-X

+

y

0

-

y
-X

+00

y

^

^

0
0 5

VI
0

• 0 6 thj: y" = 6x - 6, y" = 0
<=> X = 1 nen d6 thj eo dilm u6n 1(1, 2).
Chox = 0i:i>y = - 1 .
Bai toan 2. 8: Cho h^m so: y = x^ - 3(m - 3)x^ + 3(m^ - 3m + 5)x + 1, m la
tham s6. Tim m 6k 6b thi cua ham s6 da cho dgt eye dai, eye t i l u tai x,, X2
thoa m§n I Xi + X2 - X1X21 < 7.

,

+00

+

^

+00

^

Ham s6 d6ng biln tren khoang ( - V s ; 0), ( V S ; +

^

,^ ,,

va nghjch bi§n tren

k h o a n g ( - x ; - 7 3 ) , (0; N/3).
Ham s6 dat eye dai tai x = 0, yco = 5 va dat eye t i l u tai x = ±\/3 , VCT = - 4
• D6 thi: D6 thj ham s6 nhan Oy tai true d6i xu-ng.

Hu'O'ng din giai
D = R,
y' = 3x^ -6(m - 3)x + 3(m^ - 3m + 5)
y' = 0 o 3x^ -6(m - 3)x + 3(m^ - 3m + 5) = 0
Ham s6 eye dai, eye tieu tai Xi, X2 khi phu-ang trinh c6 2 nghi$m phan bi$t
4
Xi, X2 <=> A' = -3m + 4 > 0 < = > m <
3
Ta CO Xi + X2 = 2(m - 3); X1X2 = m^ - 3m + 5.
Do do

I Xi + X2 -

< » I m^-5m
<=>

<

X1X21 < 7

+ 11

I <7

m - 5m + 11 < 7
m^ - 5m +11 > - 7

12(m - 3)^ + 3m - 51 < 7

b) Ta CO D = R. y'= 4x(x^ - m)
y' = 0

m - 5m + 4 < 0
- 5m +18 > 0

o 1
Ket hgp thi chon: 1 < m < - .
3
Bai toan 2. 9: Cho ham s6: y = x"* - 2mx^ + 2m - 1, vdi m la tham s6.
a) Khao sat sy biln thien va ve d6 thi ham s6 khi m = 3.
b) Tim m d l d6 thj cua ham s6 da cho c6 3 d i l m eye trj ISp th^nh mpt tam
gi^e vuong.
Hu'O'ng din giai
a) Khi m = 3, ham s6 tra thSnh y = x" - 6x^ + 5

4x(x^ - m ) = 0 « > x = 0 hoSc x^ = m

H^m s6 CO 3 d i l m eye trj <=> y' = 0 c6 3 nghiem phan bi^t <=> m > 0
Khi do 3 d i l m eye trj cua d6 thj ham s6 la:

*

i

A(-Vm - m^ + 2m - 1), B(0; 2m - 1), C( Vm ; - m ^ + 2m - 1)
Vi ham s6 c h i n nen tam giac ABC can tai B e Oy, A va C d6i xtrng nhau
qua Oy.
'•>

I \

ABC la tam giac vuong <=> tam giac ABC vuong can tai B |
<=> AC = AB. N/2 « m^ = \/m O m = 1 hoSc m = 0.
V$yehQnm = 1.

1


W tr
bSi duOng

HQC sinh

gl6l

mdn

To6n

12 -

Hodnh

Phd

_____

Bai toan 2. 10: Cho ham s6: y = x" - mx^ + 2m - 1, vai m la tham so . Tim m
de d6 thj h^m so cho c6 3 diem eye tri sao cho 3 d i l m eye trj cung vb-i g6c
tpa dp la 4 dinh cua mpt hinh thoi.
Hirang din giai
Ta CO y' = 4x^ - 2mx

Cty TNHHMTVDWH

Hhong

Vi$t

tri eye t i l u ycr = 2; ham so dgt eye dgi tai ede d i l m x = ± N/2 , gid trj eye dai
yco=6
• D 6 thj: nh^n Oy Id trye doi xung
. >,? > 1?; • j^v ; u ;

^x = 0
y' = 0 <=> 4x^ - 2mx = 0 <=>
2x2-m
06 thj ham s6 c6 3 d i l m eye tri khi
nghiem phan bi^t <=> m > 0.
Khi do cac d i l m eye trj;
m
2'

chi khi phyang trinh y' = 0 c6 3

m

m + 2 m - 1 ,B(0;2m-1),C
4

m''

+ 2m-1

Vi tam gi^e ABC can tai B, AC song song Ox nen O, A, B, C la 4 dinh hinh
thoi khi va chi khi OABC 1^ hinh thoi
<=> O va B d6i xyng nhau qua AC <=>
2m-1

