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8/7/2019 BDT Hoang Thanh Thuy Logo

1/43

CHUYN

GI TR LN NHT - GI TR NH NHT

Hong Thanh Thy


2/43

Chuyn Gi tr ln nht, Gi tr nh nht 1

Mc lc1 nh ngha v cc tnh cht 2

1.1 nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Tnh cht ca GTLN, GTNN . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.1 Tnh cht 1: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.2 Tnh cht 2: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2.3 Tnh cht 3: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2.4 Tnh cht 4: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2.5 Tnh cht 5: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 Cc phng php tm GTLN, GTNN 42.1 Phng php hm s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.1.1 Ni dung phng php . . . . . . . . . . . . . . . . . . . . . . . . . 42.1.2 Cc v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.2 Phng php s dng cc bt ng thc . . . . . . . . . . . . . . . . . . . 8

2.2.1 S dng bt ng thc Csi . . . . . . . . . . . . . . . . . . . . . . 82.2.2 S dng bt ng thc Bunhiacpxki . . . . . . . . . . . . . . . . . 132.2.3 S dng bt ng thc Trbsep . . . . . . . . . . . . . . . . . . . 18

2.3 Phng php min gi tr . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.3.1 Ni dung phng php . . . . . . . . . . . . . . . . . . . . . . . . . 212.3.2 Cc v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.4 Phng php lng gic . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.4.1 Ni dung phng php . . . . . . . . . . . . . . . . . . . . . . . . . 262.4.2 Cc v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.5 Phng php hnh hc, to v vect . . . . . . . . . . . . . . . . . . . . 302.6 Cc phng php khc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

2.6.1 Phng php cn bng i xng . . . . . . . . . . . . . . . . . . . . 342.6.2 Phng php cc bin . . . . . . . . . . . . . . . . . . . . . . . . . 362.6.3 Phng php sp th t . . . . . . . . . . . . . . . . . . . . . . . . 36

3 ng dng 383.1 Gii phng trnh, bt phng trnh. . . . . . . . . . . . . . . . . . . . . . 38

3.1.1 Cc nh l . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.1.2 Cc v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.2 Tm iu kin cho tham s m phng trnh, bt phng trnh, h phngtrnh, h bt phng trnh c nghim. . . . . . . . . . . . . . . . . . . . . . 40

Hong Thanh Thy


3/43


1 nh ngha v cc tnh cht

1.1 nh nghaCho hm s f(x) xc nh trn min D. Ta ni rng M l GTLN ca f(x) trn D, nunh ng thi tho mn hai iu kin sau y:

1. f(x) Mx D2. Tn ti x0 D sao cho f(x0) = M

Khi k hiu M = maxD

f(x) Ta ni rng m l GTNN ca f(x) trn D, nu nh ng

thi tho mn hai iu kin sau y:

1. f(x) mx D2. Tn ti x0 D sao cho f(x0) = m

Khi k hiu m = minD f(x)Ch :

Khi ni n GTLN hoc GTNN ca mt hm s bao gi cng phi bit n xc nhtrn tp hp no.Cng mt hm s f(x) nhng nu xc nh trn cc tp khc nhau th ni chungcc GTLN, GTNN tng ng l khc nhau.

cho thun tin, ph hp vi chng trnh ca cc lp ph thng, trong ti liuny khi cp n GTLN, GTNN trn tp hp no , ta lun gi thit l chngc tn ti.

1.2 Tnh cht ca GTLN, GTNN1.2.1 Tnh cht 1:

Gi s A B, khi ta c:1. max

xAf(x) max

xBf(x)

2. minxA

f(x) minxB

f(x)

1.2.2 Tnh cht 2:

Gi s D = D1 D2. Khi ta c cc cng thc sau:1. max

xDf(x) = max{max

xD1f(x), max

xD2f(x)}

2. minxD

f(x) = min{minxD1

f(x), minxD2

f(x)}

Tnh cht trn cho php ta chuyn vic tm GTLN, GTNN ca mt hm s trn tp Dphc tp v vic tm cc gi tr tng ng trn cc tp D1, D2 n gin hn.Tng qut ta c th vit D thnh hp ca n tp khc nhau.

Hong Thanh Thy


4/43


1.2.3 Tnh cht 3:

Nu f(x) 0x D ta c:maxD

f(x) =

maxD

f2(x)

minD

f(x) = minD f2(x)Tnh cht trn cho php ta thay th vic tm GTLN, GTNN ca hm s f(x) v vic tmGTLN, GTNN ca hm s y = f2(x) nu bit rng f(x) 0, x D. iu ny rt haydng nu f(x) c cha cn thc hoc du gi tr tuyt i.

1.2.4 Tnh cht 4:

1. maxD

(f(x) + g(x)) maxD

f(x) + maxD

g(x)

2. minD

(f(x) + g(x)) minD

f(x) + minD

g(x)

Du bng trong (1) xy ra khi c t nht mt im x0 D m ti f(x) v g(x) cngt GTLN.Du bng trong (2) xy ra khi c t nht mt im x1 D m ti f(x) v g(x) cngt GTNN.

1.2.5 Tnh cht 5:

max f(x) = min(f(x))

Hong Thanh Thy


5/43


2 Cc phng php tm GTLN, GTNN

2.1 Phng php hm s2.1.1 Ni dung phng php

Dng o hm kho st hm s, sau lp bng bin thin (nu cn thit) t gii quyt bi ton.V chng ta ch kho st hm s 1 bin nn dng c phngphp ny i khi phi thc hin nhng php bin i thch hp lm gim s lngbin, chng hn, tnh cc bin cn li theo mt bin, t n ph.Ch : Khi s dng phng php ny nu c cc php i bin th ta phi tm li minxc nh.

2.1.2 Cc v d

V d 2.1.1. Tm GTLN, GTNN ca hm s

y = x + 1x2 + x + 1

.

Li gii. Tp xc nh ca hm s l D = R. Ta c

y =x2 2x

(x2 + x + 1)2.

Do y = 0 x = 0; x = 2. Ta c bng bin thin ca hm sx

y

y

2 0 +

0

1/3

1

0

0 0 +

T bng bin thin suy raGTLN ca hm s l max y = y(0) = 1.

GTNN ca hm s l min y = y(2) = 1

3 .

V d 2.1.2. Tm GTNN ca f(x, y) = 2x2y + xy2 trn min

D = {(x; y) : 0 x 1; 0 y 2}.

Li gii.Nhn xt: T dng ca f(x, y) ta thy nu coi mt trong hai bin l hng s th gi trca f(x, y) hon ton xc nh theo bin . Ta c

minD

f(x, y) = min

0y

2

min

0x

1

f(x, y).

Hong Thanh Thy


6/43


Xt g(x) = 2x2y + xy2. Ta c

g(x) = 4xy + y2.

Bng bin thin:

xg(x)

g(x)

0 y/4 1

g(0) g(1)

0+ +

(Ch : do 0 y 2 nn 0 y4

< 1)

T bng bin thin ta thymin0x1

g(x) = min

g(1); g(0)

= min(0; y2 2y)= y2 2y.

(V vi 0 y 2 th y2 2y 0)Suy ra

min f(x, y) = min0y2

(y2 2y) = 1.

Vy min f(x, y) = 1 khi x = 1; y = 1.V d 2.1.3. Tm GTLN, GTNN ca hm s

f(x) =

1 + 2 cos x +

1 + 2 sin x

trn minD = {x : 1 + 2 cos x 0;1 + 2 sin x 0}.

Li gii.Do f(x) 0, x D nn vic tm GTLN, GTNN ca f(x) c th quy v tm GTLN,

GTNN ca f2

(x).Xt

g(x) = f2(x) = 1 + 2 cos x + 1 + 2 sin x + 2

1 + 2(sin x + cos x) + 4 sin x cos x.

