18 CHỦ ĐỀ GIẢI TÍCH 12 - NGUYỄN TẤT THU

8/9/2019 18 CH GII TCH 12 - NGUYN TT THU

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NHM BIN SON SCH B TR g i o d c OLYMPIC

Ch bin: NGUYN TT THU - NGUYEN v n d n g

y Cc dng on rng m./ Dnh cho hc inh p 12

chng rnh C bn Nng cao. .

S Nng cao k nng lm bi.-/ Bi dng HS kh ii chn b

cho cc k hi TI\I TSH.

nh x ut bn i hd quc giah

n

WWW.FACEBOOK.COM/DAYKEM.QU

WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU

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W.DAYKEMQUYNHON.UCOZ.COM

ng gp PDF bi GV. Nguyn Thanh T


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PMU0MGPHr


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NH XUT BN I HC QUC GIA H N16 H ng Chu i Hai B Trng H Ni

in thoi: Bin tpCh bn: (04) 39714896;Hnh chnh: (04)^9714899; Tng bin tp: (04) 39714897

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ANH TH

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CNG TI ANPHA

Trnh by baSN Ki ta t lin kt xut bn

CNG TI ANPHA

SCH LIN KT

18 CH G I TCH 12

M s: 1L-538H20IOIn 2.000 c u n , kh 16 X 24 cm ti Cng t c phn in Bao b Hng PhS xu bn: 1166-2010/CXB/3-219/HQGHN, ngy 01/12/2010Quyt (lnh xut bn s': 544LK-TN/ -NXB H 0 GHNIn xong v np lti chiu qu II nm 2011.



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LI NI U

Khi mi chng ta 18 tui cng l thi gian hc lp 12, nm cui cpca chng trnh ph thng. Mi bn hc sinh u c k hoch cho tngai, m khi u l s vt qua k thi tuyn sinh vo mt trng i hc

mong c.

Nhm gp phn gip cc em hc sinh thc hin c m c , btu t vic nm vng kin thc c bn, pht trin t duy, nng co kh nngvn dng, phn tch v tng hp gii quyt vn , chng ti bin- son bsch v 18 ch ong chng trnh ton hc 12.Trong mi ch gm ba phn

I. Tm tt l thuyt: h* thng ha cc kin thc trng tm.II. V d minh ha: gm nhng v d in hnh cho mi phng php

gii ton tng ng vi ch . Cc v d ny c sp xp theo mt logic nht nh gip cc em c th tng qut ha v c k nng gii cc bi ton

ng t.III. Bi tp: gm mt h thng cc bi tp cc em c th t luyn tpnhm khc su kin thc vrn luyn k nng.Trong b sch ny, vi tinh thn tng cng trch nhim ca mi tc gi,cng nh thun tin cho vic trao i ca c gi, chng ti phn cng:

+ Tc gi Nguyn Tt Thu ph trch chnh cun 18 ch gi tch 12+ Tc gi Nguyn Vn ng ph trch chnh cun 18 ch hnh hc 12.Chng ti tin tng rng vi b sch ny, mi em hc sinh u c th

m thy nhng iu ti v v b ch.

B sch c hon ii vi s gip ca cng ty Sch - thit b gio dc Ajnpha, nhng kin ng gp qu bu ca cc bn ng nghip v sng vin ca gia nh. Cc tc gi xin chn thnh cm n!

Mc d mi tc gi dnh nhiu tm huyt cho cun sch, song s saist l iu kh trnh khi. Chng ti rt mong nhn c s phn bin vgp ca qu c gi nhng ln tibn sau cun sch c hon thinhn.

Mi kin ng gp xin gi v a ch:- Trung tm Sch gio dc Anpha

J25C Nguyn Tn 'Phng, P.9, Q5, Tp HCM.- Cng ti Sch thit b gio c Anpha50 Nguyn Vn Sng, Q. Tn Ph, Tp H Ch MinhT: 08.62676463, 38547464Email: alphabookcenter@,vahoo.comTrn ng cm n!

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T I L I U T H A M K H O

[1].Bo Ton hc tui tr ,Nh xut bn gio c.

[2]. Cc th tuyn sinh i hc v cao ng,B gio dc v o to.

[3].Sch gio khoa Ton 12,Nh xut bn gio dc, 2008.

[4]. Phan c Chnh, Phm Vn iu, Vii H, Phan Vn Hp,

Phm Vn Hng, Phm ng Long, Nguyn Vn Mu, Thanh

Sn, L nh Thnh,Mt s phng php chn lc gii cc bi ton

s cp,Nh xut bn gio dc, 1995.

[5]. Nguyn Vn Dng, Nguyn Tt Thu, Tm tt Mn thc ton THPT

(di dng bng v s ),Nh xut bn gicndc, 2010.[6]. Nguyn Vn Dng, Phng php gii ton s phc v ng dng,

Nh xut bn i hc quc gia H Ni, 2010.

[7]. Nguyn Tt Thu, Nguyn Ph Khnhy Trn Vn Thng, Phn

dng v phng php gii nhanh Gii Tch 12,Nh xut bn i hc

Quc gia thnh ph H Ch Minh, 2008.

[8]. Din n ton hoc http://www.math.vn.

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CH 1. TNH N IU

I. Tm tt l thuyt1. nh ngha:

I Gi s K l mt khong, mt on hoc mt na khong. Hm s f xc

nh trn K c gi l:e ng bin trnK nu vi mi x:, x2 K, Xx< x2 => x-L) < f(x2)I e Nghch bin trnK nu vi Vx1?x9 e K, Xj < x2 => f(Xj) > f(x2) .

2. iu kin cn hm s crn iu:Gi s hm s f c o hm trn khong Ie N en hm s f ng bin trn khong I th f' (x) > 0 vi mi X I

Neuhms fnghch binnkhonglh f'(x )< 0 vi 1 x e 3. iu kin hm s om iu:

Bnh :

Gi s I l mt khong hoc na khong hoc mt on, f l hm s lintc trn I v c o hm ti mi im trong ca I (tc l im thuc nhng khng phi u mt ca I ). Khi :e Nu f (x) > 0 vi mi X I th hm s f ng bin trn khong

: Nu f'(x) < 0 vi mi Xe I th hm s f nghch bin trn khong I Nu f'(x) = 0 vi mi Xe I th hm s f khng i trn khong

Ch : Nu hm s f lin tc trn [a;b] v c o hm f'(x) > 0 trn khong

(a; b) th hm s f ng bin trn [a; b] e Nu hm s f lin tc trn [a;b] v c o hm f (x) < 0 trn khong

(a;b) th hm s f nghch bin trn [a;b].

Ta c th m rng nh l trn nh sauGi s hm s f c o hm trn khong I. Nu f '(x) > 0 vi Vx e (hoc '(x)


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1) TX: D = R

Ta c y ' = 3x2 - 6x = 3x(x - 2) => y' = 0

Bng bin thin:_____________________

Li gii.

X = 0X = 2

0+ 0 - 0 +

+00

+00

Vy hm ng bin trn mi khong (-co; 0) v(2;+co), nghc'(0; 2).

2) TX: D RTa c: y ' = -X 3 - 3x = -x(x2 + 3)=>y = 0 o x = 0

1bin trn

X co 0 +00

y' + 0

y 00 ^ -0 0 1

Hm ng bin trn khong (-00; 0), nghch bin n(0;-hco).

3) TX: D = R \ { l } . Ta c: y' = ----- ^ - < 0 V x* l

Hm nghch bin trn tng khong xc nh.

4) TX: D = R \! - l .T a c : y = .4?2+8x' - < ; * (x +1)

y = 0 4x2 + 8x - 0 X= 0, X= -2 .

X co - 2 -1 0 +Q0

y + 0 +o

1

y c o 00

+00 +00

Hm ngfcdch bin trn cc khong: (-2 ;-l) v(-l;0).5) TX: r> = K

nr /71 ---- 7*2 _ , _ 2(x - l)(x2 - 2x - 3)Ta c: y = >/(x - 2 x - 3 ) =>y = ---- = = ^ ~ .v/(x2 - 2x - 3)2

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y ' = 0 X = 1, h m s kh n g c o h m ti X = - l , x = 3

Bng bin thin:______ '___________________________X 00 1 1 3 +00y - II + 0 - 11 +

y

+00

^ 0 ^

4 +00

Hm ng bin trn mi khong (-1; 1) v (3;+oo), nghch bin trn

(-00J-1) v (1;3).

TX: D = (-oo;3.Tac: y' = -^ 2 i^ = L = > y ' = 0 x = 2 .

Hm s khng c o hm ti cc imx = 0,x = 3.Bng bin thin:_____ ______________________________ _

X -00 0 2 3y' - II + 0 - 11y

Hm ng bin trn khong (0;2) aghch. bin trn (- 00; 0) v (2;3) -Nhn xt:e Bi ton xt tnh n iu ca hm s c chuyn v bi ton xt duca mt biu thc ( y ').

Khi tnh o hm ca hm s c dng y = |f (x), ta chuyn tr tuyt ivo trong cn thc y = v 2(x ), khi ti nhng im m f(x) = 0 th

hm s khng c o hm.d 2.1. Chng minh rng hm s y = sin 2x~2x + l lun nghch bintrn EL

Li gii.TX: D = K.. Ta c: y' - 2cos2x - 2 = 2(cos2x -1 ) v(x ) > v(xn) =>hm s nghch bin n K.



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Khr xt tnh n iu ca hm s cha hm lng gic chng ta lu l o hm ca hm s c th trit tiu ti vo hn im. Khi xt tnh on iu ca hm s n TX, ta s chuyn v xt tinhdond

mt khong cha hu hn im m ti o hm trit tiu.i vi hm a thc nu tt c cc h s khong ong thi bng 0

n ch trit tiu ti hu hn im.V d 3.1. Tm m hm s sau ng bin trn R

X3y = (m + +2)x2 - (3m - l)x + m2 .

L i gii:

Ham s xc nh trn R . Ta c: y = (m + 2)x2- 2(m + 2)x - 3m +1V o hm khng th trit tiu ti v hn im nnHm s ng bin trn R o y ' > 0 Vx e R

(m + 2)x2 -2(m + 2)x -3 m +1 > 0 Vx e R ()

c y n bi ton n iu v bi ton du tam thc bc hC th l tam thc khng i du trn E , do ta cn nhc li mt chv du ca tam thc bc hai. j

Nhc li: Cho tam thc f(x) = ax2 + bx + c, a^ o c A =b2- 4ac*N u < 0 => a.f(x) > 0 Vx e R

*Nu A = 0 => a.f(x) > 0 Vx 6 R v a.f(x) = 0 o X= -

NuA > 0 => f(x) c hc nghim x1 O o x e (- {o;x1)u (x 2;+M)

a . f( x ) < 0 x e ( x 1;x2).

T nh v du ta c ngay: f (x) > 0 (f (x) < 0) Vx e R j a > 0 < _____________ /_______ _ [A 0 lun ng vi mi X => m = -2 thbi ton

TH 2: Nen m ^ 2 khi (1) tha vi mi Xe R

^ _ f a = m + 2 > 0 J m + 2 > 0 \

a = (m + 2)(4m +1) < 0 [4m + l < 0 ^ _ 2 < m ~ ~ 4

Ket hp c hai trng hp, ta c: -2 < m < l nhng gi tr cn tm.

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d 4.1. Tm m hm s y = 2x + m sin X - 1 ng bin trn R .

Li gi.Hm s xc nh trn R .Tac: y' = 2 + mcosx

* Nu - 2 < m < 2 = > y '> 0 V x e R = > hm s ng bin trn R

* Nu m = 2 =>2 2 COSX> 0 Vx e R v y = 0 ti v hn im, do ta cha kt lun c hm s tng trn R .Tcmg t nh v d 2. ta chng minh c hm s ng bin trn R* Vi m| > 2, khi y nhn c gi tr m ln dng trn R nn hm skhng th ng bin trn R. Vy m |< 2 l nhng gi tr cn tm.

d 5.1. Tm m hm s sau ng bin n[2;+oo)

y = X3 - (m + l)x2 - (2m2 - 3m + 2)x + m(2m - 1 ) .

