SỔ TAY TOÁN HỌC CỦA THẦY NGUYỄN TIẾN ĐẠT ÔN THI TN THPT CẤP TỐC

Sta.r book- Chuyen gia sach luy'n thi

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DAY DU cAc D~NG BAI HAY THI CUA BQ GIAO DI,JC

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TOANHOC
•
A

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ON THI DAI HOC CAP TOC

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T

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THAY NGUYEN TIEN OAT

LOI NOI DAU
Chao cac em hQc sinh THPT, d~c bi~t Ia hQc sinh l&p 12 chuAn hi
thi f)~i hQc. Chi con vai thang d€m nguqc d€n ki thi THPT QG 2021,
hi~n t~i dang la giai do~n C\fC ki gdp rut, quy€t dinh trgc ti€p d€n di6m
s6 cua cac em.
V &i nhfmg yeu cdu kh~t khe cua ki thi S~p t&i, cac b~n hQC sinh l&p
12 dang g~p phai "khling hmlng" v&i qua nhiSu mon hQc va ki€n thtrc
phai ghi nh&. DiSu d6 dfin d€n tinh tr~ng "hQc tru&c quen sau" va di6m
thi thu thuemg chi dimg l~i a mtrc 5-6.
Hi6u duqc nhfmg kh6 khan do, "Bq SA tay ki~n thii'c Chftt LU'f}'llg
Nhftt" da duqc Thdy bien so~n va due k€t kinh nghi~m tU cac anh chi
d~t 26-28 di6m thi f)~i hQc, giup cac em h~ th6ng ki€n thtrc ddy du hon,
tang t6c giai ds thi va d~t di6m cao.

f)~c bi~t v&i mon Toan, cu6n sA tay Toan hqc - On thi D~i hqc
cftp tfic" se la trq thu d~c lvc nhdt giup em btrt pha di6m 8+ trong ki

thi THPT QG 2021:
1.
2.
3.

H~ th6ng ddy du cong thtrc va each giai cac d~ng bai di6n hinh
xac sudt 90% xudt hi~n trong DS thi THPT QG.
C~p nh~t phuong phap giai nhanh m&i nhdt tang gdp 2 ldn t6c
d9 giai tr~c nghi~m, myo b(rt t6c dS thi v&i Casio, Vinacal.
Trinh bay ro rang, khoa hQc, dS tra ctru va vo cimg nho gQn,

ti~n lqi khi on t~p.

Cu6n "SB tay Toan hqc - On thi D~i hqc cftp tfic" duqc thi€t k€
phu hqp v&i mQi d6i tuqng hQc sinh, ill hQc sinh gioi d€n mAt g6c dSu
co th6 t1;r tra ctru dS dang!
Hy VQng cu6n s6 tay nay se dBng hanh cfing cac Sl tU chinh phl)C
canh CUa f)~i hQC thanh cong!

Than

ai,



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eve TRI .. .................................. ....................

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s6 dQt eve t~ tgi diem

Dong 3. Bi$11 luon hoanh do eve trl
Dc;mg 4
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Dc;mg 2. 11m gtln & gtnn

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BAI 6.

BA I 7.

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Tl~M Cf\N 80 TH! HAM S6 ..........
NHAN DI~N E>O THI
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PHlJONG TRlNH TI~P TUY~N .. ...

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Dc;mg 3. Tuong gioo hom phon t'lGc .....
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eua ham s6 y=f(x) tren kho6ng (o;b).........
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ham ~6 ·•-ffv\ t•em <1onn (a; b) ...

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BAI2. PT. BPT MO VA L()GARIT ............................................................................ 46
Dc;mg l. Pruong tri:lh mu ........ .. , .. ............................................ 48
Dong 2. Phuong t·inh logar't ....................................................... 50
DO:'lQ 3. Bot phi.Jong trinh mu ... . .. .. ... .. .. . . .. .. .. . .. .. .. .. . ...................... 51
D~:mg 4. Bot phuong Mnh logartt ........................................................ 53

DQng 5. Prllor~ M~"' oot prl1ong trlnh mO & logarf+ c "'ua m .............. 54

BAI3 BAI TOAN THI,IC
CHVdNG

.J.