Hong Thanh Thy


7/43


t t = sin x + cos x th t =

2cos

x 4

. Ta c

1 + 2 cos x 01 + 2 sin x 0

cos x 12

sin x 12

6

+ 2k x 23

+ 2k(k Z)

512

+ 2k x 4

512

+ 2k

cos( 512

+ 2k) cos(x 4

) 1

6 24

cos(x 4

) 1

3 12

t

2.

Vy bi ton quy v xt hm

h(t) = 2 + 2t + 2

2t2 + 2t 1trn min

3 12

t

2.

Ta c

h(t) = 2 + 2.2t + 1

2t2 + 2t

1

> 0, vi mi

3 1

2 t

2.

Do h(t) ng bin trn D1 =3 1

2;

2

. Suy ra

min h(t) = h3 1

2

=

3 + 1

max h(t) = h(

2) = 4(

2 + 1).

Do mi t D1 u tn ti x D nnmin g(x) =

3 + 1

max g(x) = 4(2 + 1).Tm li, chng ta s dng phng php ny khi biu thc cho c th a v hm

s tnh c o hm. V xin nhc li, khi bn t n mi th iu kin ca n mi phil iu kin chnh xc, khng c ly iu kin o.

BI TP VN DNG

Bi tp 2.1.1. Tm GTNN ca hm s

f(x) = (1

x)(2

y)(4x

2y)

Hong Thanh Thy


8/43


trn minD = {(x; y) : 0 x 1; 0 y 2}.

Hng dn gii: Tng t V d 2.1.2.p s: min f(x, y) = 2 khi x = 0; y = 1.Bi tp 2.1.2. Tm GTLN, GTNN ca

f(x) =1

sin x + 4 1

cos x 4 , khi x R.

Hng dn gii: Tng t V d 2.1.3, t t = cos x sin x.p s: min f(x) =

4

8 +

2;max f(x) =

4

8 2 .

Bi tp 2.1.3. Tm GTLN, GTNN ca:a) f(x) = |1 + 2 cos x| + |1 + 2 sin x|b) f(x) =

a + cos x +

a + sin x.

Hng dn gii: Tng t V d 2.1.3.

Bi tp 2.1.4. Tm GTLN, GTNN ca

y =cos2 x + sin x cos x

1 + sin2 x.

Hng dn gii: t t = tan x.

Bi tp 2.1.5. (i hc Giao thng vn ti - 98). Tm GTLN, GTNN ca hm s

y = sin2x

1 + x2+ cos

4x

1 + x2+ 1.

Hng dn gii: t t = sin2x

1 + x2th sin1 t sin1. Khi

y = f(t) = 2t2 + t + 2.Lp bng bin thin ca hm s.

p s: min y = 2sin2 1sin 1+2 khi x = 1 v max y = f1

4

=

17

8khi sin

2x

1 + x2=

1

4.

Bi tp 2.1.6. (Hc vin QHQT - 99). Cho x, y tha mn x 0, y 0, x + y = 1.Tm GTLN, GTNN ca

P =x

y + 1+

y

x + 1.

Hng dn gii: Ta bin i

P =x(x + 1) + y(y + 1)

(x + 1)(y + 1)=

(x + y)2 2xy + 1xy + x + y + 1

=2 2xyxy + 2

.

t t = xy th 0 t 14

, xt hm s f(t) =2 2t2 + t

.

p s: min P = f

1

4=

2

3v max P = f(0) = 1.

Hong Thanh Thy


9/43


2.2 Phng php s dng cc bt ng thcNi dung t tng ca phng php:

Cho A = f(x) c min xc nh l D. tm GTLN, GTNN ca A ta s dng cc btng thc nh: Csi, Bunhiacopxki, Jensen, Trebsep... chng minh m f(x) M

trong m, M l cc hng s. Sau phi ch ra c x1, x2 D m = f(x1)

M = f(x2)

Cui cng kt lun: M l GTLN ca A; m l GTNN ca A.Phn ny ni ring v cc phn khc ni chung nu chia nh xem khi no s dng bt

ng thc Csi, khi no dng bt ng thc Bunhiacpxki,... khi no nh gi th ny,khi no nh gi th kia th qu thc s rt di dng v c khi s lm cho vn tr nnrc ri.

V vy, mi s phn chia ca chng ti ch c tnh cht tng i. i vi mi phn,thm ch mi v d chng ti s c gng trnh by mt cch d hiu nht c qu trnhsuy ngh, phn tch tm ra li gii trc khi thc hin chi tit li gii . a ra quytnh nh vy cng bi v chng ti mun hc sinh ca mnh tr thnh ch th ca mi

hot ng, ch ng, sng to trong qu trnh tm ra li gii mi bi ton ch khng phis ch l ngi c sch theo mt trnh t lp i lp li l " bi - li gii", " bi -li gii".....

2.2.1 S dng bt ng thc Csi

Cho a1, a2,...,an 0. Khi a1 + a2 + + an

n na1a2 an.

Du = xy ra khi v ch khi a1 = a2 = = an. c bit:

Khi n = 2 tha1 + a2

2 a1a2.

Khi n = 3 tha1 + a2 + a3

3 3a1a2a3.

Cc kiu vit khc thng gp:

a + b

2

ab

a, b

0.

a + b + c 3 3abc a,b,c 0.

Nhng nh gi kiu bt ng thc Csi:

(a + b)2 4aba2 + b2 2ab.

V d 2.2.1. Tm GTLN ca

y = x

1

x2.

Hong Thanh Thy


10/43


Phn tch:Mun tm GTLN ca mt tch cc tha s m mun s dng bt ng thc Csi th phinh gi tch nh hn hoc bng tng cc s hng, vi iu kin tng cc s hng ys dn ti mt hng s. Li gii.

p dng bt ng thc ab a2 + b2

2ta c

x

1 x2 x2 + 1 x2

2=

1

2.

Mt khc

y =1

2 x =

1 x2

x > 0

x2 = 1 x2

x =1

2 .

Vy GTNN ca y l1

2ti x =

12

.

Bnh lun:S 1 trong 1x2 c th thay i c. Ch nh gi quan trng nht l a2+b2 = const.Vy c th sa bi ton thnh: Tm GTLN ca

y = x

2 x2

y = x3 x2

...

Tng qut: y = x

a x2

S khc nhau l im xy ra du = .M rng:p dng bt ng thc Csi cho 3 s kiu

a3 + b3 + c3 3abc.

Tm GTLN cay = x2.

3

1 2x3y = x2.

3

2 2x3y = x2.

3

a 2x3

Hc sinh cn xc nh c u l a,b,c v tnh c a3 + b3 + c3 = const.

V d 2.2.2. Tm GTNN ca

y = x3 +2007

x2

(x > 0).

Hong Thanh Thy


11/43


12/43


Hng dn gii:a) x2 + (

2 x2)2 = 2 = const.

b) y = x(1 x)3 = 13

.3x.(1 x).(1 x).(1 x), c3x + (1 x) + (1 x) + (1 x) = 6 = const.

c) (x + 2) + (2 x) = 4 = const.Bi tp 2.2.3. Tm GTNN ca:

a) y =x

3+

15

x(x > 0)

b) y = 2x +1

x2(x > 0)

Hng dn gii:

a)x

3.15

x= 5 = const.

b) y = 2x +1

x2= x + x +

1

x2. M x.x.

1

x2= 1 = const.

Bi tp 2.2.4. Tm GTLN ca A =x

(x + 2006)2.