Li gi.

Hm s xc nh trn R.Ta c y = 3x2 -2 (m + l)x -( 2 m 2 -3 m + 2).

Hm ng bin trn [2; +oo) y > 0 Vx > 2

o f(x) = 3x 2 - 2(m + l) x (2m2 - 3m + 2) > 0 Vx e [2; +co)

V tam thc f(x) c A = 7ra2 - 7m + 7 > 0 Vm

m +1 +Nen f(x) c hai nghim: xx= --------------- ;x2 = ---------------- .

3 3

X Xo

V X! < x2 nn f ( x ) > 0 oA .

Do f(x) > 0 Vx e[2;+co)X,,


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Li gii.Hm s xc nh n M.. Ta c: y ' = mx2- 2(m - l)x + 3(m - 2).

Cch L Hm ng bin trn (2; +00) o y '> 0 Vxe (2; +00)

f (x) = mx2 - 2(m - l)x + 3(m - 2) 0, Vx e (2; +00) (1) .TH : m = 0 khi (1) ch ng vi mi X > 3 .

TH 2: m < 0 ta thy trng hp ny khng tn ti m tha mn yu cu bi ton.TH 3: m > 0, f(x) c A' = -2m z +4m +1

* Nu A' < 0 m > ^ + (do m > 0 ) => f(x) > 0 Vx e E.

* Nx A' 0 0 - + W* Nu A >00x2

=> f(x) >0, Vx > 2 x2 < 2

- < 2 y[ ' 0 m > m 3-

K_t hp vi (*) => < m < ^ + . Vy m > l nhng gi tr cn tm.

Cch 2: Hm ng bin trn (2; +co) 0, Vx e (2; +00)

o mx2- 2(m - l)x + 3(m - 2) > 0 Vx e (2; +00)

o m > 6- g(x), Vx 6 (2;+00).X - 2 x + 3

___X , V ,. . , _ (/ V 2(x2 6x + 3)Xt hm s g(x) vi X> 2, ta c: g (x) =--- ---- -jr(x - 2x + 3)

=> g '(x) = 0 o x ~ 3 + a/6 (v X 2 ) v lim g(x) = 0.X>+Q0

2Lp bng bin thin ta c maxg(x) = g(2) =

X2 3

2m > g(x) Vx e (2; +03) o m > maxg(x) = .

XS2 3Ch : Cho hm s y - f(x) Hn tc trn D

* f(x) > k Vx D minf(x) > k (nu tn ti minf(x))

* fix) < k V x e D o maxf (x) < k (nu tn ti maxf (x) ).D

V 7.1. Tm m hm s y = -x s + (m -l) x + 2 m -3 ng bin

trn mt khong c di ln hem 1.

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Hm SCxc nh n RTa c: y' = 3x2 + 6x + m - l , A' = 3m + 6

* Nu m -2 => A ' 0 =>y' 0 V xeR => hm s nghch bin trn Rnn hm s khng c khong ng bin.* Nu m > -2 y ' = 0 c hai nghim X. < v. y' 0' X e [xi;x2]

=> yu cu bi ton [x1- x 2| > 1 (Xj +X2)2 - 4xtx2 >1

4 + 4(m > 1111> . Vy m > l nhng gi tr cn tm.3 4 w 4

Bi tp.Bi 1.1. Xt tnh on iu ca cc hm s sau

1. y-=2x3 - 3 x 2 - l 2. y = -2 x4 + 4X2 3. y = Vx2 -2 x

. 2x + l . x2 - x + l c I -I4- y = _ ' 5. y = - 6. y= x + 1

Li gi i

X- 3 X 1

7. y = Vx3 -2 x 8. y=x + 2x -3 4 1 . 1 4 /9. y = x - 6 x + 8 x - l 10. y - ( x + l ) - ( 3 x - l )

11. y = >/3s inx-co sx + 2 x - l .Bi 2.1. Tm m 1. Hm s y - 2cos2x + mx - 3 ng bin trn R

2.Hm s y = (m - l)x3 - 3(m - l)x2 + 3(2m - 3)x + m nghch bin tin R .

X X3 ; , ,

3- Hm s y = - -f (m 4*l)x 4- (2m 4- l)x + m nghch bin trn (0; 3).34. Hm s y = (m 4- l)x3 - 3(m +l)x2 4- 2mx + 4 ng bin trn khong c

di nh hn 1.5. Hm s y = Xs + 3x2 - 3(m2 - l)x +1 ng bin n (1; 2 ) .

H Bi 1.1.1. Ta c y' = 6x2 - 6 x , hm s ng bin n (-00; 0) v (l;-K>o); nghch

binn (0;1).

2. Ta c y' = -8x3 + 8 x , hm s ng bin n (-00;-1) v (0;1); nghchbin trn (-1; 0) v (1; +CO).

3. TX: D = (-oo;0]u[2;+o)X 1 ' ' * '

Ta c: y 1= - == , hm s ng bin n (2;+oo); nghich bin nVX2 - 2 x

(oo;0). .

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4. Hm s nghch bin trn tng khong xc nh y5. TX: D = \ { l } .

_2x , ,Ta c y ' = .Hm s ng bin n ( x>; 0) v (2;+ o o )

' ( x - i r .Hm s nghch bin n (0; 1) v (1; 2).

x + 16. Ta c y = y/(x+ 1)2 =>y' = ' -v(x + l )2

Hm s ng bin trn (-1; +oo), nghch bin n

7. TX: D =n/2j 0J J\2

Ta c: y ' = - ^L==r . Hm s ng bin trn -W2;2vx3 - 2x 1 .

2

nghch bin n - J - ;

v (V2; +00

p , _ /712 2(x + l)(x2+ 2x - 3)8. Ta c y = y(x -j-2x-3)* =>y = -

V(x2 + 2x - 3)2

Hm s ng bin trn (-3;-l) v (1; +00); nghch bin trn, (-00; -3 ) v

(-!;!) /3cosx + siax + 2 = 2co s(x- ) + 2 0 y >0 m >4sin 2x VeKTa tm c m 4 .

2. Ta c y = 3(m - l)x2 - 6(m - l )x + 3(2m -3 )

m = l= > y = -3 < 0 Vxe IR hm s nghch bin trnR

m * lHm s nghch bin trn R

! +

3.^/3-1

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m - l< 0 [m < l< __ rn < 1

A = (m - 1)2 - (2m - 3)(m -1 ) < 0 l-m + 2>0

Vy m < l .3. Tac y = x2 + 2(m -l)x + 2m + l

X2 -2 x + lYu cu bi ton y' > 0 o x + 1 >-2m Vxe(0;3) (*)

Xt hm s f(x) =X - 2x +1

Ta c: f'(x) =

Bng bin thin

x + 1X2 + 2x - 3

,xe[0;3]

(x + l) s,f'(x) = 0x = l

X 0 1 3f - 0 +

1 1

f< r ^

(m + 3)(m +1) > 0

8m3 -

3(m +1)>0

4. Ta c: y ' = 3(m + l)x2 - 6(m 4- l)x + 2m

; m = -1 => y ' = -2 < 0 (loi).o m > -1 . Khi hm s lun c khong ng bin c di hi hn 1 m < 1.

Yu cu bi ton y' = 0 c hai nghim Xj,x2 tha |xx- x21'l

Ja ' = 9(m + 1)2 - 6m(m +1) > 0

[(Xj +x2)2- 4 x 1x2K.l

m + 3 0Vx nn hm s ng bin trn R.# - m - 1 < 1

Nu m > 0 , suy ra yu cu bi ton o 0 Vxe (a;b) (ioc f'(x) < 0 Vx e(a;b)).

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16/377

Suy ra f(b) > f(a) (hoc f(b) < f(a)).

iu ny tri vi gi thit f (a) = f(b).Vy phng trnh f (x) = 0 c t nht mt nghim trn (a;b).

T nh l ny, ta c c hai h qii sau:H q 1: Nu phng nh f(x) = 0 c mnghim th phng trnh

f'(x) = 0 c m -1 nghim.H qu 2: Cho hm s y = f (x) c o hm n cp k lin tc trn (a;b).

Nu phng trnh f(k)(x) = 0 c ng m nghim th phng trnh

(k-i) = 0 C nhiu nht l m +1 nghim.

Tht vy: Gi s phng trnh f(k_1)(x) = 0 c nhiuhn m + 1nghim

th phng trnh f(k)(x) =30 c nhiu hn m nghim, iu ny tri vi gi

thit bi ton.

T h qu 2 => nu f'(x) = 0 c mt nghim th f(x) = 0 c nhiu nhthai nghim.

II. CC V D MINH HA

V d 12.Chng minh rng : sin X< X Vx 0;2

Li gii.

Xt hm s f(x)=x-sinx, xe,-j, ta cn chng minh f(x)>0=f(0), 0


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18/377

X2 X4 ( 7Mt khc, theo cu 2: 1 - + > cosx Vx 0;

2 24 { 2J

Suy ra 1-11-j >cosx Vxe^O;j.

T : ^ i n ^ 71.

Laic: 0 < sinx < X => 0 < - 0 je^0; j .2cost 2 2 I sin

sin31 t3 - sin31 V t

nn hm s ng bin n (0; ]2

= >f(t) f g ) = l - , V t e ( 0 ; f ]

p dng vo bi ton ta c : x) + f (y) + f (z) < 3 l - \ j .

V d 4.2. Gii phng trnh: >/7x + 7 + V 7x-6 = 13.Li gii.

iu kin: x> .7

Xt hm s f(x) = V7x + 7 + sx- 6 , X>=>f(6) = 13Khi phng nh c dng: f(x) - f(6)

7 7 6M f'(x) = /_ _ + 7= >0 Vx > ~

2*j7x + 7 2\7x-6 7Nn f(x) = f (6) o X = 6 l nghim duy nht ca phng trnh cho.

V d 5.2. Gii bt phng trnh: J ~ + 7= - 3+ 2>/3 - X .1 o ~ X yj 3 X

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19/377

iu kin: --^ 0Khi ,Bpt yj2xz+ 3x2 + 6x +16 - a /4 -x < 2/3f(x) < 2V3 (*)Trong : f(x) = V2x3 + 3x2 + 6x +16 - V4-X c

fl/ 3(x2 + X+1) f (x) - = = + ~ >0 Vxe(-2;4)

V 2x 3 + 3x 2 + 6x + 16 2v 4 -X

nn f (x) l hm ng bin trn (-2; 4).

M ta c: f(l) = 2V3=> (*) f(x) < f(I) o X< 1

Ket hp vi iu kin, ta c nghim ca Bpt cho l: -2 X < 1 .

V d 7.2. Gii phng trnh: yx + 2 + yx + l= ^2x2 +1 + \fex2 .Li gii.

Nhn xt c im cc biu thc di du cn ta thy mi v biu thc

di du cn hn km nhau 1. Do nu ta t u = x + 1, v = \/2x2 iphng trnh cho tr thnh:

\/u3 + + u =yjv +1 + V o f(u) = f(y ) ., t 2

Trong f(t) = y t3 +1 + 1 , c: f'(t) = - +1 > 0 nn f(t) l hm^(t3+l)2ng bin.

Do : f (u } = f { v ) o u = v o 2 x 2 = x + l x = l , x = - - 2

Vy phng nh c hai nghim:X = l,x = .

V d 8.2. Gii phng trnh: x - X2- lOx - 2 = \!7x2+23X+12 .

Li gii.3

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20/377

Li gii.

Phng trnh (x +-2)3+ (x + 2) = 7x2+ 23x + 12 + \/7x2+ 23x + 12

f(x + 2) = f ^ x 2 + 23x +12) (1)

Trong , hm f(t) = t3+ 1 c f'(t) = 3tz +1>0 nn hm f ng bin

trn R. Do (1) X+2=yx2+ 23X+12

_ Q + / Xs - X 2 l l x 4 0 ( x - 4)(x2 + 3x + l ) = 0 x = 4 ,x = ----------- .2

Ch : S dng tnh an iu gii phng ii - bt phng trnh tathng i theo hai hng sau:

Hng 1: a phng trnh v dng f(x) = f(x0) , ong y = f(t) l mt

hm s lin tc v lun ng bin hc lun nghch bin trn tp angxt. - lm theo hng ny, chng ta cn nhm trc mt nghim caphng trnh v nhn din c tnh n iu ca hm s f.