Tf: LAI K£p ..................................................................... 56

nu.JH KHQNG GIAN .......... .............................................. 59

BAI 1. KH6t E>A DISN ............................................................................................ 59
BAI2. TH~ TfCH KH61 CHOP .... ..................................... .................................... 61
Dong 1. X6c dlnh di.JQC c hileu coo va dlen tlch day .. . .... .. .. .. . .. .. .. . 61

Dorg 2.
BAI3.

nsO Kh61 chOp ....... ~ · .. .... ... ... .. ............... .... .... .. . ..... .. .. ... . .. ..

..c.. A

MAr cAu- KH61 cAu ............................................................................... 66

BAI4.HlNHN61'l

KH61N6N .............................................................. 69

BAI5. HlNH TRV- KH61 TRV ............................................................................. .. .72
CHVdl''~"" 3: "IGilvEN HAM VA TfCf-1 PHAN ............................................. 74

BAll NGUY~N HAM ........ ....... ............................................................................. 74

Dc:;.rg 1. BOng "lguyen ham

co b6n

........ ........... ................ 74

Dc;~rg 2 Nguyen ham tl1ng phon ................................................. n

DQrg 3. 1'\Jguyem ham dc:;.rg huu tl ......................................... ......... . 18
DQrg 4. Nguyen ham dol b1en dc:;.ng 1 . .. . .. . . . . . .. . . .. . .. . .......... 80
Dana 5. Nguyen hOm d61 bien dQng 2 .. ........... ... .... ...... . .. .. .... .. .. . .. . . B1

BAI2. TfCH PHAN ............................................................................................. 86
0Qng I Tich phOn d61 b1en dc;mg ' ... ................................................. 86

Dc;mg 2. Tich phOr d o1 bien dc,mg 2. .................................................... 68

Dc:;.ng

'l.

n,_., phon tung phon . .. .. .. . .. .. . .. . ... .. . ....

. .. .. . .. . ... ... .... .. .90

BAI 3. UNG Dl,lNG T[CH PHAN ...... ....................................................................... 91


DQng 1. Dii;ln tfch hinh chang ............. ................................................. 91
D<;~ng 2. Th~ tfch khol tron xooy .................................................................. 92

CHlJdNG 5: SO PHVC ................................................. .......................................... '01

BAl l. D..;..NGD~IS6CUAs6PH0C ................................................................. 101
BAI2. BlfU DI~N HlNH HOC s6 PHCIC ............................................................... 117
DQng 1.

nm t¢p h9P d iem m bleu dien s6 phuc z thoo man dleu ki$n k . 11 7

DQng 2 Tim s6 phuc z c6 modul nho nhat. lon nhat thoo man elk k........ 119

8013. BAI lOAN U~N QUAN DfN .. .................................................................... 127
CHUONG 6: H~ TRVC TQA E>Q TRONG KHONG GIAN OXYl ................................. 128
BAll. H~TOAD¢TRONGKH0NGGIAN ......................................................... l28

BAt 2. Tl'CH CO HUONG VA CING DVNG .......................................................... 131
8AI3. PHUONG TRlNH MATCAU ....... ................................................................ l48
BAI4. PHUONG TRlNH MAT PHANG .................................................................. 15CI
BAI 5. PHUONG TRlNH ElUONG THANG ............................................................ 157
BAI6. HlNH CHI(U KHOANG CACH ........... -. -- ---...................................... 165

8AJ 7. M¢T s6 D~NG GIAI NHANH CI,IC TRl Kl-IONG GIAN ............................ 167
BA1 8. PHUONG ?HAP TOA f)O HOA ... ................. ............................................ 170



CHltdN

•. HAM

so

0

I>

1. Quy tac tinh c!qo ha : Cho u = u (X); v = v (X);
• T6ng, hi+ : (u ± v )'
•

= u

1

±V

• Dqo ham ham hqp. N~u
•

1
•

• (u.v )I = u I.v + v I.u => (C.u )I

T,.IC

0

ha

c : 1ft h~ng s6

y

= C.u I.