Hng dn gii:Ta c 2 hng gii:

Cch 1:A =

4.2006.x

4.2006.(x + 2006)2

=(x + 2006)2 (x 2006)2

4.2006.(x + 2006)2

=1

4.2006 (x 2006)

2

4.2006.(x + 2006)2

14.2006

.

Cch 2: T (x + 2006)2 4.x.2006 suy ra

A =x

(x + 2006)2 x

4.x.2006=

1

4.2006.

Bi tp 2.2.5. Cho

x, y,z > 0x + y + z = 1 . Tm GTNN ca

E = x + yxyz

.

Hng dn gii:

p dng bt ng thc Csi ta c

1 = x + y + z = (x + y) + z 2

(x + y)z

1 4(x + y)z x + y 4(x + y)2.z 16xyz x + y

xyz 16.

Hong Thanh Thy


13/43


Bi tp 2.2.6. Cho cc s dng x,y,z tha mn xyz(x + y + z) = 1. Tm GTNNca

A = (x + y)(x + z).

Hng dn gii: Nhn ph du ngoc. S dng

(x + y + z)1

x +

1

y +

1

z

.

Bi tp 2.2.7. Cho cc s dng x,y,z tha mn x2 + y2 + z2 = 1. Tm GTNN ca

A =xy

z 1 + yzx 2 + zxy 3

xyz.

Hng dn gii: iu kin x 2; y 3; z 1. Vit

A =

z 1

z+

x 2

x+

y 3

y.

Khi

z 1z

=

1(z 1)

z 1 + z 1

2z=

1

2x 2

x=

2(x 2)

2x 2 + x 2

2

2x=

1

2

2y 3

y=

3(y 3)

3y 3 + y 3

2

3y=

1

2

3.

Cng cc bt ng thc li ta tm c max A.

Bi tp 2.2.8. Tm GTNN ca

f(x, y) = x +4

(x y)(y + 1)2 (x > y 0).

Hng dn gii: p dng bt ng thc Csi cho 4 s dng

2x 2y; y + 1; y + 1; 8(x y)(y + 1)2 .

Bi tp 2.2.9. Cho cc s dng x,y,z tha mn x2001 + y2001 + z2001 = 3. Tm GTLN

caf(x,y,z) = x2 + y2 + z2.

Hng dn gii: p dng bt ng thc Csi cho 1999 s 1 v 2 s x2001 ta c

1 + 1 + + 1 + x2001 + x20012001

2001

x2001x2001.

Suy ra1999 + 2x2001

20001 x2.

Tng t cho y v z.

Hong Thanh Thy


14/43


Bi tp 2.2.10. Cho cc s dng x,y,z c tng bng 1. Tm GTLN ca hm s

f(x,y,z) =

x +1

x

y +

1

y

z +

1

z

.

Hng dn gii: S dng nh gi

1 + x = x + y + z + x 4 4x2yz.Tng t cho 1 + y v 1 + z.

Bi tp 2.2.11. Cho x > y > 0. Tm GTNN caa) f(x, y) = x +

1

y(x y)b) f(x, y) = x +

1

y(x y)2 .Hng dn gii:

a) Vit x = (x y) + y.b) Vit x = 12(x y) + 12(x y) + y.

Bi tp 2.2.12. Cho cc s dng x,y,z c tng bng 1. Tm GTLN ca hm s

f(x,y,z) =

1 x +

1 y + 1 z.

Hng dn gii: p dng bt ng thc Csi

(1 x). 2

3

1 x + 23

2=

5

3+ x

2..

Tng t cho 1 y v 1 z.

2.2.2 S dng bt ng thc Bunhiacpxki

Cho 2 dy s thc a1, a2,...,an v b1, b2,...,bn. Ta c(a1b1 + a2b2 + + anbn)2 (a21 + + a2n)(b21 + + b2n).

Du = xy ra khi v ch khi

a1b1

=a2b2

= = anbn

.

Dng thng dng: Khi n = 2 th (a1b1 + a2b2)2 (a21 + a22)(b21 + b22) Khi n = 3 th (a1b1 + a2b2 + a3b3)2 (a21 + a22 + a23)(b21 + b22 + b23). Cch vit khc ca bt ng thc Bunhiacopxki:

n

i=1a2ibi

ni=1 ai

2

ni=1 bi.

Hong Thanh Thy


15/43


Ch :Vi cc bt ng thc c iu kin, ta cn kho lo bin i nhn c biu thciu kin hoc s dng ngay biu thc iu kin bin i. Hc sinh phi nhybn trong vic nhn ra du hiu s dng bt ng thc Bunhiacopxki:

C iu kin kiu tng cc bnh phng l hng s.

Tm GTLN ca tng cc tch tng cp 2 s. Tm GTNN ca tng cc bnh phng.

V d 2.2.3. Cho

a,b,c > 0

a2 + b2 + c2 = 1. Tm GTLN ca

A = a + 3b + 5c.

Phn tch:C du hiu ca dng a1b1 + a2b2 + a3b3 vi

a2

1 = 1; a2

2 = 32

; a2

3 = 52

vb21 + b

22 + b

23 = a

2 + b2 + c2 = 1.

Li gii.

p dng bt ng thc Bunhiacopxki cho 2 b s 1, 3, 5 v a,b,c ta c

(a + 3b + 5c)2 (12 + 32 + 52)(a2 + b2 + c2) A2 35.1 = 35 A

35.


a

1=

b

3=

c

5= k > 0

a2 + b2 + c2 = 1

a = k

b = 3k

c = 5k

35k2 = 1

k =135

a =135

b =335

c =535

Vy GTLN ca A l

35.V d 2.2.4. Cho 3 s dng a,b,c v cc s x, y tha mn ax + by = c. Tm GTNN

ca

T =x2

a+

y2

b.

Li gii. Ta c

c2 = (ax + by)2 =

a

a.x

a+ b

b.

yb

2 (a3 + b3)

x2

a+

y2

b .Hong Thanh Thy


16/43


Do

T =x2

a+

y2

b c

2

a3 + b3.


ax + by = c,x

ay

b

=a

a

b

b

ax + by = c,x

y=

a2

b2

x = ca2

a3 + b3,

y =cb2

a3 + b3

Vy GTNN ca T lc2

a3 + b3.

Bnh lun:Quan trng nht l s dng c iu kin a2 + b2 + c2 = 1. V vy cc h s 1, 3, 5 thcs khng quan trng. T c th xt cc bi ton dng:Tm GTLN ca

A = ma + nb + pc (m,n,p > 0)

bit

a,b,c > 0

a2 + b2 + c2 = q(q > 0).

M rng:Nu gi thit cho hi khc l:

a, b, c > 0

a2 + 2b2 + c2 = 1

th phi xc nh cc cp s mt cch linh hot s dng c iu kin ny. Chnghn:

(a + 2b + 5c)2

12 + 3

2

2+ 52

(a2 + (

2b)2 + c2)

Suy ra

A

110

4.1 =

55

2.

V nu gi thit cho l a, b, c > 0a2 + 2b2 + 3c2 = 1

hay tng qut a, b, c > 0

a2 + b2 + c2 = 1(, , > 0)

th tnh hnh vn hon ton tng t.

BI TP VN DNG

Hong Thanh Thy


17/43


Bi tp 2.2.13. Cho x2 + y2 = 1. Tm GTLN, GTNN ca A = 3x + 4y.Hng dn gii:nh gi

(3x + 4y)2 (32 + 42)(x2 + y2) = 25.Tng qut: Cho mx2 + ny2 = p (m, n, p > 0). Tm GTLN, GTNN ca

A = x + y (, > 0).

Bi tp 2.2.14. Cho 2x + 3y = 1. Tm GTNN ca B = 3x2 + 2y2.