* nhm nghim, ta c th s dng my tanh b ti tm nghim.C th: tm mt nghim ca phng trnh f(x) = 0 ta thc hin nh

sauBc 1: Nhp biu thc f(x) (Dng phm ALPHA+ X)Bc 2: ng lnh gilphng trnh: SHIFT+CALC (SOLVE) nhp gitr ca X (nhp gi tr bt k) =.* nhn din c tnh: om iu ca hm s f, chng ta cn ch * Tng hai hm s ng bin l mt hm s ng bin* Hm s i ca mt hm s ng bin l mt hm s nghch bin.* Nu hm s y = f(x) ng bin th y = $(x) l hm s ng bin.* Nu hm s y = f(x) ng bin v nhn gi tr dng th hm s

y = l mt hm nghich bin.f(x) ~

Hng 2: Bin i phng trnh v dng: f(u) = f(v), trong u, V l cc

hm theo X.Lm theo hng ta thng p dng khi gp phng trnh cha hai phpton ngc nhau.

V d 8.2. Cho n+1 s thc alJa2,...,an,a khng ng thi bng khng v n + 1 s thc b1}b2,... ,bn,b. Chng minh rng nu phng trnh

f (x) = 0 c nghim th c nhiu nht l hai nghim.

Trong f(x) = ,$aX + b - (ax + b) v k > 2 l s t nhin chn.i=l

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21/377

= ak-l

Li gii.

Ta c: r w - t - j - A i=ik.(aix+bi)

V -} a a2=>f "(x)= - - i y - - r - -phng trnh f(x) = 0 c nhiu nht hai nghim (pcm).V d 9.2. Chng minh rng phng trnh sau c nghim duy nht

X5 -X 2 - 2 x - l = 0. >Li gii.

Ta c phng trnh X = (x + 1)2

Suy ra nu x0 l nghim ca phitg trnh, th x >0 x0 > 0

=> (x0 + 1)2 > 1 => Xq> 1 => x0 > 1

Do ta chi gii phng trnh, vi X>1.Xthms f(x) = x5-X2 - 2 x - l, X>1

C f (x) = 5x4- 2 x - 2 = 2x(x3-l )+ 2 (x 4- l ) + x5>0 . M f(l).f(2) < 0 => phng trnh cho c nghim duy nht.

V d 10.2. Gii h phuong trmh: ^ = x - y[x2 - 12xy + 9y2 + 4 = 0 . (2)

Li gii:

iu kin: X, y > .2

T phng trnh (2) ta thy nu h c nghim (x; y) th x. > 0 (*)

(1)o-s/2x + l - X = 2y +1 - y (3) .

Xt hm s f(t) = >/2t + l - t , ta c: > 1 - =0 t = 0v2t + x . .

=>hm f (t) ng bin n ( -;0) v nghch bin trn (0;+co).Do (*) nn ta c cc trng hp sau:

TH1: X,y e [ - i ; 0) f(x) = f(y) o X = y (o f (t) ng biii).

TH 2: x,y e[0;+oo) => f(x) = f(y) o x = (d f (t) nghch bin).

Tm li c hai trng hp u dn nx = y , tc l (1) X = y iay vo

(2) ta c 2x2 = 4 o X = V2 (do X>2

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22/377

*-*vm

T phng trinh th nht d dng suy ra X > 0,

t - 2y = 2t > 0, ta c y = .

Thay vo phng trih u ta c: x(4x2 +1) = t(4t2 +1)

Vi hm f(u) = u(4u2 +1) ng bin trn R+ nn a c X= t , suy ra

5 - 4x2y= 2 _ 9 (5 - 4x2)2 I----- -

Thay vo phng trnh th hai ta c: 4x 4-----------

+ 2V3 - 4x = 7.( 31 ' (5 4x2 )2 /------

Trong khong 0, - , hm s g(x) = 4x2+ ------ + 2V3 - 4x c:

g'(x) = 4x(4x2 - 3) r = = < 4x(4x2 - 3) < 0.V3-4X

Mt khc li c g 1= 7 v vy phng trnh g(x) = 7 ch c mt

nghim duy nht l X = , suy ra y = 2.t

Vy h cho c nghim duy nht1

X = 2-

y = 2

V 12.2. Gii h phng nh :

x2 + x - y - l = 0

y2 + y - z - l = 0

z2 + z t 1 = 0

t 2 + t - x - l ~ 0

H

hi giif(x) = y (1)

f(y) - 2 (2) Trong hm s f(u) = u2 + U -1 ngf(z) = t (3) .f(t) = x (4)

1bin trn ( ;+0) v nghch bin trn (-co; .

2 ^

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23/377

T h => x ,y ,z , t > f ( - ) = - v f ( - ) = - 2 4 4 6

* Nu X> => f(t) > > f ( - ) => t > (V < t .2 2

T ag is : x = m ax{x,y,z,t}=>x> y=> f(x)>f(y)=>y >z => f(y) > f(z) => z > t => f(z) > f(t) = > t > x = > x > y > z > t > x =>x = y = z = t . Thay vo h =>x = y = z = t = l .

1 1 '* Nu X< - => y,z, t < - (V nu c mt trong bn s > -

2 2

cn li cng > - )

Gi s X> z f(x )< f(z) =>y < t f(y)> f(t) = > z > x= > x= z .

X = z; y = t

Tng t => y = t => h tr thnh : f (x) = y

f(y) = x

=>x + f(x) = y+ f (y )o x 2+ 2 x - l = y2+ 2 y - l

(x - y)(x + y + 2) = 0 X = y hoc X + y + 2 = 0.

Vi X= y =>X2 = l o x = y = - l (do X< )2

Vi x + y + 2 = 0=>y = -2~ x= > x2 + x - l = - 2 - x o x = - l= > y = - l .

Tm li h c hai cp nghim : X = :=z = t - 1 v X = y = z - 1 = -1 .

Bi 1.2. Chng minh cc bt ng thc sau, s in a sin b , . - , 7t1. > v i 0 < a < b <

a b 2

3 tan X 3 ^ w _ /n 71112. ----- ------ < < --------- --------- vi Vx 3x Vxe[O:^02

l4. sin(cos x) < cos(sin x) Vx (0;)2

5- a - b + ca > (a -b + c)a vi a > b > c > 0 ,a > l. Bi 2.2. Gii cc |*hng nh bt phng trnh sau

2) ' i 5x - + n/2x^1 + X = 4

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24/377

3. a/(x + 2)(2x- 1 ) - 3>/x + 6 < 4 -yj(x +6)(2x- 1 ) + 3Vx + 2

4 . 3x( 2 + > /9x2 + 3) + (4 x + 2) ( 1 + V l + X + X2 ) = 0

5. >/6 x + l = 8 x 3 - 4 x - 1 6. J ^ - + J = 6V 3 - X V2 - X

7. siis 3xsin(x + ) + cos(3x + )cos3x = 04 4Bi 3.2. Gii cc h phng trnh. h bt phng trnh sau

| x 2 + 5 x + 4 < 0 2 f x3 + 2 = 3 y

x3+ 3x2 - 9x - 1 0 > 0 y3+ 2 = 3x

X7 + X = y + 4 I s i n 2 x + 2 x = s in 2 y + 2 y

' [4x5 = 5y4 * [X2 + 4xy + 5y2 = 5

5.

x = y 3 + y 2 + V - 2 f _ . _ ,-------

3 3 0 - 12(2x 4"1) + 2x +1 = (2 y- 3)%/ -2y = z + Z + z - 2 6. \ ____ __3 2 y 4 x + 2 + V 2 y + 4 = 6

z = x + x + x - 2 ^ .

Bi 4.2. Cho phng trnh X3+ ax2 + bx + c = 0 (a * 0) (1) c ba nghim

phn bit Chng minh phng, trnh sau ch c hai nghim thc phnbit:

4(xs + ax2 + bx + c)(3x + a) = (3x2 + 2ax + b)2 (2).

Bi 5.2. Chng minh, rng phng trnh : X5 + - , x------2008 = 0 c ng

Vx2 - 2hai nghim dng phn bit.

Hng dn giiBi .2.

1. Chng minh hm s f(x) = s-mx nghch bin trn (0;- )X 2

2. Ta chng minh hm s f(x) = tanx + 2sm x -3 x ng bin n [0;^) v

hm s g(x) = 4 sinX

- sin 2x - 3x nghch bin n [0; ).2 2

3 . X h m s f ( x ) = t a n X + s in X - 2 x n g b i n tr n [ 0 ; ^ )

4. Chng mii cos(sinx) > cosx > sin(cosx)

5. Xt hm f(x) = aa - ba + x - (a - b + x) vi X e [0;b]

Bi 2.2.1. v tri l mt hm ng bin. S : X = 12. v i l mt hm ng bin. S : X = 1'

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25/377

3. iu kin: X> . 2 '

Kh : Bpt (Vx+2 + Vx + 6)(a/2x -1 - 3) < 4 (*)

* Nu - J 2 x - 1 - 3 < 0 X < 5 =>(*) lun ng.

* Nux > 5, ta xt hm s f(x) = (y]x + 2 + -Jx-r 6 ) ( j2 x - lT 3) c:

pt/ 1 1 \/ / ~T o\ vx + 2 -hVx--6 _ w f(x ) = ( 7= =) (V 2 x - 1 -3 ) + -------------- >0 Vx>52Vx + 2 2 Vx + 6 > /2 x-l

nn f (x) l hm ngbin n (5;+oo) v f(7) t=4 nn

(*) o f ( x ) < f (7) o X < 7. Vy nghim ca Bpt cho :< x < 7 .2

4. Bin i phng trinh v dng f(-3x) - f(2x + 1)

Vi f (t) = 2t + Wt2 +3 vi t > 0 . S: X=

, 55. Bin i phng trnh v dng f(^6x + l) = f(2x)

Trong f( t) = t 3 + 1 . S : X = COS, X = COS , X = COS9 9 9

6. v i l hm s ng bin. S: x==.

7. Trc ht, ta thy rng cosx = 0 hoc sin3x = 0 khng ia mnphng trnh-Xt: cos X. sin 3x * 0.

sin(x + ). sins(3x) + cos(3x + ).coss X= 04 4 .

o (sin X + cos x). s in3 (3x) + (cos 3x - sin 3x). COS3 X = 0

(sin X + cos x) __(s in 3x-c os3 x)

cos3x sin?3x(sin X + COS x) _ sn(i - 3x) + COS (71 - 3x)

o ------------------------------ T~------ ---------cs X sin (t i- 3 x )

o (tanX + 1)(1'+ tan2 x) = (ct(i - 3x) 4-1)(1 4- cot2 (71 - 3x)).

Xt hm s:f(t) = (t + l)( t2 +1) = t3 + 12 + 1 +1, t R =>f (t) = 3t2 + 2t +1 > 0, vt nn y l hm ng bin.T bi, suy ra: f(tan x) = f(cot(7 - 3x)) tanX = cot(n- 3x)

o tanx = tan(3x - ) o x = 3 x - - + k.rc o x = - k - , k e Z .2 2 4 2

Bi 3.2.1. Xt hm s f(x) = X3 + 3x2 - 9x 10 trn (-4; -1 ). S: ~-4


26/377

2. Tr hai phng trnh: X3 + 3x = y3 + 3y o X= y

(Do hm s f(t) = t3 + 3t ng bin trn R)

S: H c nghim X= y = -2 ,x = y = 1.

3. Hm s f(t) = t7 + 1 ng bin nn t phng trxih th nht ta c X= y

5i vo phng nh th hai ta gii c X = 0, X = .