= f (u), u = u ( x) => y~ = y~ .u~

cap 2. /"(x)=[/ (x)J'
1

+ Y nghia ea h<;>e:
+ Gia t6e me thai eua ehuy6n d9ng s = f(t) t~;Li thai di6m t0
la: a(t0 ) = f"(t 0 ).

+ V~n t6e me thai eua ehuy6n d9ng s = f (t) t~;Li thai di6m t0
la:
• D(jiO

v(t0 )=f'(t0 ).

h m cap cao f("l(x)=[f"- 1l(x)J' ,(nEN, n;::::2).

2. Bang cong fhuc finh c!qo ham:

9


DfO ham cua ham SO' c

(f(x))'=f'(x)
r----·

(C la h~ng s6).

(c)' = o

( xa )' = a.xa-1

e-J

(J;)' =-1
2J;

( sinx )' =cosx

(cos x )' = -sin x

X

(

2
=-_!_
X

tanx)

I

1

= -2COS

X

(a x )' = ax .ln a

(e x)' =ex
(

1

I

1

I

(cotx) =--.sm 2 x

1

I

Inlxl) =-

(Iog a lxl) = xlna

X

fO ham c a am hqp (f(u))'=u'.f'(u)
'

'

(K.u)'=K.u'(Klahangso).

( ; } =-

(

-r--

(a)'a-1

u
-a.u .u

(~)' =__!{__

~~

2~

'T

srnu =u .cosu
•

)'

I

I

U

I

(cosu)' =-u 1.sinu

I

I

( cotu )' = --.-usm 2 u

(tanu) = -2cos u
( e" )' =u 1.e"

I (a")' =u .a".lna

[u [)'

I (log aiu I)' = _!!____
u.lna

( ln

1

= !!..._

3. Cong

u

U'c trnh nhanh dc;wo ha

(~

I

ax+b

10

)

ad -be
= (cx+d/.

an thU'c:


1i)
TOAN TIIAY lll,IT

0

Dfnh nghia
• Ham s6 y = f (x) duQ'c gQi la d6ng bi~n tren mi~n
D

Â:>

ã Ham s6 . y
D

Dlnh

¢:>

va

Vxp Xz ED

XI< Xz :=:;. f(xl) < f(xz)·

f (x) . duQ'c gQi la nghjch bi~n tren mi~n

=

va XI< Xz :=:;.

Vxp Xz ED

f(xl) > f(xz)·

ly

Gia sir y = f (x) c6

d~o

ham tren khoang (a; b), thi:

• N~u f'(x)>O, VxE(a;b):=:;. hams6 f(x) dbngbi~ntren
khoang (a;b).
N~u f'(x) < 0, Vx

E

(a; b):=:;. ham s6 f(x) nghjch bi~n tren

khoang (a;b).
• N~u f(x) dbng bi~n tren khoang (a; b):=:;. f'(x)
N~u f(x) nghjch bi~n tren khoang (a; b):=:;. f'(x) :-::;

o,

c. o,

Vx

Vx

(a; b).

E

E

(a; b) .