Hng dn gii:Ta c

1 = (2x + 3y)2 = 2

3.

3x +2

2.

2y2

23

+3

2

(3x2 + 2y2)

=

43

+ 94

.B.

Tng qut: Cho x + y = . Tm GTNN ca mx2 + ny2 (m, n > 0).

Bi tp 2.2.15. Cho2

x+

3

y= 6. Tm GTNN ca C = x + y.

Hng dn gii:Nu s dng

(a1b1 + a2b2)2 (a21 + a22)(b21 + b22)

v cn tm GTNN th x + y phi v phi, do a1 =

x; a2 =

y v cn c

a1b1 + a2b2 = const

nn b1 =

x; b2 =

y

. Mt khc, s dng c iu kin2

x+

3

y= 6 th

b21 + b22 =

2

x+

2

y=

2

x+

3

y= 6

suy ra =

2; =

3.

Bi ton c gii quyt!

Bi tp 2.2.16. Cho 3x 4y = 7. Tm GTNN ca D = 3x2

+ 4y2

.Hng dn gii:Tng t Bi tp 2.2.2.

Bi tp 2.2.17. Cho x2 + y2 = 1; u2 + v2 = 1. Tm GTLN, GTNN ca

E = u(x y) + v(x + y).Hng dn gii:Ta c

[u(x y) + v(x + y)]2 (u2 + v2)[(x y)2 + (x + y)2]

E

2(x2 + y2) = 2.

Hong Thanh Thy


18/43


Bi tp 2.2.18. Cho xy + yz + zx = 4. Tm GTNN ca B = x4 + y4 + z4.

Hng dn gii:Theo bt ng thc Bunhiacopxki th

16 = (xy + yz + zx)2

(x2 + y2 + z2)2

x4 + y4 + z4.

M(x2y2 + y2z2 + z2x2)2 (x4 + y4 + z4)2

nn16 3.(x4 + y4 + z4).

Bi tp 2.2.19. Gi s x0 l nghim ca phng trnh

x4 + ax3 + bx2 + cx + 2 = 0.

Tm GTNN ca C = a2

+ b

2

+ c

2

.

Hng dn gii:T x40 + ax

30 + bx

20 + cx0 + 2 = 0 suy ra

x40 + 2 = (ax30 + bx20 + cx0).

Theo bt ng thc Bunhiacopxki ta c

(ax30 + bx20 + cx0)

2 (a2 + b2 + c2)(x60 + x40 + x20).

T ta tm c GTNN ca C theo x0.

Bi tp 2.2.20. Cho

xy 0x2 + y2 = 100

. Tm GTLN ca

D = x

3 + y + y

3 + x.

Hng dn gii:

p dng bt ng thc Bunhiacopxki ta c

D2 = (x3 + y + y

3 + x)2

(x2 + y2)(6 + x + y)

D2 100.(6 + x + y) 100.[6 +

2(x2 + y2)]

D2 100.(6 +

200).

Bi tp 2.2.21. Cho cc s m, n tha mn16

m2+

9

n2= 1. Tm GTNN ca

S =

m2 + n2.

Hng dn gii: p dng bt ng thc Bunhiacopxki cho 2 b s16

m2; m2 v

9

n2; n2.

Hong Thanh Thy


19/43


Bi tp 2.2.22. Cho

(x 1)2 + (y 2)2 + (z 1)2 = 1.

Tm GTLN ca T = |x + 2y + 3z 8|.

Hng dn gii: rng

x + 2y + 3z 8 = (x 1) + 2(y 2) + 3(z 1).

Bi tp 2.2.23. Cho x2 + y2 + z2 4x + 2z 0. Tm GTLN, GTNN ca

F = 2x + 3y 2z.

Hng dn gii: iu kin ca x,y,z tng ng vi

(x 2)2 + y2 + (z + 1)2 5.

Vit li F thnhF = 2(x 2) + 3 2(z + 1) + 6.

Sau s dng bt ng thc Bunhiacopxki.

Bi tp 2.2.24. Cho x(x 1) + y(y 1) + z(z 1) 43

. Tm GTLN, GTNN ca hm

sf(x,y,z) = x + y + z.

Hng dn gii: Ta c

x(x 1) + y(y 1) + z(z 1) 43

3(x2 + y2 + z2) 3(x + y + z) + 4.

Mt khc theo bt ng thc Bunhiacopxki th

(x + y + z)2 3(x2 + y2 + z2).

T ta d dng tm c GTLN, GTNN ca f(x,y,z).

2.2.3 S dng bt ng thc TrbsepBt ng thc Trbsep: Cho hai dy s sp th t ging nhau

a1 a2 anb1 b2 bn

Khi ta c

(a1 + a2 + + an)(b1 + b2 + + bn) n(a1b1 + a2b2 + |anbn).

Hong Thanh Thy


20/43


V d 2.2.5. Cho

a, b, c, d > 0

ab + bc + cd + da = 1. Tm GTNN ca

S =a3

b + c + d+

b3

c + d + a+

c3

d + a + b+

d3

a + b + c.

Li gii.Khng mt tnh tng qut, gi s a b c d. t

A = b + c + d, B = c + d + a, C = d + a + b, D = a + b + c

th ta c1

A 1

B 1

C 1

D.

Theo bt ng thc Trbsep ta c

S =a3

A +b3

B +c3

C +d3

D

14

.(a3 + b3 + c3 + d3) 1

A+

1

B+

1

C+

1

D

1

16.(a + b + c + d)(a2 + b2 + c2 + d2)

1A

+1

B+

1

C+

1

D

S 1

48.(A + B + C+ D)

1A

+1

B+

1

C+

1

D

(a2 + b2 + c2 + d2).

Mt khc theo bt ng thc Bunhiacopxki ta c

(A + B + C+ D) 1

A +

1

B +

1

C +

1

D 16

a2 + b2 + c2 + d2 ab + bc + cd + da = 1.

Do S 1

48.16.1 =

1

3.

Du = xy ra khi v ch khi a = b = c = d =1

2.

V d 2.2.6. Cho cc s dng x, y. Tm GTLN ca

S = (x + y)(x3 + y3)(x6 + y6)

x10 + y10.

Li gii.Gi s x y. Khi

x3 y3; x6 y6.Theo bt ng thc Trbsep ta c

(x + y)(x3 + y3) 2(x4 + y4)(x4 + y4)(x6 + y6)

2(x10 + y10).

Hong Thanh Thy


21/43


T 2 bt ng thc trn ta c

(x + y)(x3 + y3)(x6 + y6) 4(x10 + y10).

Do

S

4(x10 + y10)

x10 + y10= 4.

Vy S c GTLN l 4.

BI TP VN DNG

Bi tp 2.2.25. Cho a1, a2,...,an l cc s thc dng thay i. t S =ni=1

ai. Tm

GTNN ca

T =ni=1

ai

S ai

.

Bi tp 2.2.26. Cho a1, a2,...,an l cc s thc dng thay i. t S =ni=1

ai. Tm

GTNN ca

T =ni=1

ai

S 2ai

.

Bi tp 2.2.27. Cho tam gic ABC c 3 gc nhn A,B,C. Tm GTLN ca

T =a + b + c

a

A+

b

B+

c

C

.

Hong Thanh Thy


22/43


2.3 Phng php min gi tr2.3.1 Ni dung phng php

Ta c y0 l mt gi tr ca hm s y = f(x) trn min D khi v ch khi h

f(x) = y0x D

c nghim.Trong nhiu trng hp iu kin c nghim y sau khi bin i s a v dng

y0 .V y0 l mt gi tr bt k ca f(x) nn t thu c

min f(x) =

max f(x) = .