4. Hm s f(t) = sin 2t + 2t ng bin n R. S: X - y - -J= .

5. Hm s f(t) = t3 + 12 + 1 - 2 ng bin trn R. S: X= y = z = 1.6. Phng trnh th nht ca h tng ng vi

. 2(2x + l)3 +2x + l =2A/(y -2 )3 + y-2

o 2 x + l = ^ y - 2 => 4 x + 2 + V x2 + 8 x + 1 0 = C 5 X =

Bi 4.2. Gif(x) = Xs + ax2 + bx + c . Khi (2) c vit di dng:

2f(x).f"(x) = [f'(x)]2 o g (x ) = 2f(x )f"(x)-[f,(x)f = 0 .

Ta c: g '(x) = 2f (x)f (3)(x) ( f(3) l o hm cp ba ca hm f).Gi Xj < x2 < x3 l ba nghim phn bit ca f(x), ta c:

g'(x) = 12(x - xx)(x - x2)(x - x3) => g'(x) c ba nghim xl ,x2,x.i

X 00 x2 Xg +C C

g ' ( x ) - 0 + 0 - 0 +g ( x ) +00 ,^g(x2)

" ^ g ( x xX g (x 3)

+00

V f(xi) = 0=>g(xi) - - [ f ' ( x i)]2 pcm.

Bi 5.2. iu kin: X>\2 (do X > 0).

Xt hm s : f (x) = X5 + x - - 2008 vi X > V2 .

= > f (x ) = 5 x 4 - 1 x> f "(x ) = 2 0 x 3 + 3 x . - > 0 Vx > V2Vcx2- 2)5

V .f(x) = 0 v nghim =>f'(x) = 0 c nhiu nht mt nghim

:=>f(x) = 0 c nhiu nht'l hai nghim.

M: lim f(x) = +00; ( S ) < 0; lim f(x) = +00=> f(x) = 0 c . hai nghim

1 e (y2;-JS)vx2 > V3 .

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27/377

:H 3. c c TR HM s 3. i'

THUYTrc tr hm s :

s xc nh trn tp hp D (D cR ) v x0 e D

1) x0c gi l mt im cc i ca hm s f nu tn ti mt khong

(a;b)cD (a;b)cha im xnsao cho: f 0 Vx => Hm s khng c cc .

2) TX:D = RX = 0

X = 1

Ta c:y' = -4 x3+4x = 4x(x2 -l )= > y ' = 0

X 00 1 0 1 +O0

y + 0 - 0 + 0

y2

" N

2

00

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29/377

rj'im Hm s t cc i ti cc im- X = 1 vi gi cc i ca hm s l

y(l) = 2 v hm s t cc tiu ti im X = 0 vi gi tri cc tiu ca

hm s l y(0) = 1. t3) Tp xc nh: D = R

Ta c: y = -4 x3 + 1 2 x - 8 = ^ 1(x - 1 ) 2 (x + 2 )

, ^ Fx=i > y = 0 oX = -2

Bng bin thin:X co 2 0 +00

y 1o+

0 -

y V - 00-Hm t cc i ti X = -2 vi gi tr cc i ca hm s l y(-2) = 25,

hm s khng c cc tiu.4) TX: D = M\ {1}

"x = 0

( x - l)z - V IX = 2 _ , , X2 - 2x ,Ta c: y =-Y=>y' = 0

X 00 0 2 +00

y + 0 - 0 +, '

y

- 1

00 . 00

+00 +00

Hm s t cc i ti im X= 0 vi gi tr cc i ca hm s ly(0) = 1 v hm s t cc tiu ti im X= 2 vi gi cc tiu ca

hm s l y(2) = 3. .

5)TX: D = [-2;2]

^ y ' = 0 X 2,= 42 .y '= = = = = = > y ' = 0 X = -~v2,x =yJ2.

Hm s khng c o hm ti cc im X = 2 .

X - 2 -~Ji & 2y ' '11 - 0. + 0 -

y0 .

^ - 2

^ 2 ^

^ 0

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30/377

Hm s t cc i ti im X =y/2 , y(V2) = 2 v t cc tiu ti im

X = J2, y ( W 2 ) - - 2 .

2) TX:D = (-oo; v/3]u[>/3;--oo).

T'- ' r. X 2v X2 - 3 - X , r~z7Ta c: y ' = 2 . ~ = -----===== ^ y ' = 0 o 2Vx2 - 3 = X

-v/x^-3o X = 2 v hm s khng c o hm ti X = V3 .

x>0

4(x2 - 3) = X2

Bng bin thin: ?

Hm s t cc tiu ti im X= 2, y(2) = 3, hm s khng c cc i.

C : Cho hm s y = f(x) xc nh trn D.im X = x0 e D l im cc tr ca hm s khiv ch khi hai iu kin

sau y cng tho mn:Ti X= x0 o hm trit tiu hoc khng tn t

o hm i du khi X i qua x0.

V d 2.3. Tm cc tr ca cc hm s sau1) y = 3- 2c os x -c o s2 x 2) y = 2 sin 2 x- 3.

Li gii.1) Hm s cho xc nh v lin tc trn R .

Ta c y = 2s inx + 2sin2x = 2sin x(l + 2cosx)

s i n X = 0 X = kc

=>y' = 0 o 1 2t 2n i n _ , k e Z .cosx = = COS X= + KZ7U

2 3 L 3y" = 2cosx + 4cos2x =>y " ~ +k27ij = 6cos - - 3 < 0 .

Hm s t cc i ti X= + k2t, y + k2nI= .a 1 . 3 ) 2

y u(k) = 2cosk + 4> 0 ,V keZ .

Hm s t cc tiu ti X= krc , y (krc) = 2 (1 - COSkt).

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31/377

2) Hm s cho xc nh v lin tc trn R .

Ta c : y' = 4cos2x=>y' = 0cos2x = 0 o x = + k^-,k eZ4 2

Q o ^ "n 1 * _ o n 1 i - F 8 klli k=2ny = -8sin2x => y + k = -8 sin -r+-k7c =y= l> 0 Vx e R => hm s khng c cc tr

* Nu m * 0. Khi y l mt tam thc bc hai nn y = 0 c nghim

v i u khi qua cc nghim o y ' = 0 c hai nghim phn bit hay

A = 12m2 - 3m > 0 o m < 0 v m > - .4

Vy m < 0 V m > l nhng gi tr cn tm.____________ _____ 4____________ '__________________________

Nhn xt: Hm s y = ax3+ bx2 + cx + d, a * 0 c cc tr khi v chi khi

y ' = 0 c hai nghim phn bit.

V d 43. Cho hm s y = X4 + 4mx3 + 3(m + l)x2 +1. Tm m 1. Hm s c ba cc tr2. Hm s c cc tiu m khng c cc i.

Li gii.Ta c y = 4x3 + 12mx2 + 6(m + l)x = 2x(2x2 + 6mx + 3(m +1))

"x - 0=>y' = 0 o

f(x) = 2x + 6mx + 3m + 3 = 0

Tathy:o Nu f(x) c hai nghim phn bitx15x2 * 0 , khi y s i diu khii qua ba im 0, , x2 khi hm c hai cc tiu v 1 cc i

30



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32/377

Nu f(x) c 1 nghim X= 0, khi y' ch i du t - sang + khi iqua mt im duy nht' nn hm ch c mt cc tiu. Nu f(x) c nghim kp hoc v nghim thi y ' ch i du t - sang +khi i qua X= 0 nn hm t cc tiu ti X = 0 .T trn ta thy hm s lun c t nht mt cc tr.

1) Hm s c ba cc tr khi v ch khi f(x) c hai nghim phn bit khc 0

If(0) ^ 0 ' 1k ' ' m ^ - 1

2) Theo nhn xt n ta thy hm ch c cc tiu m khng c cc i^ l L i ' 1 - J ____ 1 + Vo hm s khng c ba cc tr o ------ < m < -.

X = 0

4ax2 + b = 0.(1)Ch : y' = 4ax3 + 2bx - x(4ax2 + b )= > . '= 0o

y = a x 4 + bx 2 + c (a 90)1. i vi hm trng phng

Ta c

e Hm c ba cc O: (1) c hai nghim phn bit khc 0 b * 0

ab 0; hm c hai cc i, 1cc tiu khi a < 0.Hm c mt cc khi v ch khi (1) c nghim kp hoc v nghim

A < 0

hoc c 1 nghim x=.0 0 v ch e cc i khi a < 0.

b>0

b = 0 . Khi hm chi c cc

2. i vi hm s bc bn y = ax * bx + cx + d .

_ 0 0 x=0Ta c: y ' = 4ax + 3bx + 2cx = > y '= 0 o

4ax + 3bx + 2c = 0 (2)

Hm s c ba cc tri khi v ch khi (2) c hai nghim phn bit khc 0

9b ^ ac > ^ . Khi hm. c hai cc tiu, mt cc i khi a > 0;

c * o hm c hai cc i, 1 Gc tiu khi a < 0.a Hm c mt cc -khi v ch khi (2) c nghim kp hoc v nghim

A < 0

= 0hoc c 1 nghim X = 0

c cc tiu khi a > 0 v ch c. cc i khi a


33/377

Td 5.3. Tm m hm s sau c cc tr .mx - 1

Li gii. *:

Vi m = ta c y = - X 2 + X - 1 , ta thy hm s t cc i ti X = .

Suy ra m = 0 tha yu cu bi ton

m^O, hm s xc nh vi mi X * -.771

~ , mx2 -2 x + l- 2 mTa c: y = ---------- -T-----

(mx -1 )

Suy ra y = O o mx2 - 2 x + l - 2 m = 0 (*) j

Hm s a cho c cc tr khi v ch khi phng trnh (*) c hai nghim

phn bit khcm

1 2m2 - m +1 > 0 ng vi mi m.1+1 - 2m *0 & V

m }

I A = l -m ( l- 2 m ) > 0

I+1 - 2m * 0L m

Vy hm s cho lun c cc t vi mi m .

hn xt: Hm s y - am * 0 c cc tr kh v ch khi y 1= 0mx + n '

c hi nghim phn bit.

d 6.3. Tm m hm s y = + PQ3C+ 1 CCtiu ti X= 1 .x + m

Li gii. .Hm s xc nh vi mi X 5*-m \

Ta c: y = x + - =>y = l - 1X + m (x + m)

V hm s c o hm ti mi im X-m nn hm t cc tiu ti

X = 1 th trc ht y '(1) = 1 -------- r = 0 o m = 0;m = -2 .(1 + m)

M y = ---- T nn(x + m)

* m = 0 =>y"(l) = 1 > 0 = > X = 1 l im cc tiu =>m = 0 tha yu cubi ton.



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34/377

* m = -2 => y '(1) = - 1 < 0 => X = 1 l im cc i = > m = -2 khng thayu cu bi ton.KL: m = 0.

Nhn st:9 Nu trnh by li gii theo s sau

Hm s at cc tiu tai X= 1 o y = i --------yx + m (x + m)

i X fy'(i)=oHm s at cc tiu tai x = l o \ y M(l)> 0

V d 7.3. Tm m hm s y = -2x + 2 +mVx2 - 4 x + 5 c cc i.Li gii:

Hm s xc nh trn R .T ' x - 2 t mTa cy = -2 4-m; ;y -

Vx2 - 4x + 5 ( x 2 - 4 x + 5)3/2

* Nu m = 0 th y ' = -2 nn hm s khng c cc tr.

**Nu m * 0 v du ca y"ch ph thuc vo m nn hm c cc i

th trc ht y


35/377

V d 83. Tm m hm s y = (x - m)(x2 - 3x - m -1 ) c cc i v cc

tiu tho x c d .Xqt I= 1.