Khoang (a; b) dUQ'C gQi chung la khoang don di~U cua ham s6.

n

-

Tim cac khoang ddn
-

-

di~u (dQng khong chua m)
-

al min: Tim ca.c khoang don di~u (hay khao sat chi~u bi~n
thien) cua ham s6 y = f(x).
ng phap )
• Bu6'c t. Tim t~p xac djnh D cua ham s6.
11


Tinh d~o ham y' = f'(x) . Tim cac di~m

•
xi' (i

= 1, 2, 3, ... , n) rna t~i d6 d~o ham b~ng 0 ho~c khong xac dinh.
S~p x€p cac di~m

•

X i

theo thu n.r tang dAn va l~p

bang bi€n thien.
Neu k€t lu~n v~ cac khoimg dbng bi€n va nghich bi€n
•
dua vao bang bi€n thien.
~ D~ (

1.1·21

Tim tham

Bal toan

so di ham so ddn di~u tren miin d
on dl~u trin mien xac d!nh cua no

y = f(x ;m)

~ PhU'dng phap J
cr>

xet h

= f(x) = ax 3 + bx 2 +ex+ d .
- Buuc . T~p xac dinh: D = R
' 2 Tinh d~o ham y' = f'( x) = 3ax 2 + 2bx +c.

+ D~

y

f(x) dbng bi€n tren

JR.
af'(x)

<=>y'=f'(x);:::-:0, VxElR<=>

+ D~

{

~ J'(x) =

~

a f'(r)

0, \fx E lR <=>

Dftu cua tam thuc b~c hai f(x)

{

= 3a

· _

~

~m?

0

~ J'(x)

-

, f(x)~O, VxElR<=> {a>O ·
,

~~0

f(x)~O,

<0
2

4b -12ac

= ax 2 + bx + c.

De

De

12

2

4b -I2ac

f(x) nghich bi€n tren

lR <=> y' = f'(x)

•

= 3a > 0

VxElR<=>

{a~~0

~

~ m

0

?


1i)
TOANTNAY DI)T

({) Xet ham sA nhftt bl
Btr&c
- Btruc

ax+b
y=f(x)=--·
cx+d

T~p xac dinh: D = ~ \ {- ~} ·

Tinh dao ham y' = f'(x)
·

= a.d- b.c2 ·
(cx+d)

+ DS f(x) d6ng bi~n tren
D<;::::>y'=f'(x)>O,

VxEDGa.d-b.c>O~m?

+ DS f(x) nghich bi€n tren
DG y' = f'(x)
VxEDGa.d-b.c
D6i v6i ham phiin thuc thi khong c6 dfru

Baltoan

y=f(x;m)

D CO th

l, PhU'c:ln
dt'

n

11

=

11

xay ra t~i vi tri y'.
D

rongd6

(-oo;a), (a;+oo), (a;f3), ( a;f3 ], [ a;fJ),

pha-;-)
Ghi di~u ki~n dS y = f(x;m) don di~u trenD.

Ch~ng h~n:
D~yeucfiu y =f(x;m) d6ngbi~ntren D<;:::?y'=f'(x;m)';?.O.
D~ yeu cfiu y = f(x;m) nghich bi€n tren D <;:::? y' = f'(x;m) -s, 0.

D(>c l~p m ra khoi bi€n s6 va d~t v~ con l~i la g(x)
m 2. g(x)
duac:
·
[
·
m ~ g(x)

Khao sat tinh don di~u cua ham s6 g(x) tren m~n
xac dinh D.

13


Dva vao bang biSn thien kSt lu~n:

riTe

Khi m ~ g(x) ::::::> m ~ mgxg(x).
[ Khi m ~ g(x) ::::::> m ~ ming(x)
D

Bai toan 3

y

= f(x;m) = a'x3 + b'x 2 + c'x+ d

trin kh

Tinh y' = f'(x;m)

- Brrtlc

Ham s6 don di~u tren (x,;x2 )

Brrt1c

ng h H~m p hAan b"A
lttt
"A

= l ¢::> lx, - x2 1= l
S2

true

- Btrtlc

¢::>

{L1>0
a :t0

(z")

4P = 12 .

-

¢::>

(x1 +

x

2)

2

-

(ii)

Giai (ii) va giao v6i (i) dS suy ra gia tri m dn tim.

y

1

=f

1
(

x; m) = ax 2 + bx + c .
-

4ac Ta se giai ra duqc m
0

Thaynguqc m vaopt y 1 =f'( x;m)=ax 2 +bx+c =0.
NSu ra 2 nghi~m thi ch9n m d6.