2.3.2 Cc v d

V d 2.3.1. Tm GTLN, GTNN ca

f(x) =2x2 + 10x + 3

3x2 + 2x + 1.

Li gii.Gi y0 l mt gi tr ca hm s cho th phng trnh sau c nghim:

2x2

+ 10x + 33x2 + 2x + 1

= y0 (1).

V 3x2 + 2x + 1 > 0, x nn phng trnh (1) tng ng vi2x2 + 10x + 3 = y0(3x

2 + 2x + 1)

(3y0 2)x2 + (y0 5)x + y0 3 = 0 (2).

Ta xt hai trng hp:

Trng hp 1: Nu 3y0 2 = 0 y0 = 23

th y0 5 = 0 nn phng trnh (2) cnghim. Do f(x) nhn gi tr 2

3vi x0 no .

Trng hp 2: Nu 3y0 2 = 0 y0 = 23

th phng trnh (2) l phng trnh bc

hai, do (2) c nghim khi v ch khi

= (y0 5)2 (y0 3)(3y0 2) 0 2y20 + y0 + 19 0

1

152

2 y0 1 +

152

2, v y0 = 2

3.

Hong Thanh Thy


23/43


Kt hp hai trng hp ta c

1 1522

y0 1 +

152

2.

Vy GTLN ca f(x) l1 +

152

2; GTNN ca f(x) l

1

152

2.

V d 2.3.2. Tm GTLN, GTNN ca hm s f(x, y) = x2 + y2 xt trn min

D = {(x; y) : (x2 y2 + 1)2 + 4x2y2 x2 y2 = 0}.

Li gii.Gi t0 l mt gi tr bt k ca hm s f(x, y) trn min D. Khi h phng trnh sauc nghim

x2 + y2 = t0

(x2

y2 + 1)2 + 4x2y2

x2

y2 = 0

(I).

Ta c

(I)

x2 + y2 = t0

(x2 + y2)2 3(x2 + y2) + 1 + 4x2 = 0

x2 + y2 = t0

t20 3t0 + 1 + 4x2 = 0

y2 +t20 + 3t0 1

4= t0 (1)

t2

0 3t0 + 1 + 4x2

= 0 (2)

Phng trnh (2) c nghim n x khi v ch khi

t20 3t0 + 1 0 3 5

2 t0 3 +

5

2.

Li c

(1) y2 + t20 t0 1

4= 0.

M

t20

t0

1 < 0,

t0 nn phng trnh (1) lun c nghim n y,

t0.

Nh vy 3 52

t0 3 + 52

l iu kin cn v h (I) c nghim. Do

max(x;y)D

f(x, y) =3 +

5

2; min

(x;y)Df(x, y) =

3 52

.

V d 2.3.3. Tm GTLN, GTNN ca T = x2 + y2 trn tp

D = {(x; y) : (x y)2 = x + y xy}.

Hong Thanh Thy


24/43


Li gii.Gi T0 l mt gi tr ca T. Khi h phng trnh sau c nghim

T0 = x2 + y2

(x y)2 = x + y xy (I).

Ta c

(I)

T0 = x2 + y2

x2 + y2 = x + y + xy

T0 = x2 + y2 (1)

T0 = x + y + xy (2)

T (1) suy ra

T0 = x + y + xy 2(x2 + y2) + x2 + y22

T0

2T0 +T02

2T0 2

2T0 + T0

T20 8T0 0 0 T0 8.

Mt khc T(0, 0) = 0; T(2, 2) = 8. Vy ta c

max(x;y)D T = 8; min(x;y)D T = 0.

BI TP VN DNG

Bi tp 2.3.1. Cho a,b,c l cc s tho mna + b + c = 5

ab + bc + ca = 8

Tm GTLN, GTNN ca a.

Hng dn gii:Ta c b + c = 5 a v

bc = 8 a(b + c) = 8 a(5 a).

Mt khc bc (b + c)2

4=

(5 a)24

nn

(5 a)24

8 a(5 a).

S dng tam thc ta c 1 a 73

.

Hong Thanh Thy


25/43


Bi tp 2.3.2. Tm GTLN ca hm s f(x, y) = |x y| trn min

D = {(x; y) : x2 + 4y2 = 1}.

Hng dn gii:Ph du gi tr tuyt i, a v hai h v tm iu kin hai h c nghim.

p s: max(x;y)D

f(x, y) = 52

.

Bi tp 2.3.3. Tm GTLN, GTNN ca

f(x) =3 + 4x2 + 3x4

(1 + x2)2

trn R.

Hng dn gii:

a v phng trnh(y0 3)x4 + 2(y0 2)x2 + y0 + 3 = 0.

Nu y0 = 3 th phng trnh c nghim. Nu y0 = 0 th phng trnh c nghim khi v ch khi

(y0 3)t2 + 2(y0 2)t + y0 + 3 = 0t 0

p s: minxR

f(x) = 52

;maxxR

f(x) = 3.

Bi tp 2.3.4. Cho hm s

f(x) =x2 + px + q

x2 + 1(x R).

Tm p, q max f(x) = 9; min f(x) = 1.

Hng dn gii:

Gi y0 l gi tr bt k ca f(x) th phng trnh

y0 =x2 + px + q

x2 + 1(1)

c nghim. Ta c(1) (y0 1)x2 px + (y0 q) = 0.

Khi y0 = 1, bi ton qui v tm p, q phng trnh 0 c nghim.p s: (p = 8; q = 7) hoc (p = 8; q = 7).

Hong Thanh Thy


26/43


Bi tp 2.3.5. Tm GTLN, GTNN ca hm s

y = x +

2x2 + x + 1.

Hng dn gii: Gi s y0 l mt gi tr no ca hm s. Khi

y0 = x + 2x2 + x + 1 (y0 x)2 = 2x2 + x + 1 y20 2y0x + x2 = 2x2 + x + 1 x2 + (1 2y0)x + (1 y20)2 = 0.

Xt iu kin c nghim x ca phng trnh ny tm ra min gi tr ca y0.

Bi tp 2.3.6. Tm GTNN ca

y = x +x2 + 1x

, vi x > 0.

Hng dn gii: Gi y0 l gi tr tu ca y trn min x > 0. Khi h sau c nghimy0 = x +

x2 +

1

xx > 0

0 < x y0(y0 x)2 = x2 + 1

x

0 < x y0y20 2y0x =

1

x

0 < x y02y0x

2 y20x + 1 = 0

p s: minx>0

f(x) = 2.

Bi tp 2.3.7. Cho x2 + y2 > 0. Tm GTLN, GTNN ca

f(x, y) =x2 (x 4y)2

x 4y2 .

Hng dn gii:Nu y = 0 th min f(x) = max f(x) = 0.Nu y = 0 th ta vit

f(x, y) =

x2y2 x2y 2

2

x2y

2+ 1

.

tx

2y= t v xt g(t) =

4t 4t2 + 1

.

Hong Thanh Thy


27/43


2.4 Phng php lng gic2.4.1 Ni dung phng php

Phng php ny nhm thay i hnh thc ca bi ton dn n vic tm GTLN, GTNNca hm s lng gic. Phng php ny c bit t ra hiu qu i vi cc hm i snhiu n vi dng thng gp nht l khi c iu kin

x

2 +y

2 = 1. Khi ta tx = sin t

y = cost, t [0;2].

Trong trng hp khng c iu kin ca n s, thng t

x = tan t, t

2

;

2

.

2.4.2 Cc v d

V d 2.4.1. Tm GTLN ca hm sy = (1 + x)2006 + (1 x)2006, vi x [1;1].