Li gii.Hm s xc nh n M.Ta c : y ' = 3x2 - 2(m + 3)x + 2m -1

=>y' = 0 o 3 x2 - 2(m + 3)x + 2m -1 = 0 (1)Hm s c hai im cc tr tha mn x c d -Xc t = !(!) c hai nghim

x1Jx2 tha mn:

A' = m2 + 7 > 0 rxr x2 1=1 y' = 0 3x2 -6 x + m2 =0 (1).hm s c cc tr (1) c 2 nghim phn bit x1?x2

o A ' =3(3 - m2) > 0 o -V3 < m < V ., 2 1

Chia y cho y ' ta c phn d l: y = (m2 - 2)x + m2 + m

3 3Nn phng trnh ng thng ' i qua cc im cc tr l :2 1

y = (m2 - 2)x + m2 + m => cc im cc tr l :3 3

A^K^Im 2-2)X! + |m 2+ 3mj, B^x2;(-|m2-2)x 2+ m2+3mj .

Gi I l giao im ca hai ng thng d v dT,2m2 --6m +15 ll m 2 +3m -3 0 ,

=>(---------5 ;-------- 9---- ).15 -4 m . 15-4m

, , 2 * A v B i xng qad th trc ht d_Ld m2 - 2 = - 2 o m = 0

Khi 1(1;-2) v A(x1;-2x1); B(x2;-2x2)

=> l trang im ca AB =>-A v B i xng nhau qua d -Vy m = 0 l gi t cn tm.

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37/377

V ,d 123. Tim m hm s y= X4 -2m 2x2 +1 c 3 im ce tr l 3 nca mt tam gic vung cn. V

Li gii:Hm s xc nh n K .Ta c y' = 4x3 - 4m2x = 4x(x2 - m2)

Hm s c ba cc tr O' m * 0 .Khi ta cc im cc ca th hm s l:

A(0;1), B(m;l-m4), C(-m;l-m4).

Ta thy AB = AC nn tam gic ABC vuiig cno AB + AC2 = BC2 2(m2 + m8) = 4m2 o m = 1Vy m = 1 l nhng gi tri cn tim._____________________________

Ch : i vi hm s trng phng y = ax4+ bx2 + c, a *0 ba icc tr ca th hm s (nu c) to thnh mt tam gic cn c n

nm n trc Oy._______

V d 13.3. Tm m th hm s y - X4 - 2mx2 + 2 c ba im cc t

to thnh mt tam gic nhn gc ta lm trc tm.Li gii.

_ , ' > |"x = 0Tac y' = 4x -4m x=>y' = 0 o

|_x = m

Hm s c ba cc tr o m > 0. Khi to ba im cc tr ca thm s l A(0;2),B(-\/m;-m2 + 2),C(%/m;-m2 + 2)

V tam gic BAC cn ti A nn 0 A 1B C nn o l trc tm tagic ABC c> OB _LAC o OB-C = 0 (*)OB = (~4m;-m 2 + 2), AC - (Vm; -m 2)

Nn (*) -m + m2(m2 - 2) = 0 m3 - 2m -1 = 0

(m + l)(m2 - m -1 ) = 0 o m = l gi tr cn tm.

V d 14.3. Cho th (C) :y = x4 -6 x 2 +2x. Chng minh rng (C) cim cc phn bit khng thng hng. Vit phng trinh ng tr

qua 3 im cc tr .Li gii.

Trc ht ta c y' = 2(2x3 6x + l)=>y' = 0 o 2x3- 6x +1 = 0

Xt hm g(x) = 2x3- 6x +1 lin tc trn R. v cg( -2).g (-l) = -9 < 0 , g(-l) .g( l) = -9 < 0 s g(l).g(2) = -15 < 0

Do phng trinh g(x) = 0 c ba nghim phn bit hay hm s c cc tri phn bit

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38/377

) ^ JGi M(x0;y0) l mt im cc tn no . ** . .

( 1 'VU 2'

Do x= 3x0 - i => y0 = Xo - 6x* + 2x0 = -3x^ +1 x0

3Suy ra c ba im cc tr u nm trn Parabol y = -3x2 + X nn n

J 2khng ing hng.

Mt khc ta li c y - (-3xJ + x0)2 = 9xJ - 9x + Xq /

=9x0(3x0- ) - 9 ( 3 x 0 - ~ ) + ^ x 2

117 2 63 19.= , x0 X0 +774 2 x + 2

Suy ra Xq+ yj = 0 - ^ x 0 + 9 ( x 0 - * ) - 5 * , + sJ 0 J0 4 0 2 , 2 42 322

_ 2 2 131 121 9 _Xft + Yn + -----xn+ ----- yn- = 0 .

0J0 8 0 12 0 2

T T A 2 2 131 121 9 _Vy cc diem cc tr nm trn ng n X + y + X+ - y - = 0.8 12 2

V d 15.3. Cho hm s y = X4- (3m - l)x2+ 2m +1 (1). Tm m thhm s (1) cbaim cc tr A, B, c cng vi im (7;3) nitipc mt ng trn.

Li gii.

Ta c: y' = 4x X -2 3m -1

1Nu m =>y' = 0x = 0 khng tha yu cu bi ton.

3

e Nu m > => y = 0 X = 0,x = .( m 3 V 2

th hm s c ba im cc tr: A(0; 2m 4-1)

A {B

^ '3m - 1 -9 m 2 + 14m + 3, c

3m - 1 -9m +14m + 3

V.1 y V. /V tam gic ABC cn ti A nn tm I ca ng trn ngoi tip tam gicABC nm trn Oy, suy ra 1(0; a ) .

__ ..A , . ^ 1 |.3m- 1 -9m 2 + 22m + 7Gi M l trung im ca AC, suy ra M (-J----- ;---------------------).

37



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39/377

/

Ta c:

IA2=ID2

VLC = 0

(a - 2m - 1)2 = 49 + (a - 3)2

3m -1 9m - 22m - 7 \

8+ a

9m 6m +1

(1)

=0 (2)

T () =>a =4m + 4m 57

thay vo (2) ta c c:

= 0

4(m-1)3 m - l 9m2 -22m - 7 4m2 + 4m -57^19m2 6m + 1

4 8 + 4(m -1 ) J 4o 8(m -1 ) + (3m - l)(9m3- 23m2 + 23m -107) = 0

27m4- 78ra3+ 92m2 - 336m + 99 = 0o (m - 3)(27m3 + 3m2 + 101m - 33) = 0 (*)

Do 27m3+3m2 +10 lm -3 3 = (3m -l)(9m 2 + 4m + 35)+ 2 > 0

(do m > ) nn (*) o m = 33Vy m = 3 l gi tr cn tm.

V d 16.3. Gi (Cm)l th hm s y = + + + m + ^x + 1

rng vi mi m , th (Cm)lun c cc i, cc tiu v khong cch

gia hai im bng V20.Li gii.

TX: D - R \ { -1}

Ta c y = X + m + ^

y= l -

x + 1

-;y' = 0x = 0;x = -2(x + 1)2

th (Cm) lun c im cc i l M(-2;-3 + m), im cc tiu l

N(0; m +1) vi mi m .

MN = (0+ 2)2 + (m +1 + 3 - m) =y2.

Ch : Gi s thi hm s y = at cc tri tai im (x0;y0) . tnhv(x)

y0 ta c hai cch sau:

I Cch : Thay trc tip x0 vo hm s

Cd 2: Ta c: y0 =v'(x0)

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40/377

mx +3mx + 2m^-lV d 17.3. Tm m thi hm s: y =

X"1cc i, cc tiu v hai im nm v hai pha vi trc Ox.

Li giiHm s xc nh trn M.. /

_ , , mx2 - 2mx - 5m -1Taco. y = ---------------- T--------

( x - 1)2 ..=> y' = 0 mx2 - 2 m x - 5 m - l = 0 ( x 1) (1)

Hm s c hai im cc tr o (1) c 2 nghim phn bit x1?x2 * 1

c hai im

m * 0o 0

6m -15* 0

6 .

m >0

Hai im cc tr ca th hm s nm v hai pha trc Oxy(x1).y(x2) < 0 (*).? '

Ta c: (xt)= 2m(x1-1 ) , y(x2) = 2m(x2 -1) . c /nvvrn ^ %' => y(x1).y(x2) = 4m2C(x1x2 ~(x1+ x2) +1] = 4m(-2m - 1 ) .

1(*) 4m(-2m -1) < 0

m < 2 .

m > 0

* Vy1

m < ~ , . 1 ,2 l nhng gi tr cn tm.

m > 0

V d 18.3. Chng t rng ch c mt im A duy nht trn mt phng to

sao cho n l im cc i ca th f (x) = m m +..-x - m

ng vi mt gi tr thch hp ca m v cng l im cc tiu ca thng vi mt gi tr m thch hp khc. Tm to ca A .

higii.Hm s cho xc nh trn D = R \ {m}.

rp , >. X2 - 2m x + m 2 - 1

Tac f (x) = --------------- 5------ , x * m^ ' (x -m)Tam thc g(x) = x2- 2mx + m2-1 c A = 1 > 0,Vm.

Do f '(x) = 0 = m -1

x2 = m +1

fOxj = -m 2 + m -2 =>M (m -l ;- m 2 + m -2 )

f(x2) = -m 2 + m + 2 => N(m + l;-m 2 + m + 2)

4 V/- Wt" n A f Hc> rq + (cT -* =2^pii> 3 9

,A " H - 1 * " * b , * ' -\ )0 ^ ^ ------/**.* 4-



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41/377

t A(x0;y0). Gi s ng vi gi tr m = m1 th A l im cc i v

ng vi gi tr m - m 2 th A l im cc tiu ca th hm sx0 = rti -1

Ta c:

Theo bi ton, ta c :

x0 =m2 + 1

= + - 2 j y 0 = -m 22 + m2 + 2 '

-1 = m2 +1

-m 12 + m1- 2 = -m 22 + m2 + 2

J m1- m2 = 2 Jm1- m2 = 2

[(mj - m ,) (m-j; + m,2- 1) = -4 m1+ m2 = - 1

0 => hm s t cc tiu ti x = l .

Vy m = 1 l gi trcn tm.5. T>: D = R\{-m}

i-p , , 3 - mTa c y = X-1 +

x + m, , m - 3 2(m - 3)

y =1 + 7 = (x + m)

Hm s t cc i ti X= - l ' ^ y ,(-l)= '0 ! +

(x + m)m -3

= 0(m-1)2

o m - m 2 = 0 m = - l m = 2 .

m = - => y "(-1) = - < 0 =>X = -1 l im cc i

a m = 2=>y"(-l ) = 2>0=>x = - l l im cc tiu.Vy m = -1 l gi tri cn tm.

Bi 3.3. -1. Ta c: y ' = X2 + 2(m + l)x + 6 - 2m

Yu cu bi ton 0

Khi : X1 + x2 = 2 => hai im cc tr cch u ng thng X = 1.

3. Ta c y' = 3(x2 - 2mx + m2 -1 ) ' = 0 Xj = m + l,x2 = m - lDo Xj x2=> hm s lun c hai cc tr

y1=m(m2 + 3), y2 = (m - l)(m2 + m 4- 4)Yu cu bi ton o y1.y2 < 0 m(m -1) < 0


45/377

Vy 0 < m < 1 l nhng gi tr cn tm.4. Ta c y' = -3 x2 + 6(m + l)x - 3m2 - 7m + 1

=> y ' = 0 o - 3x2 6(m + l)x + 3m2 4- 7m = 0 (1)

Hm c cc tr A = -3m +12 > 0 m < 4 , khi (1) c hai nghim3(m +1) - Vl2 -3 m 3(m +1) + Vl 2 - 3m

X = -------------------------- Xn = -------------------------- ., 3 3Ta thy iin s t cc tu ti Xi, do vy yu cu bi ton tha mn

m 0 m < 1.

3m 2 + m 4 < 0

Bi 4.3.

Ta c y ' = 6[x2 + (m - l)x + m(l - 2m)] =>y' = 0 o

Hm s c cc tr o l - 2 m 5m o m .3

xx = l - 2m

Xo = m

Chia y cho y ta c: y = - (2x + m - l ) y ( 3 m - 1)2X - m ( l - 2m)(m -1)6

:=>y = -(3m - 1)2X - m(m - 1)(1 - 2m) l phng trnh ng thng

qua cc im cc tr ca th hm so.

N X f-(3 m -l)2 = -4 ___'1) Yu cu bi ton < m = l .