14

= 12

4x1.x2

Ap dl,mg cong thuc IL1 = (azt I v6i L1 = b 2

- Brruc

<=> y' =0 c6 2

Ham s6 don di~u tren khoang c6 d() dai

B -6'

¢::>

= ax 2 + bx +c.


li)
TOR~.!!!~.! Dill

Bill toan 1. xac dfnh tin ddn di'u cua ham so y
thong qua do thl y

• N~u db thi ham

=

s6

=

f (X)

f'(x)

y = f' (X) n~m

a phia

tren Ox tren D

-+/'(x) > O.Khid6 y=f(x) dbngbi~ntrenD
• N~u db thi ham

s6 y = f' (X) n~m a

---+ f'(x) < 0. Khi d6 y

Bill toan 2.
hams

lPh&tdn
Xetha
- B l1

g '(x ) =

=

phi a du6i Ox tren D

f ( x) nghich bi~n trenD

ho doth. f'(x). Hoi khoang ddn di'u cua

J[u(x)J.

~
g(x)=f[u(x)J.
Tim TXD va D;;to ham

(![ u (x )])' = u '( x ).f'[ u( x )]

u'(x) = 0
- Bwc 2. Xet g'(x) = 0 ~ [ /'[ u(x)] =

0

~ x= xi. i={1 ;2;3 ... }

- Brrtfc 3 L~p BBT.
- Bmk K~t lu~n.

15


1i)
TOll! TN4Y II(IT
0

,,

I>

a· c

• .:!

Cho ham y = f(x) xac dinh va lien tt.Ic tren (c6 th~ a la
la +oo) Va

X0 E

-

h;

b

(a;b):

N@u t6n t~i s6
X E (X 0

-CXJ,

X0

h

+h) Va

sao cho

X :;z: X 0

f(x) < f(x 0 )

v6i mQI

thi ta n6i ham SfJ j(x) d~t C\fC

d~i t~i di~m xo.

N@u t6n t~i s6 h sao cho f(x) > f(xJ v6i mQi
X E (X0

-

h;

X0

+h) Va

X :;z: X 0

thi ta n6i ham SfJ j(x) d~t

qrc ti~u t~i di~m xo.

ca
*

Gh\ su y = f(x) lien tt.Ic tren khoang

K = (xo -h; xo +h) va c6 d~o ham tren K ho~c tren K\{xo},

v6i h > 0. Khi d6:
N@u f'(x) > 0 tren khoang (xo- h; xo) va f'(x) < 0 tren
khoang (xo; xo +h) thi xo lam9tdi~mcl,lcd~icuahams6 f(x).
N@u f'(x) < 0 tren khoang (xo- h; xo) va f'(x)

>

0 tren

khoang (xo; xo +h) thi xo lam9tdi~mcl,lcti~ucuahams6 f(x).
x

f'(x)
f(x)

16

.,·,+It

x0 - lt

0

+

A"

f'(x)
f(x)

~·.-It

x0 +h

x,
+

0

-

~fco ~

J


c·
d6i diu tir m ano

khi . x. di qua di~m

lfO'Il

(theo chi€u tang) thi ham s6 y = f(x) d~t ClJC ti~u t~i di~m

X0

•

khi x di qua di~m xo

a

N€u f'( x ) d6i diu tir

X0

(theo chi€u tang) thi ham s6 y = f(x) d~t ClJC d~i t~i di~m xo.
Khi d6 di~m M(x0 ;f(x0 ) ) g9i la di~m ClJC

tti (ClJC

d~i ho~c ClJC

ti~u) cua ham s6 v&i Yo = f(xJ gQi la gia tri ClJC tri cua ham s6.

tnh
(xo - h;

Gia sir y = f (x) c6 d~o ham dp 2 trong khoang

'
X0

+h), v6i h > 0. Khi d6:

N€u y '(x0 ) = 0, y"(x0 ) > 0 thi

X0

la di~m ClJC ti~u.