Li gii. V x [1;1] nn ta c th t x = cos t, t [0; ]. Khi hm s tr thnhy = (1 + cos t)2006 + (1 cos t)2006

=

2cos2t

2

2006+

2sin2t

2

2006= 22006.

cos4012

t

2+ sin4012

t

2

22006. cos2 t

2

+ sin2t

2

= 22006.

Vy GTLN ca y l 22006, t c khi

cos4012t

2= cos2

t

2

sin4012t

2= sin2

t

2

, chng hn

sin t = 0

cos t = 1 th x = 1.


y =1 + x4

(1 + x2

)2

.

Li gii. t x = tan t, t

2

;

2

. Khi hm s c chuyn v dng

y =1 + tan4 t

(1 + tan2 t)2=

1 +sin4 t

cos4 t1

cos4 t

= sin4 t + cos4 t = 1 12

sin2 2t.

V 0

sin2 2t

1 nn

Hong Thanh Thy


28/43


GTNN: min y = 1 12

=1

2, t khi

sin2 2t = 1 cos2t = 0Chng hn t =

4th x = 1.

GTLN: max y = 1 0 = 1, t khisin2 2t = 0 sin2t = 0

Chng hn t = 0 th x = 0.


u = 2x + 3

3y + 2, vi 4x2 + 9y2 = 16.

Li gii. T gi thit ta c x22

+3

y42

= 1.

tx

2= cos ; (

3y

4= sin , [0;2]. Khi hm c chuyn v dng

u = 4 cos + 4

3sin + 2.

Ta c

u = 8(1

2cos +

3

2sin ) + 2 = 8 sin

+

6

+ 2.

V

1

sin +

6 1 nn ta c

GTNN: min u = 8 + 2 = 6, t khi

sin

+

6

= 1 = 4

3

x =12

y =3

2

GTLN: max u = 8 + 2 = 10, t khi

sin

+

6

= 1 =

3

x =1

2y =

32

V d 2.4.4. Tm GTLN, GTNN ca hm s

u =4xy 4y2

x2 + y2.

Li gii. Ta xt hai trng hp:

Trng hp 1: Nu y = 0 th u = 0.

Hong Thanh Thy


29/43


Trng hp 2: Nu y = 0 th chia c t s v mu s cho y2 ta c

u =

4x

y 4

x2

y2+ 1

.

tx

y= tan t, t

2;

2

. Khi hm tr thnh

u =4tan t 4tan2 t + 1

=

4sin t

cos t 4

1

cos2 t= 4sin t. cos t 4cos2 t= 2 s i n 2t 2(1 + cos 2t)= 22sin2t

4 2.

V 1 sin

2t 4

1 nn ta c

GTNN: min u = 22 2, t c khi

sin

2t 4

= 1 t =

8 x

y= tan

8.

GTLN: max u = 2

2

2, t c khi

sin

2t 4

= 1 t = 3

8 x

y= tan

3

8.

Nhn xt: Trong trng hp khng c n s, php lng gic ho c xc nh theohai hng sau:

Hng 1: Nu c th s dng c n ph t = g(x, y) chuyn hm ban u vhm mt n theo t. Khi tu thuc min gi tr ca g(x, y) ta la chn t

g(x, y) = sin hay g(x, y) = tan .

Hng 2: Trong trng hp cn li, php lng gic ho thng c s dng l

x = tan v y = tan ; ,

2

;

2

.

V d 2.4.5. Tm GTLN v GTNN ca

u =(x + y)(1 xy)(1 + x2)(1 + y2)

.

Hong Thanh Thy


30/43


Li gii. t

x = tan

y = tan , ,

2;

2

th hm s chuyn v dng

u =(tan + tan )(1 tan . tan )

(1 + tan2 )(1 + tan2 )

=

sin( + ). cos( + )

cos . cos . cos . cos 1

cos2 .

1

cos2

=1

2. sin2( + ).

V 1 sin2( + ) 1 nn ta c GTNN: min u = 1

2, t c khi

sin 2( + ) = 1 + =

4 tan( + ) = 1 x + y

1 xy = 1.Chn x = 0; y = 1.

GTLN: max u = 12

, t c khi

sin2( + ) = 1

+ =

4 tan( + ) = 1 x + y

1 xy = 1.Chn x = 0; y = 1.

BI TP VN DNGBi tp 2.4.1. Cho cc s x, y tho mn x2 + y2 = 1. Tm GTLN ca

u = x1 + y + y

1 + x.

Hng dn gii: t x = sin ; y = sin .

p s: GTLN: max u = 2 +

2, t khi x = y =

2

2.

Bi tp 2.4.2. Tm GTLN v GTNN ca

a) y =x2

1 + x4

b) y =1 + x6

(1 + x2)2

. Hng dn gii:

Hong Thanh Thy


31/43


2.5 Phng php hnh hc, to v vectNi dung phng php

tm GTLN, GTNN bng phng php ny ngi ta thng s dng cc tnh chtsau y:

Trong tt c cc ng gp khc ni hai im A, B cho trc th ng thng niAB l ng c di b nht. Trong mt tam gic tng hai cnh lun ln hn cnh th ba. Trng hp xy ra du

bng khi tam gic suy bin.

Cho M = d. Khi ng thng vung gc k t M xung d ngn hn mi ngxin k t M xung ng thng y.

Trong cc tam gic cng ni tip ng trn th tam gic u c chu vi v din tchln nht.

Chng ta s s dng phng php ny khi m trong ni dung cc bi ton tim n yut hnh hc m c th ban u ta cha nhn ra n.

c bit cn nh cc cng thc sau:

* Trong mt phng:Khong cch gia hai im A(x1; y1), B(x2; y2)

AB =

(x2 x1)+(y2 y1)2

Khong cch t im M(x0; y0) n : Ax + By + C = 0

d = |Ax0 + By0 + C|A2 + B2

Phng trnh ng trn tm I(a; b) bn knh R

(x a)2 + (y b)2 = R2.

* Trong khng gian:Khong cch gia hai im A(x1; y1; z1), B(x2; y2; z2)

AB = (x2

x1)+(y

2 y1)2 + (z

1 z2)2.

Khong cch t im M(x0; y0) n mt phng : Ax + By + Cz + D = 0

d =|Ax0 + By0 + Cz0 + D|

A2 + B2 + C2

Phng trnh mt cu tm I(a; b; c) bn knh R

(x a)2 + (y b)2 + (z c)2 = R2.

Hong Thanh Thy


32/43


33/43


Bnh lun: Nh vy l da vo cng thc tnh khong cch gia hai im v vic sdng hp l bt ng thc tam gic m chng ta gii c bi ton. Quan trng nhtl vic nhn ra bng dng ca tng hai on thng trong biu thc.

Mt cu hi nh t ra l ti sao li chn im A(1

2;

3

2) m khng l A(

1

2;3

2);

B(1;

1) m khng l B(1;1). Mc d cc biu thc tnh khong cch AC,BC khng hthay i. Ta chn nh trong li gii nhm cho A, B nm khc pha ca nhau so vi trchonh; C Ox. Nu ly im B nm trn cng khng sao nhng li gii s di hn bi tm c min(AC+ CB) khi vn phi ly B i xng B qua Ox tc B(1; 1). Vychi bng ta chn lun B B ngay t u.M rng: Cc h s ca cc biu thc liu c phi l bt k khng? Nu thay x2 2x + 2bi biu thc x2 2x th sao? Vit x2 2x =

(x 1)2 1 liu c lm tip c

khng? Trong khi cng thc tnh khong cch l:

(x2 x1)+(y2 y1)2. Du + chkhng phi l du .