1- m ( l - 2u a)(m - 1 ) - 0

2) Yu cu bi ton o -(3m - 1)2 = - l om = 0

2 -m =

3Bi 53.

Ta c: y = m + -^-=> th hm s c hai cc tr m/=m) I

V -m w ~ m JSuy ra AB2 = + l6 (-m ) > 2 j -.16(-m) = 16=> AB>4

-m V-m

ng thc xy ra o = -16m o m2 = m = - .. -m 4 2

Bi 6J .Ta c: = 4x[x2 - m(m -1)] .Hm s c cc tr o m e( -oo;0 )u (l;+ co ).

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47/377

AB song song vi ng thng d, nn -2m = 1 m = (loi).2

Tam gic ABC nhn o lm trong tm j Xl 2 3 f,ly i+y2+y3=o

ihay: - (x^+X2 +X3) + 3(X1 +x2 +x3) + 3m = 0 (*) i'

2 - sDo: Xj + +X3= (xx + x 2 + x3)2 - 2 ( x1x 2+ x 2 3+X 3X J = m e-

2

nn c: (*) -----+ 3m = 0 o m = 0( loai),m = 6(nhn) i2 " 6Vy m = 6 l gi cn tm. |-

B 113. TX: D - K 1'Ta c: y' -4x(x2- m ) . Hm s c ba cc tr o m >0 . IKhi th hm s c ba im cc tr I

A(0; m -1 ), B(Vm; m - m2 -1 ), C(-V m; m - m2 -1 )

Gi H trung im ca BC. \AXJ 1

Gi thit ta C AB = 2.sinC = 2. - => AB2= 2AH \AC

Do m + m4= 2m2m(m - l)(m2+ m -1 ) - 0

y ^Kt hp vi iu kin ta c m = 1 , m = tha mn.

Bi 12.3. * ^ ^ ^ IHm s c hai cc tr tri du th hm s ct Ox ti ba im phn Ibit. Phng trnh honh giao im: IX3 - (2m + l)x2 + 3mx - m = 0 (x - l) (x2 - 2mx + m) = 0

X -- 1X2 - 2mx + m = 0 (*) | |

Yu cu bi ton o (*) c hai nghim phn bit khc 1

| a - m2- 1>0 f , fHay _ o m e (-co;0)u(;+oo). I

[1- m 0 IBi 13.3. TX: D = R I

Tnh y ' = 3(x2- m). Hm s c cc tr o m > 0 IGi Afcpyj) v B(x2;y2) l hai im cc tr ca (C), khi :

- mx1- 3mXj +1 - 3m = -2m x1 +1 - 3m Iy2 = mx, - 3mx2+ 1 - 3m = -2m x2+1 - 3m. IV A, B cch u ng thng d : X- y = 0 nn ta c cc trng hp

oTrung im 1(0; 1 -3m)ca AB thuc d o l - 3 m = 0 o m = - . 3

46



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49/377

CH 4. GI TR LN NHT- GI TR NH NHATc a H M S

I. TM TT L THYT1. nh ngha.* Cho ham s xc nh trn D

i) S M gi l gi tr ln nht (GTLN) ca hm s y = f (x) trn D' f (x)MVxeD nu-T Z ' , ta k hin m = mnf(x).

[3x0 e D : f(x0) = ra v XeD w

2. Phng php tm GTLN, GTNN ca hm s

Phng php chung: tm GTLN, GTNN ca hm s y = f (x) trn Dta tnh y ', tm cc im m ti o hm trit tiu hoc khng tn tiv lp bng bin thin. T bng bin thin ta suy ra GTLN, GTNN.Ch :* Nu hm s y = f(x) lun tng hoc lun gim trn [a;b] th

maxf(x) = max{f(a),f(b)}; minf(x) = min{f(a),f(b)} .

. * Nu hm s y = f (x) lin tc trn [a; b] thi lun c GTLN, GTNN trn

on v tm GTLN, GTNN ta lm nh sauBl: Tnh y' v tm cc im XJS x2,.. .,xn m ti y it tiu hoc hms khng c o hm.

B2: Tnh cc gi tr f(xn),f(a), f(b). KM max f (x) = max {f ( X i f (xn), f (a), f (b)}

^ f (x ) = mi i {f (x ) ,.. . f (xn) ,f (a ) ,f (b )} .Xa;b]

. * Nu hm s y = f (x) l hm tun hon chu k T th tm GTLN,

GTNN ca n trnD ta chi cn tm GTLN, GTNN trn mt on nm

trong D c di bng T.* Cho hm s y = f (x) xc nh trn D. Khi t n ph t = u(x), ta tm

c t E vi Vx e D , ta c y = g (t) th Max, Min ca hm f n D

chnh l Max, Mii ca hmg trn E * Khi bi ton yu cu tm gi tr ln nht, gi tr nh nht m khng nin tp no thi ta hiu l tm GTLN, GTNN trn tp xc nh ca hm s.* Ngoi phng php kho st tm Max, Min ta cn dng phngphp min gi tr hay Bt ng thc tm Max, Min,

48



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50/377

IlvGC v x BUSMENS HA; V 1.4.Tm GTLN v GTNN ca cc hm s sau

_/0 \ 2 2 0 x 2 +1 0 x + 31) y = (3 -x ) V 5 -x : . 2) y v 3x2 +2x + l

3) y = yjx2 + X + 1 -yjx2 - x + 1 trn [2;3]

4) y = x + V 4 -x 2 ( Khi B - 2003 ).

5) y = V-x2 + 4x + 21 - V-x2 + 3x +10 (Khi - 2010 ).Li gi:

1) iu kin : 5 - x 2 > O o x e uv/5; >/53 .

, , 2(x2 - 4) , nTa c y = 7= = ^ = > y ' = 0x = 2

V f ) = 0, f (2) = 1, f (-2) = 5

nn maxf(x) = f(-2) = 5; min f(x) - f(V) = 0.2) Ta thy hm s xc nh vi mi x e R .

10x2 + 22x + 4Ta c: y' =

(3x2+ 2x +1)2

=>y, = 0 < 5 x + ll x 4 -2 = 0 o x = - - , x = 25

10Mt khc: lim y = nn ta c bng bin thin sau:

3

X 00 2 +005

y + 0 - 0 +

y

7

H/3 2

Da vo bng bin thin suy ra max f(x) - 7, mini (x) = .

Ch :Vi bi ton n ta c th gii bng phng php min gi tr nh

sau: y = ^0? +-- (20 - 3y)x2 + 2(5 - y)x + 3 - y = 0 (1)3x + 2x + X

20* y = - => (1) c nghim-

' 2 0* Nu y => (1) c nghim

3

49



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51/377

2

-sv T, , 2x + l 2x - l3) la CO y = , -. -

o A' = -2y 2+ 19y -3 5 >0 < y < 7 .2

Vy maxf(x) = 7, minf(x) = .

5

5

2 V x 2 + x + 1 2 v x 2 - x + 1

:=>y ' = 0 (2x + l)Vx2 - x + 1 = (2x - l)-s/x2 + x + 1 (1)

Bnh phng hai v ta c phng trnh h qu(2x + l)2[(2x - 1)2+ 33 = (2x - 1 ) 2 [ ( 2 x + i f + 3]

(2x + l)2 = (2x - 1)2 X = 0

ihay vo ( 1) a thy 1= 1 v l.Vy phng trnh y' = 0 v nghim hay y r khng i du trn R,

my(0) = 1 >0 =>y >0 VxeR .

Vy max f (x) = f(3) =; min f (x) - f (-2) - -JE - yj .[-2;3) t-2;3)4) Hm s xc nh trn D = [-2; 2]

T ' , 1 x V 4 - X 2 - X . _ r Ta c : y = l ~ r= = . y ' - 0 V4-X =xV 4 - X 2 V 4 x 2

[0< X < 2 r -c * r o X = V2 .

[4- X2 = X2

Do y(-2) = -2;y(2) = 2 ;y (^ ) = 2V2 nn:ma x y = y(\ /2) = 2 72; m in y = y(-2 ) = -2 .

5) iu kin: < o -2 < X< 5-X2 +3 x + 1 0 > 0

-X + 4x + 21>0

Xt trn min -2 < X< 5, ta c: y = -y/25 - (x - 2): , nn

= 0

H x - | j \ / 2 5 - ( x - 2 ) 2H

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52/377

y ' = 0o u - .417

Lp bng bin thin ta c c: min y = -2 SI21- sin 1+ 2; max y = .8

3) t t = cotx

_ . _ 2tanx 2cotx 2t _ t - =>sin2x = ~; cos2x = -

1 + t a n X 1 + c o t X 1 + t t + 1

_ t 2 + 2t - l

. . _ (sin4x-i-2 sin2X - l)(eos4X + 2COS2 X -1 )

g (sin4 X + l)(cos4 X+1)

sin4 xcos4 X + 8 sin2 xcos2 X - 2 u 2 + 8 u - 2 T . .g(x) = " L-----~ ~ = h(u) .

sin XCOS x -2 s in xcos x + 2 u -2 u + 2

trong u - s in 2 X COS2 x; 0 u < .4

, -5u2 + 4u + 6 15 ------r- >0 Vue[0;

(u -2 u + 2) 4hm s h(.u) .lun tng trn E0;] nn max h(u) = h() =

4 ,10:1/4] 4 25mmh(t) = h(0) = - l .

Vy maxg(x) = ming(x) = -1 .25

4) Vi | V x T T j + |V s ^ x j = 4 => tn ti s thc t e [0;l]



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54/377

sao choVxT =

V s - X =

4t

i + t 2 -T7-1 - J., 2 t 2 4 t 6f / 4 . \ '*.Tn_ . K m:y = ------ = f(t) vi t e 0;12(1 1 ) t - 8t - 3 1 j

1 + t2

Ta c: f (t) = } - 362 0,y > 0 V X + y = . 4

Tm GTNN ca biu thc p = + .X 4y

Li gii5 4 1

Tac x + y = :=>4y = 5 - 4 x =>p = +4 X 5 - 4x

Xt hm sf (x) = + - ,xef o;v ' X 5 - 4 x ^ 4

= > f ' ( x ) * ~ + 7 ^ - f , (x ) = 0 o x = 1 '

v 7 X2 ( 5 - 4 x)2Lp bng bin thin ta c: min f (x) = f (l) = 5, t suy ra min p = 5

K)t c khi X = l;y = .4

1ng thc xy ra khix = l,y = T- .

4Ch : Bi ton n c th gii theo cch sau

T ( \ T> 17 L x s l o _ 2i 'Ta c: (x + y).p = + + >_ + 2 = ~4 X 4y 4 4

f 5x + y = 4 1Suy ra p > 5. ng thc xy ra khi < A x = l,y =

4 y _ X 4

[ X _ 4y

Vy maxp = 5.

53



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55/377

V d 4-4- Cho X, y e [-3; 2] tha X 3 + y3= 2 . Tm GTLN, GTNN ca biu

thc: p = X 2 + y2.Li gii.

T gi Ihit ta suy ra c X = /2- y 3 thy vo p ta c

P = V(2-y3)2 +(y3f =^(2 - t ) 2 + $/F = f(t)

Trong ta t t = y3.

V x e [-3; 2] => X3 e [-27;8] => -27 2 - y3 < 8

6 < y3 < 29 ,do y3e [ -2 7 ;8 ] = >t e [6;S].

Xt hm s f(t) n D = [-6;8], ta c:

= ------1 = => f (t) = 0 o ^ 2 - t = ^ o t = l .3%/t 3.v2 - 1 ______ ____

t -6 0 1 2 8

f'(t) - II + 0 -

f(t) \ ^ \ ^Da vo bng bin thin ta c c: min p = min f(t) = f(0) = f(2)=3/4

t c khi (x,y) = (0,\/2) hoc hon v.

max p = max f(t) = f ( - 6 ) = 4 + \SQ .

t c khi (x,y) = -$3;2j hoc hon v.