N€u y'(x0 ) = 0, y"(xJ < 0 thi

X0

Ia di~m ClJC d~i .

DANG·.

;1 -.
;

Tim

CCC

di~m

.

Cu'C

dai
. &

.

Cu'C

fi~U

CUC

ham SO

Phltdng phap )
•

Buuc I Tim t~p xac dinh D cua ham s6.
~
Tinh d~o ham y' = f'(x). Tim cac di~m xi' (i = 1,2,3, ... ,n)

rna t~i d6 d~o ham b~ng 0 ho~c khong xac dinh.
S~p x€p cac di~m X; theo th(r tlJ tang dftn va l~p bang
bi€n thien.
Tir bang bi€n thien, suy ra cac di~m ClJC tri (dlJa vao
n9i dung dinh ly 1).

17


'MIJlUP!It

:r '1lM Vat ft
. ~II'

m

Tim tham

SOd~ ham SO dQt CI/C trj fQi di~m Xo

l, PhU'dng phap

)

• Btruc I . Tim t~p xac dinh D cua ham s6.
• Butte 2 Tinh d~;to ham y' va y".

• B~rO'c 3. Dva vao yeu c~u bai toan, ghi diSu ki~n va giai h~ tim
tham s6. C1,1 th€:
o

,
:..
Ham so d~;tt

o

:..
,
Ham so

o

,
:..
. . .;(
Ham so d~;tt eve tn t~;tt dtem x

.

C\fC

d~;tt C\fC

. .;(

d~;tl t~;t1 dtem x = xo ~

{y'(xJ = 0
y "(xJ < ·
0

.;(
. .;(
{y'(xJ = 0
tleu t~;tt dtem x = xo =>
,
·
y (x0 )>0

= xa

{y'(xJ = 0

~ y"(xJ

* 0·

c 4 V&i m vira tim duqc, th~ vao ham s6 va thir l~;ti (ve
bang bi~n thien va nh~n, lo~;ti) .

N~u dS bai yeu c~u tim gia tri eve tri tuong wg, ta seth~
X=

xo, m =? vao y = f(x). Con n~u dS bai yeu du xac dinh t~;ti

do la di€m

C\fC

d~;ti hay

C\fC

ti€u, ta th~

gia tri y"(xJ > 0 =:>X= X 0 la di€m

C\fC

y"(xa) < 0 => x = xa la di€m eve d~;ti.

18

X

= X m = ? VllO y",
0

,

ti€u Va n~U

n~U


D~NO

Bi~n

3

lu(ln hoanh d() cl!c trj ham b(jc 3

Ta c6: y' = 3ax 2 + 2bx +c.
Di~u ki~n

b2

-

3ac ~ 0. K~t lu~n : Ham s6 khong c6 qrc

b2

-

3ac > 0 . K~t lu~n: Ham

tri.

s6 c6 hai di~m eve tri.

» Di4u kl'n di ham s6 c6 dleni CI/C trl

cung dau.

trai dau.
Phlldng phap )
Ham

s6 c6 2 di~m eve tri tnii d~u

<=> phuang trinh y' = 0 c6 hai nghi~m phan bi~t tnii d~u
<=> A.C = 3ac < 0 <=> ac < 0.
Ham

s6 c6 hai di~m eve tri cling d~u

<=> phuong trinh y ' = 0 c6 hai nghi~m phiin bi~t cung d~u

Ham

s6 c6 hai di~m eve tri cung d~u duang

<=> phuong trinh y' = 0 c6 hai nghi~m duang phiin bi~t

19


B

<=> S = x1 + x2 = -- > 0 . Ham s6 e6 hai di~m eve tri eung d~u am
A

c >0

P = x1.x2 = A

<=> phuong trinh y' =0 e6 hai nghi~m am phan bi~t

B

<=> S=x1 +x2 =--<0
A

c

P=x1.x2 =->0
A

, Tim dleu kl9n di ham

so co hai cl!c

man:
X1
)