Vy mi h s ca biu thc di cn l tu nhng phi tho mn x 0. Nh vyl cc em cng c th t ra cho mnh v bn b nhng biu thc n gin. Chng hn:Tm GTNN ca

f(x) =

x2 4x + 5 +

x2 6x + 10V h s ca x2 trong hai biu thc di cn c nht thit phi bng nhau khng? Nukhng bng th sao?V d: f(x) =

x2 4x + 5 + 2x2 4x + 6

BI TP VN DNGBi tp 2.5.1. Tm GTNN ca

1. f(x) =

x2 + x + 1 +

x2 + x 1

2. f(x) = x2 x + 1 +

x2 3x + 1Hng dn gii: Ging v d.

Bi tp 2.5.2. Tm GTLN ca:

f(x) =

x2 6x + 45

x2 6x + 10Hng dn gii: S dng AB AC BC.

Bi tp 2.5.3. Tm GTNN ca:

f(x) =

2x2

2x + 1 + 2x2 + (3 + 1)x + 1 + 2x2 (

3 + 1)x + 1

Hng dn gii: Tt c u c 2x2. Lm th no dng c cng thc tnh khongcch:

2x2 2x + 1 =

(x 1)2 + x2

2x2 + (

3 + 1)x + 1 =

(x +

3

2)2 + (x +

1

2)2

2x2 (

3 + 1)x + 1 =

(x

3

2)2 + (x 1

2)2

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Bi tp 2.5.4. Tm GTNN ca: f(x, y) = x2 + y2 trn min

D = {(x, y) : x 2y + 8 0; x + y + z 0; 2x y + 4 0}.

Hng dn gii: Hy xc nh min D trn mt phng Oxy ri xem ngha ca f(x, y) =x2 + y2 biu th ci g?

Bi tp 2.5.5. Tm GTLN, GTNN ca: f(x, y) = 4x + 3y trn min

D = {(x, y) : x2 + y2 + 16 = 8x + 6y}.

Hng dn gii: Xc nh min D chnh l ng trn. Tnh

f(x, y) = 4x + 3y =1

2

8x + 6y

=

1

2

x2 + y2 + 16

Cn tm GTLN, GTNN ca x2 + y2. Xem x2 + y2 biu th ci g?

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2.6 Cc phng php khc2.6.1 Phng php cn bng i xng

Ni dung phng php:Phng php ny thng c s dng nu iu kin rng buc cc biu thc v biu

thc cn tm GTLN, GTNN c tnh i xng vi cc bin th ta thng d on GTLN,GTNN xy ra khi cc bin t gi tr bng nhau.

Sau dng cc bt ng thc chng minh d on ny.

V d 2.6.1. Cho xy + yz + zx = 1. Tm GTNN ca M = x4 + y4 + z4.Li gii. Ta c: x4 + y4 + z4 x2y2 + y2z2 + z2x2 2

3(xy + yz + zx)2

Suy ra M 13

M =1

3 x = y = z = 1

3.

Vy GTNN ca M l

1

3 .

V d 2.6.2. Cho

a, b > 0

a2 + b2 = 4. Tm GTLN ca

T =ab

a + b + 2.

Phn tch: Phi to c ra: a +b +2, ab t gi thit. Mun vy phi phn tch a2+b24theo a + b + 2, ab v nhn t no khc na v s nh gi nhn t ny chng hn.Li gii.

a2 + b2 4 = 0 (a + b)2 4 = 2ab (a + b + 2)(a + b 2) = 2ab T = ab

a + b + 2=

a + b 22

M

(a + b)2

2(a2 + b2)

(a + b)2 8 a + b 2

2.

Vy T =a + b 2

2 2

2 22

=

2 1.Du bng xy ra khi v ch khi a = b =

2.

BI TP VN DNG

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Bi tp 2.6.1. Cho

x, y,z > 0,

x + y + z = 3. Tm GTNN ca

T =

x2 + xy + y2 +

y2 + yz + z2 +

z2 + zx + x2.

Hng dn gii: Ta c:

x2 + xy + y2 =3

4(x + y)2 +

1

4(x y)2

34

(x + y)2

Tng t ta c:

y2 + yz + z2 34

(y + z)2

z2

+ zx + x2

3

4 (z + x)2

Bi tp 2.6.2. Cho

a, b > 0,

a + b = 1. Tm GTNN ca:

(a +1

b)2 + (b +

1

a)2.

Hng dn gii: Dng Csi dng x2 + y2 2xy.Bi tp 2.6.3. Tm GTLN ca:

4a + 1 +

4b + 1 +

4c + 1

bit a,b,c > 0; a + b + c = 1.Hng dn gii: s dng c a + b + c th phi bnh phng cc cn thc. Sau dng Bunhiacpxki.

Bi tp 2.6.4. Cho

a,b,c > 0,

a + b + c = 1. Tm GTNN ca

1

a2 + 2bc +1

b2 + 2ac

1

c2 + 2ab

Hng dn gii: S dng bt ng thc:

(x + y + z)(1

x+

1

y+

1

z) 9.

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2.6.2 Phng php cc bin

Ni dung phng php:Ta bit, nu hm s y = f(x) lin tc trn on [a, b] th GTLN, GTNN ca f(x)

trn on [a, b] hoc l f(a) hoc l f(b), hoc l gi tr cc tr ca f. Nh vy l khi tmGTLN, GTNN ca A(x1, x2,...,xn) nu xi [a, b] th ta nn lu ti gi tr ca A khixi = a hay xi = b.Ch ti cc bin i thng dng sau:

xi [a, b] suy ra (xi a)(xi b) 0. Du bng xy ra khi xi = a xi = b. x,y,z [a, b] suy ra

(x a)(y a)(z a) 0(x b)(y b)(z b) 0

V d 2.6.3. Cho a,b,c,d [0;1]. Tm GTLN ca

abcd + 1

+ bacd + 1

+ cabd + 1

+ dabc + 1

.

Li gii. V a,b,c,d [0;1] nn bcd abcd. Li c (a 1)(b 1) 0 nn ab + 1 a + b.T suy ra

P aabcd + 1

+b

abcd + 1+

c

abcd + 1+

d

abcd + 1

P a + b + c + dabcd + 1

1 + ab + 1 + cdabcd + 1

2 + ab + cd

abcd + 1 2 + 1 + abcd

abcd + 1 3 + abcd

abcd + 1 3 + 3abcd

abcd + 1= 3.

Vy GTLN ca P l 3, xy ra khi a = 0, b = c = d = 1.

BI TP VN DNG

Bi tp 2.6.5. Cho

x1, x2,...,x10 [1;3],x1 + x2 + + x10 = 15

Tm GTLN ca S = x31 + x32 + + x310.

Bi tp 2.6.6. Cho

x,y,z [0; 3],x + y + z = 5

.

Tm GTLN ca S = x2 + y2 + z2

2.6.3 Phng php sp th t

Ni dung phng php:Nu vic sp th t li cc hng s, cc bin s khng lm mt tnh tng qut ca bi

ton th nn thc hin v chng cho ta thm gi thit tm GTLN, GTNN. V khi sp xp li cc hng, cc bin ta nn ch ti cc phn t Max, min ca chng.

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V d 2.6.4. Cho cc s x1,...,x10 thay i nhng lun tho mnx1x2...x10 0,x1 + x2 + + x10 = 2006

Tm GTLN ca S =9i=1

xixi+1.

Li gii. Gi s xk = max{x1, x2,...,x10}, ta c

S =k1i=1

xixi+1 +9i=k

xixi+1

xkk1i=1

xi + xk

9i=k+1

xi

xk9i=1 x

i xk= xk(2006 xk) 2006

2

4.

Du bng xy ra khi

x1x2 =

2006

2,

x3...x9 = 0. Vy GTLN ca S l

20062

4.