° ^ ^ = y^

m' + 2 m - 1 » m - 4 m + 2 = 0
4

2

m = 2 ± V2 (thoa m§n). V^y m = 2 ± >/2 .
Bai toan 2.11: Cho h^m s6: y = - x** - 2mx^ + m^ + m, vd-i m Id tham s6
a) Khao sat sy b i l n thien va ve d6 thj cua ham so khi m = - 2 .
b) Tim m d l d6 thi hdm so c i t tryc hodnh Ox tgi 4 d i l m phdn bi^t A, B, C, D
sao cho AB = BO = OC = CD.
Hirang din giai
a) Khi m = - 2 ham so tra thanh y = - x " + 4x^ + 2
• T i p xac dinh D = R, ham s6 chSn
• Sy b i l n thien: y' = = -4x^ + 8x = -4x(x^ - 2)

0
+

y'

0

-

0

+00

+

0

+00

Hdm so d6ng biln tren moi khoang (-00; - N / 2 ) va (0; V2); nghjch b i l n tren
m6i khoang (-\/2

A' = m 2 + m 2 + m > 0

2m2+m>0

S ^ -2m > 0

m<0

P = -m^ - m > 0

m^ + m < 0

m >0V m<—
2
<=>-1 < m < —
m<0
2
-1
; 0) vd ( N/2 ; +«) . Ham s6 dgt eye t i l u tgi d i l m x = 0, gia

= 2 ^ » t j = 4ti.

Theo dinh li Viet ta c6 t, + t2 = - 2 m , tit2 = - m ^ - m.
Do do

y
+00'

Ddt t = x ^ t > 0 thi PT : - 1 ^ + 2mt - m^ - m = 0
Do thi est trgc hoanh tgi 4 d i l m phan bi?t khi phyang trinh b^c 2 c6 2
nghi^m dyang phan bi^t ti < t2.

BO = OC = CD khi va chi khi ^

Bang biln thien
-00

- x" - 2mx^ + m^ + m = 0

Vi d6 thj d l i xyng qua true tung nen 4 giao d i l m A, B, C, D thoa m i n AB =

y' = 0c=>x = 0 v x = ±\/2
X

b) Cho y = 0 o

5t, = -2m
4t? =-m^

-m

4.4m^ = 25(-m2 - m)

42m'' + 25m = 0 c : > m = 0 h a y m = Ta chpn m = -

25
41

25
41


10 trpng diS'm bSi duOng HQC sinh gioi m6n To6n 12 - LS Hodnh

Phd

S6 nghi^m cua phu-ang trinh
Bai toan 2. 12: Cho ham so y = - x " - 2x^ + 3 .
4
a) Khao sat sy bi^n thien va ve d6 thj cua ham s6
b) Tim m d4 phyang trinh I x" - 8x^ + 6 1 = m c6 8 nghi$m phan bi^t
V
Hiwyng d i n giai
a)y = - x ^ - 2 x 2 + 3 .
4

m
(C) va du'O-ng thSng y = — .

.c

Bai toan 2. 13: Cho ham s6: y = - x " - (3m + 1)x^ + 2(m + 1), vo'i m la tham

4 Stf bi§n thien: y' = x^ - 4x = x ( x^ - 4)

4
s6.

y' = 0 c=> X = 0 hay x = ± 2
Bang biln thien
X
-2
0

2

0
+

0

-

+CC

0

y

Tim m d^ d6 thi ham s6 eo 3 d i l m eye trj lap thanh mpt tam giac c6

trpng tam la g6c tpa do.
HLPang d i n giai

+

3

-ac

-1

; -<

0< — <1<:=>04

• Tap xac dinh D = R. Ham s6 chin.

-

la giao didm cua d6 thi

Dya vao d6 thj, phyang trinh c6 8 nghi$m phan bi?t khi va chi khi:

f

y'

-x^-2x^ +3 = —
4

y ' = x^ - 2(3m + 1)x = x[ x^ - 2(3m + 1)]
+

00

y ' = 0 « x = Ov x ' = 2(3m + 1)

^

Ham s6 d6ng bidn tren moi lmoi lt i l u tai X = +2, yci = - 1 .

Ham s6 da cho c6 3 dilm cue tri c> 3m + 1 >0t o m > - —
3
Khi do 3 d i l m eye tri cua d6 thi la: A(0; 2m + 2),
B(- v^6m + 2 ; -9m^ - 4m + 1) va C(

+ 2 ; -9m^ - 4m + 1)

Vi ham s6 c h i n nen tam giac ABC can tgi A thupc tryc Oy, B, C doi xung

• 0 6 thj: 0 6 thi (C) ham s6 nhan Oy la true d6i xCeng

y

nhau qua Oy.

j .