\

x1 < x 2 a
Hai eve tri x1 , x2 thoa man x1 < a < x 2

<=> ( x1 - a) ( x2 - a) < 0 <=> x1 .x2 - a ( x1 + x2) + a 2 < 0
Hai eve tri Xp x2 thoa man xl < x2 < a

2
<=>{(x1-a)(x2 -a) >O<=>{x1.x2 -a(x1+x2)+a >0
x1 + x2 < 2a
x1 + x 2 < 2a
Hai eve tri Xp x2 thoa man a < xl < x2

20


1i.)

TOAN THAY l)f:IT

~{(x1 -a)(x2 -a) >O~{x1 .x2 -a(x1 +x2 )+a
x1 + x2 > 2a

2

>0

x1 + x2 > 2a

dol glii'a

em

dU'bng th&ng:

ph~
Cho 2 diem

A( xA;yJ, B( xB;yB)

va dm'm.g th~ng ~ : ax+ by+ c = 0.

NSu (axA +byA+c)( a.xs +bys +c) < 0 thi hai diem A, B n~m v~ hai
phia so v&i dm'm.g th~ng ~.

NSu (a.xA +byA +c)(a.xB +byB +c) > O thi hai diem A, B n~m cung
phia so v&i dm'm.g th~ng ~.

~ M9t

so tntbng hqp d(ic bi+t:
\

+ Cham s6 c6 2 eve tri cung d~u
~phuang trinh y' = 0 c6 hai nghi~m phan bi~t cilng d~u

+ Cac diem C\fC tq cua d6 thi n~m cung v~ 2 phia d6i v&i ttvc Oy
<::::> ham sf> c6 2 eve tq trai d~u
<::::> phuang trinh y' = 0 c6 hai nghi~m trai d~u

+ Cac diem C\fC tri cua d6 thi n~m cung v~ 1 phia d6i v&i ttvc Ox
<::::> phuang trinh y'

= 0 c6 hai nghi~m phan bi~t va Yc• ·Ycr > 0

Dac

+ Cac diem C\fC tri cua d6 thi n~m cung v~ phia tren d6i v&i ttvc Ox
21


¢::?

phuong trinh

y' = 0

c6 hai nghi~m phan bi~t va

Yc•·Ycr > 0
{

Yc •

+Ycr >0

Cac di~m ClJC tri cua d6 thi niim cung v~ phia du6i d6i v6i ttvc Ox
Â::?

phuong trinh

YcãÃYcr > 0
{

Yc•

y' =

o

co hai nghi~m phan bW va

.

+Ycr < 0

+ Cac di~m ClJC tri cua d6 thi n~m v~ 2 phia d6i v6i tf\lc Ox

¢::>

phuong trinh y' = 0 c6 hai nghi~m phan bi~t va Yc• ·Ycr < 0

(ap d\lng khi khong nh~m duqc nghi~m va vi~t duqc phuong
trinh dm'mg thllng di qua hai di~m ClJC tri CUa d6 thi ham s6)
Ho~c: Cac di~m ClJC tri cua d6 thi n~m v~ 2 phia d6i v6i ttvc Ox
¢=>db thi cit ttvc Ox t~i 3 di~m phan bi~t

¢=> phuong trinh hoanh d() giao di~m

f ( x) = 0

c6 3 nghi~m

phan bi~t (ap d\lng khi nh~m duqc nghi~m)

y'.y"

PhU'O'Ilg trinh darimg thing qua de di~m e'le t i y d = y- ~
Khoing deb giiia hai cli~m qre trj cua
AB =

22

~ 4e +16e3
a

, •

etA thj ham sA b'e 3 Ia:
b

2
-

3ac

vat e = - - -

9a


-- - -

~

IAI

-~

Ct/C trj cua ham trung phu'dng

Mc)T s6 KiT

au.\ CAN NHtJ

+ Ham s6 c6 m<)t qrc tri <::} ab ~ 0.
+ Ham s6 c6 ba eve tri <=> ab < 0.
+ Ham

,
so, c6 dfulg m<)t ClJC tri va ClJC tri la ClJC tieu

a=O
{b>O
,
Ham so c6 dung m<)t

{a >O

<=> b ~ 0

hoac
.