BI TP VN DNG

Bi tp 2.6.7. Trong tam gic ABC, tm GTLN ca

T =a + b + c

aA + bB + cC.

Hng dn gii:

Gi s a b c khi A B C. p dng bt ng thc Trbsep ta c

(A + B + C)(a + b + c) 3(aA + bB + cC)

3 aA + bB + cC

a + b + c

3

a + b + caA + bB + cC

.

Du bng xy ra khi v ch khi ABC u.

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3 ng dng

3.1 Gii phng trnh, bt phng trnh.3.1.1 Cc nh l

Vn tn ti nghim ca mt phng trnh, bt phng trnh c iu kin thng linquan cht ch ti vic tm GTLN, GTNN ca hm s. Mi lin h y th hin qua ccnh l di y.

nh l 1. Xt phng trnh, f(x) = x D (1). Gi thit tn ti

M = maxxD

f(x), m = minxD

f(x)

Khi phng trnh (1) c nghim khi v ch khi m Mnh l 2. Xt bt phng trnh f(x)

x

D (2). Bt phng trnh ny c nghim

khi v ch khi M .nh l 3. Xt bt phng trnh f(x) x D (3). Bt phng trnh ny c nghimkhi v ch khi m nh l 4. Bt phng trnh (2) ng x D m .Bt phng trnh (3) ngx D M .

3.1.2 Cc v d

V d 3.1.1. Gii phng trnh

x 2 + 4 x = x2 6x + 11

Li gii. t f(x) =

x 2 + 4 x. Xt trn min 2 x 4.Theo bt ng thc Csi ta c:

f2(x) 2(x 2 + 4 x) = 4

Do f(x) 0 nn f(x) 2.Ta thy f(3) = 2. Vy max

2x4f(x) = 2.

t g(x) = x2

6x + 11 = (x 3)2

+ 2. Suy ra g(x) 2, g(3) = 2. Vy min2x4 g(x) = 2.Vy phng trnh cho tng ng vi h sauf(x) = 2

g(x) = 2

x 2 + 4 x = 2(x 3)2 + 2 = 2

D thy h ny c nghim duy nht x = 3. chnh l nghim ca phng trnh cho.

V d 3.1.2. Gii phng trnh:

3xs2 + 6x + 7 +

5x2 + 10x + 14 = 4

2x

x2.

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Li gii. t

f(x) =

3xs2 + 6x + 7 +

5x2 + 10x + 14

=

3(x + 1)2 + 4 +

5(x + 1)2 + 9

f(x)

5.

x

D =

{4

2x

x2

0}

.f(x) = 5 x = 1.t g(x) = 4 2x x2 = 5 (x + 1)2 5, g(x) = 5 x = 1.Ta c g(x) 5 f(x) Vy

f(x) = g(x)

3xs2 + 6x + 7 +

5x2 + 10x + 14 = 5 (1)

4 2x x2 = 5 (2)

(2) x = 1. M x = 1 cng tho mn (1). Vy x = 1 l nghim duy nht caphng trnh cho.

BI TP VN DNGBi tp 3.1.1. Gii phng trnh:

x2 2x + 5 + x 1 = 2

Bi tp 3.1.2. Gii bt phng trnh:x

x2 1 +

x +

x2 1 2

Bi tp 3.1.3. Gii h phng trnh:

6x2

x3 6x + 5 = (x3 + 4)(x2 + 2x 6)x +

2

x 1 + 2

x2

Bi tp 3.1.4. Gii phng trnh:

2

7x3 11x2 + 25x 12 = x2 + 6x 1


2x4 + (1 2x)4 = 127


2 x2 +

2 1

x2= 4 (x + 1

x)

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3.2 Tm iu kin cho tham s m phng trnh, bt phngtrnh, h phng trnh, h bt phng trnh c nghim.

V d 3.2.1. Tm m phng trnh sau c nghim:

20x2 + 10x + 3

3x2 + 2x + 1= x2 + 2(2m

3)x + 5m2

16m + 20

Li gii. t f(x) =20x2 + 10x + 3

3x2 + 2x + 1. Dng phng php min gi tr ta suy ra: max f(x) =

7.t g(x) = x2 + 2(2m 3)x + 5m2 16m + 20. Theo tnh cht ca hm s bc hai th:

min g(x) = g(3 2m) = m2 4m + 11 = (m 2)2 + 7

T ta c: min g(x) > 7, m = 2 v min g(x) = 7 khi m = 2.Vy:

+ Nu m = 2 th maxxR

f(x) < minxR

g(x), suy ra phng trnh cho v nghim.

+ Nu m = 2 th maxxR

f(x) = minxR

g(x) = 7 v phng trnh cho tng ng vi

h:

20x2 + 10x + 3

3x2 + 2x + 1= 7 (1)

x2 + 2x + 8 = 7 (2)

(2) tng ng vi (x + 1)2 + 7 = 7 x = 1. Thay x = 1 vo (1) ta c:

V T(1) =

13

2 = 7 suy ra h trn v nghim.

Vy h cho v nghim.

V d 3.2.2. Tm m bt phng trnh sau ng vi mi 2 x 1.

m2x + m(x + 1) 2(x 1) 0

Li gii. Bt phng trnh cho c th vit li l:

f(x) = (m2 + m

2)x + m + 2

0 (1)

(1) ng vi mi 2 x 1 th phi c min2x1

f(x) 0.V f(x) l hm bc nht nn ta c:

Nu m2 + m 2 0 th min2x1

f(x) = f(2) = 2m2 m + 6. Ta c h

m2 + m 2 02m2 m + 6 0 1 m

3

2

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Nu m2 + m 2 < 0 th min2x1

f(x) = m2 + 2m. Ta c h

m2 + m 2 < 0m2 + 2m 0 0 m < 1

Vy 0 m 32

l nhng gi tr cn tm.

BI TP VN DNG

Bi tp 3.2.1. Tm m phng trnh sau c nghim:2x2 2(m + 4)x + 5m + 10 + 3 x = 0

Hng dn gii: Ta c

(1) 2x2 2(m + 4)x + 5m + 10 = x 3

2x2 2(m + 4)x + 5m + 10 = (x 3)2x 3

x 3f(x) =

x2 2x + 12x 5 = m

H ny c nghim khi v ch khi

minx3

f(x)

m

maxx3

f(x)

Sau dng phng php min gi tr gii quyt bi ton. p s: m 3.Bi tp 3.2.2. Tm m h sau c nghim

x5 (x 3)5 = m0 x 3

Hng dn gii: ti bi ton ph: "Cho hm s f(t) = tn + (1 t)n, n N" th

max0t1

f(t) = 1; min0t1

= 12n1

Chuyn bi ton cho v dng ny!

p s:243

16 m 243.

Bi tp 3.2.3. Tm m bt phng trnh sau ng vi mi 4 x 6(4 + x)(6 x) x2 2x + m

Hng dn gii: S dng phng php iu kin cn v ta i n p s: m

6.

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Bi tp 3.2.4. Tm m h bt phng trnh sau c nghim2x2 7x + 3x2 mx + m 0

Hng dn gii:Vit li h cho di dng:

f(x) = x2 mx + m 01

2 x 3

H ny c nghim khi v ch khi min1

2 x 3

f(x) 0, tm i tm min1

2 x 3

f(x). T

i n p s:

k

1

2hoc k

4.

Bi tp 3.2.5. Tim m bt phng trnh sau c nghim

x2 + 2|x m| + m2 + m 1 0

Hng dn gii: Bt phng trnh c nghim khi v ch khi minxR

f(x) 0Xt cc kh nng:

+ m > 1

+

1

m

1

+ m < 1

p s: 1 m 12

.

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