O la trpng tam cua tam giac ABC «
o

yA + ye + yc = 0

2m + 2 + 2 ( - 9 m ' - 4m + 1) = 0
2

<=> 9m^ + 3m - 2 = 0

«

m=

.
Chpn gia trj m = —
3

1
m=—
3
Bai toan 2.14: Khao sat sy biln thien va ve d6 thi cua ham so
a) y = - x * - 2x' + 5
b) Ta e6 phyang trinh
| x ' ' - 8 x ^ + 12| = m C3>

-x'-2x2+3

m

b)y= ^
' 2
Hiro-ng d i n giai

a) y = - x ' - 2x' + 5.
• T$p xac djnh D = R. Ham s6 chin
• Sy biln thien lim y = - x va lim y = - Q O

0 6 thj (C) cua ham s6 y = - x " - 2 x 2 + 3
4

dyo'c suy ra ty d6 thj (C) bing

each giy nguyen phin n i m phia tren Ox, con phan n i m phia dyo'i Ox thi
l i y d6i xyng qua Ox.

y = -4x^ - 4x = -4x(x^ + 1), y' = 0 o X = 0.

3
+ x2__,
2


BBT

X

-00

+

y'

+00

0

=x+2
| | | | | ^ TTaa CO
^ 6 y^ y=^ ,
x^ - 2x

-

0
5

y

+4^^^ n§n TCX: y = x + 2.
x''-2x

b)D = R ^ H ; | } s u y r a 2 T C D : x = - 1 v a x = | .

> ' ''

-00

H^m s6 dong bi4n tren khoang {-nghjch bi4n tren khoang (0; +«).
Ham s6 dat eye dgi tgi d i l m x = 0: yco = 5.
• D 6 thj: y" = -12x^ - 4 < 0, Vx nen do thj khong eo di^m u6n.

Taco lim y = - - n 6 n T C N : y = - - .

—+X

2

a)y = x + - ^
N/x

b y = N/X^ - 4 x + 3 .
•Hu'O'ng din giai

a) D = (0; +oo). Ta c6 lim y = +cc nen TCD: x = 0 (khi x -> 0*)

3
- - .

2
2
• T$p x^e dinh D = R: H^m so chin.

3
Ta e6 lim (y - x) = lim

-j=r

n^n TCX: y = x (khi x -> +oo).

• S y bien thien: lim y = +«.
X->±aC

y' = 2x^ + 2x = 2x(x^ + 1), y' = 0 o x = 0.
BBT
-00
+00
X
0

-

y'
y

0

+00

+

b) D = (-oc: 1] u [3;
D6 thi kh6ng c6 TCD.
Gpiy = ax + b l a T C N , TCXthi:
r
,.
y
,, \ / x ' ' - 4 x + 3
,.
, 4
3
,
a i = lim - = lim
= lim , 1 - - + — = 1;
x-»fA,

+00

X

x->^^*y

X

x-»+o-.

X

X

bi = lim (y - x) = lim [Vx^ - 4x + 3 - x)

"^-3/2"

Ham s6 d6ng bi§n tren khoang (0; +«), nghjch bi4n tr6n khoang (-«); 0)

CO d i l m u6n.

trye hoanh (-1;0) v a ( 1 ; 0 ) .
Bai toan 2.15: Tim c^c du'ang ti?m can cua do thj m5i h^m s6 sau:

x^ - 2x

x^ + X + 1
b)y =
' '
-5x^ - 2x + 3

Hirang din giai
a) D = R \; 2} suy ra 2 TCD: x = 0
x = 2.

AC

X

^ = — = -2
+i

x^

X

V$y ti0m c^n xien: y = x - 2 (khi x + x ) .
a2=

3
I
Giao d i l m vai trye tung (0; - - ) , giao d i l m vb'i

x3+2

^

V

• Do thi: y" = 6x^ + 2 > 0, Vx nen d6 thj khong

a)y =

-4x + 3
..
= lim
= — = lim ,
''-'^"^Vx^-4x + 3 + x

LiTT

3
dgt eye tieu tai (0; - - ) .

...

Bai toan 2.16: Tim c^c du-b-ng ti^m cgn cua d6 thi moi ham so sau:

Cho y = 0 =:> X = ± VN/B - 1
b) y =

^'x^^/.J,

,.
y
,. \ / x ^ - 4 x + 3
lim - ^ ^ lim
X ^ - ^

X

X-*-=r

X

1 ^ 3
- X . X1 - X
- + x^


= hm—^
X--•-=..

^2 = lim (y + x) = lim
X-f-OD

iim
x->t«^

x

X—>+ac ^

L

4

3
,
+ _ = -1

X

7x^ - 4 x + 3 + x

X

:


Tài liệu bạn tìm kiếm đã sẵn sàng tải về

Tải bản đầy đủ ngay

×