Nhn xt:* Cch gii trn ch i hi chng ta k thut kho st hm s. Ci khca bi ton trn l iu kin hn ch ca x,ye [-3;2]! Nen x,y khng

b rng buc bi iu kin ny th bi ton tr nn om gin v ta c thgii bi on trn theo cch chuyn qua.tng v tch ca x ,y .

fa3 -3ab = 2t a = x + y,b = xy:=>^|a2 > 4b

fb ^ S- 2

3a2 _ a 3 - 2

a > 4 -------1 3a

a3- 8=>-------


56/377

Xt hm s f(a) vi a e D = (0;2] c: f'(a) = ~ ^ ^3 3a 3a

:=>f'(a) = 0 a = \/2 .

Lp bng bin thin ta c ngay min p =f(y/2) = y4

t c khi x,y e |o ,W |.

lim f(a) = +00=> p khng c GTLN.a-0+

* Khi gp bi ton m cc biu thc c trong bi ton l cc biu thci xng hai bin th ta c th chuyn v bi ton ca tng v tch hai bin vi lu Ss > 4P ___________________.

V d 5.4. Cho a}b l cc s thc dng tha mn: ab + a + b = 3. TimonpT \T TI 3a 313 b 2 12GTLN ca: p = - + - + - a - b .

b+1 a+1 a+bLi giLNhn iy cc biu thc'C ong bi ton l cc biu thc i xng haibin a,b n n ta tt = a + b=>ab = 3 - t v

a2 +b2 =t2 - 2 ( 3 - t ) = t2 + 2 t - 6

V (a +'b)2 4ab => t2 > 4(3 - 1) o t2 + 4t - 1 2 > 0 t > 2 (do t > 0 )

Khi : p = 3(-?2+ b2>+ 3(a + b) + - g ~ ( a 2 +b2)(a + l)(b + l) a + b v

3t2 +6 t-18 + 3t 3 - t . 1 . 0 12. 1 ----------- ;--------- + ---------- r - 2 t + 6 = - ( - r + t + - ) + - .4 t 4 t 2

Xt hm s : f(t) - - t 2 + 1+ vi t > 2.t

_ 12Ta C: f'(t) = ~2t + l- -^ -< 0 Vt>2

t2

=>f( t) < f(2) = 4 Vt >2 p < . ng thc c o a = b = l .2

Vy maxP = . 's2V d 6.4. Cho cc s thc X, y thay i v tho mn (x + y)s + 4xy > 2. Tm

gi nh nht ca biu thc : A = 3(x4+y4 + x2y2)-2 (x 2+ y2) + l .Li gii.

, (x + y)3 + 4xy >2 3 2Ta c: < _=>(x + y) +(x + y) - 2 > 0 =>x + y >1.

(x + y) -4 xy > 0

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57/377

A = 3(x4 +y4 +x2y2-2(x2 + 5 ) + l ^ [ ( x 2+y - z zy ~- (x? + yz) + l

>3 (x2+y2)2_(s2+y2)24

-2(x2 +y2) + l = (x2 +y2)2 2(x2-+y2) + l4

t t = X2 + y2 > X+ y- > => t > v A > t2 - 2t +1.2 2 2 4

, 9 0 1Xthms: f(t) = t - 2 t + l , t > C

4 2

f ' ( t ) = 1 - 2 > 0 V t > = > f ( t ) > f ( ) =

2 2 2 169 1

=> A > . ng thc xy ra X= y =.16 5 J 2

9Vy gi tn nh nht ca A bng .

V d 7.4. Cho iai s thc a, b > 0. Chng minh: a4 + b4 : a3b + b3a (1).Li gi i:

* Nu mt trong hai s a,b bng 0 th (1) lun ng.* Vi a 0, t b = ta .Khi (1) tr thnha4( l + 14)>.a4(t + t3) o t 4 - t 3 - t + l > 0 .

Xt hm s (t) = t4 - 13 - 1 +1, c :

f (t) = 4t3 - 3t2 -1 = (t - l)(4t2 + 1 + 1) =>f (t) = 0 o t = 1.Lp bng bin thin, t suy ra f(t) f(0) = 0 pcm.

Nn x t: * Bi ton n ta ch cn bin i trc tip l Cc kt qu* Cch gii trn c trnh, by lu vi chng ta v mt tnh cht l tnh cht ca biu thc ng cp hai bin.C th: Biu thc f(x?y) c gi l biu thc ng cp bc k nu:

f(mx, my) = mkf (x, y) .' f (x )

Khi gp bi ton chng minh BT hai bin c ang : - > p , trong7 ^ f g(x,y)

f(x,y) v g(x?y) l nhng biu thc ng cp bc k hai bin, ta c

th t X= ty (y * 0). Khi BT cn chng mii tr thnh:

ffi' > p y l BT mt bin. chng minh BT ny ta c th..sg(t,l)ng phng php kho st hm s.* Khi gp biu thc ng cp ba bin a,b,c ta c th t b = xa5c = yav chuyn v bi ton hai bin.__ __ ___

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58/377

V d 8.4. Cho hai s thc x,y thay i v tha mn h thc X2 + 2 = 1 .

Tm GTLN, GTNN ca biu thc p = x + 6xy HQr 3 _ 2008).. 1 + 2xy + 2y

Ta c: p =

* 6* * , ,1 + 2xy + 2y X + 2x + 3y* Nu y - 0 => p = 1.* Nu y50 th t: X = ty, t e l

t y + 2ty + 3y t + 2t + 3

Ta c :f'(t)= 4t ^ Gt + 1f ,f'(t) = 0 o t 1= 3,t2 = limf (t) = l. (t2 +2t + 3) 2 t~>*0

Lp bng bin thin ta c: f - j < f () < f (3), Vt =>-3 < f (t) 0 a< 2

* Nu a = 0x = y = 0=>p = 0 (1)

*Nu a * 0, ta gi s y 5t 0 . t X = ty

^ p . > g k , i = i i a f(t)a X + xy + y t + 1 +1

Kho st hm s f(t) tac: f'(t) = y^--- t J" f'(t) = 0 t = ( t + t - r 1 ) 2

T ta c c min f (t) = f (" ) = ----r^ j

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59/377

7+ 2V7 _ 7-2>/7 ^7 + 2>/7max fit) = f( r) = --------- ==>----- r--- < < ---------

2 3 3 a 3_ 7 -2^ 7 _ 7 + 2V7 /-=>---- a < p < a < 7 + 2V7 .

3 3

Vy min p = 0 ; max p = 7 + 2V7.

V d 10.4. Cho X, y,z l s thc tha mn X2+ y2 + Z2 = 2.Tim gi tr ln

nht, nh nht ca biu thc: p = Xs + y3 + zs - 3xyz.Li gii:

T cc ng thc : X2 + y2 + z2 + 2(xy + yz + zx) = (x + y + z)2

X3 + y3 _|_23 _ 3xyz = (x + y + z)(x 2 + y2 + z2 - xy - yz - zx)

v iu kin ta c:

(x + y + z)2- 2p = (x + y + z)(x + y + z - xy - yz zx) = (x 4- y + z) 2

2

j.2__2

|3t t = x + y + z=> -y/ f'( t) = 0 o t = y 2

max f(t) = f (>/2) = 2>/2; min f(t) = f(-V2) = - 2V2" [->/6;V]

Vy maxp = 2V2 t c khi X = >/2;y - z = 0

min p - -2%/2 t c khi X = -%/2;y = 2 = 0.V d 11.4. Cho cc s thc khng m a,b,c tha mn: a + b + c = 1. Tmgi tr nh nht ca biu thc :

p = 3(a2b2+ b2c2 + c2a2) + 3(ab + bc + ca) + 2Va2 + b2 4-c2 .

= ab + bc + ca => 0 < t < (a + k + c) _ 1.3 3

Khi : 3(a2b2 + b2c2 + c2a2) ^ (ab + bc + ca)2 = t2

a2 + b2 + c2 = (a +b + c)2- 2(ab + bc + ca) = 1 - 2tSuy ra p > t2 + 3t + 2y j l -2 t = t2 + 3t + 2 (\/ l- 2 t -1 ) + 2 = f(t)

Xt hm s f(t) vi te 0 ;-3

ta c:

f Yt) = 2t + 3 r ^ = = > '"(t) = 2 - = L =


60/377

Suy ra f'(t) l hm gim trn0;i

f tng => f(t) f(0) = 2, vt e

T tac c rinP = 2.V 12.4. Cho x,y l cc s thc thay i. Tm gi tr nh nht ca biu

thc: P = ( x - l) 2 +y2 +^(x + l)2 +y2 +y -2 |.

Li giLTrong h ta Oxy, xt u (x -l; y ) ,v (x + l;y) .

Do u + V u + v nntac:1 11 1 1 i V

^ (x - l ) 2 4-y2 + x + l f +y2 >yj4 + y = 2>/l + y2 .

Do A ^2-y/l + y2 +y-2 = f(y)

* Vi y < 2 =>.f (y) = y j+ y2 + 2 - .

Lp bng bin thin suy ra ngay f (y) > f ( - i ) = 2 + a/3 .V 3

* Vi y >2 =>f(y) = 2 ^ * + |y 2|>2\/+*->2>/5 >2 + >/3

Vy gi tr nh nht ca p bng 2 + V. ; _____________

Lm : Khi gp biu thc ong cn c dng tng hai bnh phng ta lintng n phng php hnh hc vi nh gi quen thuc sau:

Cho & vc t UijUaj.-jUjj, khi ta c: X ui - 2 j Uii=i 1=1

4V d 13.4. Ch.0 a, b, c l cc s thc dng tio ab + bc + ca > .

3

Tim GTNN ca biu thc: p = ja2+ + b2 + ~ + c + S-.i (b+1)2 V (c+l) (a+1)Li gii.

Xt ba vc t sau:u = (a;!), V = Ib; 1, m = Ic; I.b + 1 V c +1 y \ a +

p dng BT u + V -4- m > u + V + m ta c:

p >J(a + b + c)2 +1 -h + I > J(a'-i-b + c)2 +b+1 c+1 a+1

t t = a + b + c= > t>yj3(ab + bc + ca) = 2

81( + b + c +3)

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61/377

Xt hm s f (t) = t2 + , tk 2 .(t + 3)

Tac :f(t) = 2 t ---- = 1691(t + 3) (t+3)

Trong: g(t) = (t -2 )( t3 + l l t 2 +4 9t + 125) + 169

=>g(t) >0 v t> 2 =>f'(t)>0 v t> 2 =>f(t)>f(2)= v t >225

0 \/8=> p > --

5

ng thc xy ra a = b = c = .

V y m i n P ^ .

V d 14.4. Cho cc s thc dng a, b,c tha mn (a + b + c)3 =32abc.

Tm GTLN, GTNN ca biu thc: p = a4 + b4 + c4

(a + b + c)

Li gii:Khng mt tnh tng qut, ta gi s: a + b + c = 4 =>abc = 2

Khi: p = (a4 +b4 +c4V= Q.256' ' 25.6

t t = ab +. bc + ca . Ta c :

Q = (a2 +b2 +c2)2 - 2 ( a V + b2c2+ c V )

= (a + b + c)2 - 2 (ab 4- bc + ca) j -2 | (ab + bc + ca)2 - 2abc (a + b + c) j

= (42 - 2t)2 - 2 (t2 -1 6) = 2(t2 - 32t +144).

2 2t = ab + bc + ca = a(b + c) + bc = a (4 -a ) + = -a 2+ 4a +

a a

M (b + c)2> 4 b c o (4 -a )2 > o ( a - 2 ) (a 2 -6 a + 4]>0

o - 3 - \ / 5 a < 2 ( v 0 < a < 4 )

Xt t = -a 2+ 4a 4-, a 5 t


62/377

Vy maxP = t c chng hn khi X = 2;y = z = 1

minP = 383~165>/5 (Jc chng han khi X = 3 - V5; y = z = 1 + .256 2

V d 15.4. Cho cc s ng a, b, c vi a + b + c < 1.

Tm GTNN ca biu thc : p = 3(a + b + c) + 2^+ + j .