+

hoac
.

+

Ham

ClJC

tri va

ClJC

tri la

ClJC d~i

{a
<=> b s 0

a=O
{b < O

, va m('>t ClJC d~i <=> {ab >
O
so, c6 hai ClJC tieu
<0.
,

,

{a
+ Ham so c6 m<)t ClJC tieu va hai ClJC d~i <=> b > 0 .
Mc)T s6 C0NG THU'C GtAI NHANH
Gia sir ham s6

y = ax 4

+ bx 2 + c c6 3cvc

tri t~o thanh tam giac

ABC thOa man dfr ki~n: ab < 0
Dti' kifn

Congthllc
thoa min ab < 0

Tam giac ABC vu6ng can tl;l.i A

23


Tam giac ABC d~u

b3 =-24a

Tam giac ABC co di~n tich SWJc

=S0

Tam giac ABC co di~n tich max(S0 )

Tam giac ABC co ban kinh du(mg tr(m n9i tiSp

Tam giac ABC co ban kinh duoog tron ngo~i tiSp

RMBC =R
Tam giac ABC co d9 dai c~nh BC
Tam giac ABC co d9 dai AB

= m0

= AC =n0

-+

am; +2b = 0

l6a n~- b

2

4

+ 8ab = 0

Tam giac ABC co cgc hi B,C E Ox

b2 = 4ac

Tam giac ABC co tr<;mg tiim 0

b 2 = 6ac

+8a- 4ac = 0

b3

Tam giac ABC co trgc tiim 0

Tam giac ABC cung diSm 0 t~o thanh hinh thoi

b2 = 2ac

Tam giac ABC co 0 la tiim du(mg tron n9i tiSp

b3

-

8a- 4abc = 0

Tam giac ABC co 0 la tam du(mg tron ngo~i tiSp

b3

-

8a- 8abc = 0

Tam giac ABC co c~nh BC

= kAB = kAC

Tam giac ABC co diSm C\JC tri each d~u tn,Ic hoanh

24

b3

.e - 8a(e - 4) =0
b 2 = 8ac


1i)
TOANTHAY ~T

Bat toan 1. Cho
ham

cf6 thl f'(x).

Oi SO dlim CI/C frl

CUD

so f(x)

Phlldng phap )

• Db thj cua f '(X)
C\l"C

c~t va bang qua ttvc hoanh ~i n diem nen c6

trj.

• Db thi ham s6

y=

f' (x) c~t ttvc Ox t~i

X0

,c~ttit dudi lin lrin

---+ f'(x) d6i d~u tir am sang duang. Khi d6 ham s6 y
d~t

C\l"C

ti€u t~i

=

f(x)

X0

• Db thi ham s6 y =

f' ( x) c~t ttvc Ox t~i

X0

,c~t tit tren ~uang

dutYi ---+ j '(X) d6i d~U tlr duang sang am . Khi do ham sfJ
y = j

(X)

X
0

ho cf6 thl f'(x) . Hoi

Bal toan 2.
ham

d~t C\l"C d~i t~i

so dlim CI/C tr! cua

so f[u(x)].

PhU'dng p ap )
Xet ham g(x) =f[ u(x)J.
Buth: 1. Tim TXD va D~o ham
g'(x) =(![ u(x )])' = u '( x ).f'[ u(x )]

,

,

- BU'O'c2. Xet g (x) = 0 ¢::>

[

u'(x)=O

/ ' [ u(

x)] = 0 ¢::> x =X;·

. ._ . .
Vmi-{1,2,3... }

- Buth: 3. L~p BBT.
- Buth: 4. K~t lu~n.

25