9

Li gii:

Ta c: (a + b + c)f+ - + - ! > 3%/bc 3?/ =9=>\ a b c j Vabc

Do : p > 3(a + b + c) + = 3^t + - j = 3f(t)

6Trong O - + zL+ ->-

a b c a+b+o

Ta c : f T(t) = 1J =2 < 0,vt e (0;1], nn hm s nghch bin trn "t

t2 - 6

t 2

(0;1] => f ( t )> f ( l) = 7,v te (0 ;l].

1Vy minP = 21 t c khi a = b - C = .

. 3

V d 16.4. Cho a,b,c > 0 v a2 + b2+ c2 = 1. Tm GTNN ca biu thc:

p = + + (a + b + c).

a b c L i gii.

t t = a + b + c 0 f(t) ) - V , v t (0;Js].

Vy minP = 2V3 t c khi a = b = c = -J=.

V d 17.4. Cho 4 s thc a,b,c,d io mn: a2 + b2 = 1; c - d = 3 .

Tm GTLN ca biu thc: F = ac + bd - cd < 9 +



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63/377


64/377

1 2 3t: x = ,y = ,z = =>x,y ,z>0.

a b cKhi : 21ab + 2bc + 8ca f(3) = .

ng thc xy ra khi:

X = 3

2x + 2 x y - 7 _ + X

2x 2xy - 7

2x + 4y

2xy - 7

X = 3

5y =

2z = 2

a =

3

b = .53 .

c

2

Bi 1.4. Tm gi tr ln. nht, gi tr nh nht (nu c) ea cc hm s sau:Ly = 2x3 - 3x2 - I2x +1 trn on [-1;4]

63



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2. y = -- + X2 H-1 trn on [-1; 3]

3. y = ^X-^ n oan [-3;-2l3 x - 2 '

A _ 2 x 2 - 3 x + 3 a . [3 94. y = ------------ trnQn

X1 . [2 2.5. y = 2x - Vx2 + 1 trn on {-2; 5]

6. y = x n khong (0;-t-oo], ....

7. y = COS2 X. sin X trn [0; tt]

8. y = |xs - 3x + 2| trn [-1;3]. .

9. y =[xs +3x2 -7 2x + 901 trn [~5;5] .

sinx + 110. y - 2 3 - c o s X

- fx>0, y >1Bi 2.4. Cho hai s tic X, y io mn: { . Tm gi tri nh nht,

[x+y=3

gi tr ln nht ca biu thc: p = Xs + 2y2+3x2 + 4xy 5x -

r f xy + 3 = 0Bi 3.4. Cho X > 0 v s thc y tha mn: .Tm gi tr ln

[2x + 3y tha mn X2 + y2 + z2 =1. Tm gi tr nh nht ca

biu thc p = - + -- + + 2xyz.X y

Bi 6.4. Tm tt c cc gi tri ca a v b tho mn iu kin: a > - v f -> 12 b2a3+1 '

sao cho biu thc p = -------T- at gi tri nh nht Tm gi tri nh nhtb(a-b)

-Bi 7.4. Cho bn .s nguyn a, b, c, d thay i tha: l< a < b < c < d < 5 0 .

a c

X4

Tm GTNN ca biu thc p = + ( bi - 2002).b d

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Bi 8.4. Cho tam gic ABC khng t. Tm GTLN ca biu thc:p = cos2A + 2^ (c osB + cosC) (Khi A 2004 ) .

Bi 9.4. Cho cc s thc dng x?y. Tm gi tr ln nht ca biu thc:

p _ . 4xy2

(x + -n/ x 2 + 4y2 )3

Bi 10.4. Tu theo gi tr ca tham s m, hy tm GTNN ca biu thc:p = (x + my - 2)2 + [2x + (m - 2)y - 1]2.

Bi 11.4. Cho cc s thc m,n,p,a,b>0 sao cho: -

5a2- 3ab + 2

______1m + n + p = mnp -

a____b

mn + np + pm = a

Tm GTNN ca: p -a2(b-a)

Bi 12.4. Cho cc s thc a, b,c tha a2+ b2 + c2 = 1. Tm GTLN ca biuthc: p = 2ab + bc + c a .

Bi 13.4. Cho cc s thc ng a,b,c tha a + 2b + 2c-a b c. Tm gi trnh nht ca p = a + h +c .

Bi 14.4. Cho cc s thc khng m a,b,c c tng khc 0. Tm gi tr ln

1. ^ T> a3+b3 +16c3nht, gi tr nh nht ca biu thc: p -----------.(a + b + c)

Hng dnBi 1.4.1. maxy = y(4) =33 ,min = y(2) = -19

- l ;41 [ - 1:4]

2. max y = y(V2) = 2 , min = y(3) =[ - 1:3] E -1;3I 4

3. m ax y= y(-3) = , min = y(-2) = 372 8

6 0 . / 0 \ _ c4. max y = y() = , min - y(2) = 59 yy2 7 rl22 22

5. maxy = y(5) = 10 - J , min = y(-2) = -4 - y[E2:5 2:5

6. miny = y(>/3) = 2V3, khng tn ti GTLN.(0;+co)

7. miny = y(0) = y(7t) = 0 ,[0;tc]

maxy = t c khi X= a r c s in - i .[0;] 9 ' V3

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8. miny = y(l) = 0, maxy = y(3) = 20.(1;3J

9. max y = y(-5) = 400 .[-5,5}min y = 0 vi x0 e (1; 2) l nghim ca phng trnh:[-5;5]

X3 + 3x 2 - 72 x + 90 = 0 .

1+V10. miny = 0, maxy = -

Ta c y=3-x^l=>xxe[0;2]

Khi : p = X3+ 2(3 - x)2+ 3x2 + 4x(3 - x) - 5x

= X3 + X2- 5x +18 = f (x)

Xt hm s f(x) trn [0; 2] ta c: f'(x) = 3x2 +2x-5=>f'(x) = 0 o x = l

Vy max p = max f(x) = f(2) = 20, min p = min f(x) = f(l) = 15.[0 2 [0 ;2

Bi 3.4.

X2 +3y =

T gi thit suy ra x =>5x2-14x + 9< 0 o 1 x a2 + b2 + c2 > 3\/a2b2c2 => abc < .4 8

p dng BT C si, ta c: (a + b)(b + c)(c + a) 8abc

\ + + ^P^ Sabc + -^ - .a b c abc abc1 _ 3

t t = abc => 0 < t < v p > 8t + - f(t)8 t

X t h m f (t ) c f '( t) = 8 < 0 => f (t ) > f() = 25 v t e ( 0 ; 3t 8 2

=> p > 25 . ng thc xy ra a = b = c = .

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Bi 5.4.

Ta c: - + + - > => p > + 2xyz = + 2t3 =f(t)X y z %jxyz yjxyz t

Vx2 +Y2Trong , ta t t = lxyz=> 0 < t

+ 2T _ _1_

Xt hm s f(t), tac: f'(t) = 6t1 = ^ , ^ < 0 v t e ( 0 ; i ]t t V 31 l_29>/3

Vy minP =29>/3

Bi 6.4. T gi thit, ta suy ra a * 0 y b(a - b) > 0.

Ta c : 0 < b(a - b) < v 2a3+1 > 0 nn p > = f(a).4 a 2a3- 2

Xt hm so f(a), a > c f '(a) = -----3 => f(a) = 0 a = 1

Bng bin thin:X - 1 / 2 0 1 +00

f ' + 0 +

f3 ^ 3. "

T bbt f(a) > 3 Va > => p >2 2

ng ic xy ra o

Vy minP = 3.

a = fa - 1^ hoc \ I .1 ]b =

b = 7 l 24

Bi 7.4. V 1 a < b < c < d < 50 v a,b,c,d l cc s nguyn nn c b +1.c _ a c l b+1 ,Suy ra + + = f(b).

b d b 50

D thy 2 < b < 48 nn ta xt hm s : f (x) = + x + ,x g [2;48]

Tacf'(x) =\ - + =>'x) = 0x = 5\2.y J X2 50 w

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Do 7 v 8 l hi s nguyn gn 5V2 nht v vy:

min f (b) = mn{.f (7) ;f (8)} = = -5?-.[2;48] v ' I w W J 175 200 J 175

V yG TNN P= .175

Bi 8.4. Ta c A < 90 => cos2A = 2cos2 A -1 2cos - 1 = 1 - 4 sin2

ng thc xy ra COS2A = COSA (1).n r _ 0 . c B - C . c

cosB + cosC = 2sm .COS------- 0 < t < . Ta C: p < -4 t2 + 4rj2,t+1 = f(t)2 2

/2 /2Xt hm s f(t), te (0 ;], c f'(t) = -8t + 4>/2 =>f'(t) = 0 o ' t =

J oLp bng bin thin ta c: f(t) < f( ) = 3 => p < 3.

2

A = 90ng thc xy ra o J

[B = c = 45

Vy maxP = 3.

Bi 9.4.

Lp bng bin thin ta c minf (x) = f 5V2)

Xt f(t)=- ----- 4t t > 0 => f ' (t) = - - 3t) (t + Vt2 +4)3 \/t2 + 4 ( tW t 2 + 4) 3

=>f'(t) = 0 o ^t2 -f 4 = 3t t = - i - .s/2

Lp bng bin thin ta c msta f (t) = f =.

Bi 10.4.

, X + my- 2 = 0 ^Xt h: < (*) c D ='-m - 2 .

[2x + (m -2 )x -1 = 0

Nu m * -2 => (*) c nghim duy nht (x0; y0)

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Khi p > 0, p = Oo x = :

=>minP = 0.=y *a 3(m n+ np+ pm)

m + n + p > 3^/mnp

X ,h . s

Suy ra f(b)gimtrn (0; ] => f(b)>f^ J

Xt hm g(a), via 6 ( 0 ; - ^ ], g'(a) = l 0 ^


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Da vo BBT ta suy ra f(t) < f( t0) =1 + V3 1+V

ng thc xy ra o

2ab + bc + ca > 0

a2 + b2 + c2 = 1

a = ba = b =

3+V12

c=3 - V 3

6

1 + V

c =3 - V 3

rp< _2(b + c) jhcT gi thit => a =

b c - 1 b c - x, 4Vb . 4-s/bc

=> a 4- D+ c > ---- hb + c -------b c-1 be -1

Xt hm s f(t) vi t =yfbc> 1 .

Ta c: f (t) = 4(t27 1)~ 8t2 +2 = 2(t 4t +1)

(ta c be > 1)

2 ^ = - - + 2t = f(t).t - 1

(ta - i r (t* -

=>f(t) = 0 l) .

Lp bng bin thin => f(t) > f(t0) :=>a + b + c > f(t0)

"b = c = t0

2t0 . Vy min(a + b + c) = f(t0) .ng thc xy ra

a = t2 -1 Lq J.

Bi 14.4.a 3 4- "h3+ 1fir*3

f(a;b;c) = ---------- 3' . V f (ka;kb;kc) = f (a;b;c), Vk > 0 nn(a + b + c)

khng gim tnh tng qut ta gi thit a +1) + c = 1, khi

f(a;b;c) = a3 +b 3 +16c3 . Ta c a3 + b3 > (a + b)3 = ( l - c ) 3 nn

f(a;b;c) > ( l - c ) 3 + 16c3 = ( l- c ) 3 + 64c3 = g ( c ) ,0 ^ c lKho st hm g(c), c [0; 1] ta c

/ \ 84 e , , \ 1 16 . d 168 1 ^ (a ; ; c ) 4 g ^ 81 81'

Mtkhc: f(a;b;c) = a3 +b3+c3+15c3


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9. Tip tuyn: Gi I l im un. Cho M e(C)* Nu M = I th ta c ng mt tip tuyn i qua M v tip tyn ny ch s gc nLnht (nu a > 0), ln nht (nu a< 0).* Neu M khc th c ng 2 tip tuyn i qua M.

n. CC V D MINH HAV d 1.5. Cho

18 CHỦ ĐỀ GIẢI TÍCH 12 - NGUYỄN TẤT THU

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