1.3.gia tri lon nhat - nho nhat cua ham so

1

Chng I. Hm s Trn PhngBi 3. Gi tr ln nht, nh nht ca hm s

BI 3. GI TR LN NHT, NH NHT CA HM S

A. GI TR LN NHT, NH NHT CA HM S I. TM TT L THUYT

1. Bi ton chung: Tm gi tr nh nht hoc ln nht ca hm s

Bc 1: D on v chng minh

Bc 2: Ch ra 1 iu kin

2. Cc phng php thng s dng

Phng php 1: Bin i thnh tng cc bnh phng

Phng php 2: Tam thc bc hai.

Phng php 3: S dng bt ng thc c in: Csi; Bunhiacpski

Phng php 4: S dng o hm.

Phng php 5: S dng i bin lng gic.

Phng php 6: S dng phng php vct v h ta

Phng php 7: S dng phng php hnh hc v h ta .II. CC BI TP MU MINH HA:

Bi 1. Tm gi tr nh nht ca P(x, y) = x2 + 11y2 ( 6xy + 8x ( 28y + 21 Gii. Bin i biu thc di dng P(x, y) = (x ( 3y + 4)2 + 2(y ( 1)2 + 3 ( 3

T suy ra MinP(x, y) = 3 (

Bi 2. Cho x, y > 0. Tm gi tr nh nht ca: S =

Gii.

S

S .

Vi x = y > 0 th MinS = 2

Bi 3. Tm gi tr ln nht ca hm s

Gii . =

S

S. Vi , (k(() th Bi 4. Tm gi tr nh nht ca biu thc

Gii.

Vi , th

Bi 5. Cho . Tm gi tr nh nht ca biu thc:

S = 19x2+ 54y2 +16z2 (16xz ( 24y +36xyGii. Bin i S ( f(x) = 19x2 ( 2(8z (18y)x + 54y2 +16z2 ( 24y

Ta c ((x = g(y) = (8z (18y)2 ( (54y2 +16z2 ( 24y) = (702y2 +168zy ( 240z2

( ((y = (84z)2 ( 702.240z2 = (161424z2 ( 0 (z(R ( g(y) ( 0 (y, z(R

Suy ra ((x ( 0 (y, z(R ( f(x) ( 0. Vi th

Bi 6. Cho x2 + xy + y2 = 3. Tm gi tr ln nht v nh nht ca biu thc: S = x2 ( xy + y2Gii Xt y = 0 ( x2 = 3 ( S = 3 l 1 gi tr ca hm s. Xt y ( 0, khi bin i biu thc di dng sau y

vi

( u(t2 + t + 1) = t2 ( t + 1 ( (u ( 1)t2 + (u + 1)t + (u ( 1) = 0 (*)

+ Nu u = 1, th t = 0 ( x = 0, y = ( u = 1 l 1 gi tr ca hm s+ Nu u ( 1, th u thuc tp gi tr hm s ( phng trnh (*) c nghim t ( ( = (3u ( 1)(3 ( u) ( 0 ( .

Vy tp gi tr ca u l ( ; Max u = 3

Min S = 1 ( ( t = 1 (

Max S = 9 ( Maxu = 3 ( t = (1 (

Bi 7. Cho x,y(R tha mn iu kin

Tm gi tr ln nht, nh nht ca biu thc S=

Gii. Bin i

( (

Do (4x2 ( 0 nn (

Vi x = 0, y = , th . Vi x = 0, y = , th

Bi 8. Tm gi tr nh nht ca hm s

Gii. Gi y0 l 1 gi tr ca hm f(x) ( tn ti x0 sao cho y0 =

(

( g(x0) = . Ta c g(x) = 0 c nghim x0

( (( = =

Do y0 = nn

(( ( 0 ( 2y0 ( 1 ( 0 ( . Vi x = th Minf(x) =

Bi 9. Cho Tm cc gi tr ca m sao cho

Gii. Ta c

Gi (P) l th ca y = f(x) ( (P) = (P1) ( (P2) khi (P) c 1 trong cc hnh dng th sau y

Honh ca cc im c bit trong th (P): Honh giao im (P1), (P2) xA = 1; xB = 4 ; Honh nh (P1): .

Nhn vo th ta xt cc kh nng sau:

( Nu xC ([xA, xB] ( m([ (3, 3] th Minf(x) = Min(f(1), f(4)(.

Khi Minf(x) > 1 ( ( 1 < m ( 3 (1)( Nu xC ([xA, xB] ( m([ (3, 3] th Minf(x) = =

Khi Minf(x) > 1 ( (2)

( Kt lun: T (1) v (2) suy ra Minf(x) > 1 (

Bi 10. ( thi TSH 2005 khi A)

Cho ;. Tm Min ca S

Gii: S dng bt ng thc Csi cho cc s a, b, c, d > 0 ta c:

Bi 11. ( thi TSH 2007 khi B)

Cho . Tm Min ca S

Gii: S dng bt ng thc Csi cho 9 s ta c

S

Bi 12. Cho Tm gi tr nh nht ca S =

Gii:

Mt khc, S = = =

Suy ra 2S ( ( ( ( MinS =.

Bi 13. Cho x, y, z > 0. Tm Max ca: S =

Gii: S dng bt ng thc Csi v BunhiaCpski ta c 3 nh gi sau:

;. T suy ra

Bi 14. ( thi TSH 2003 khi B)

Tm gi tr ln nht, nh nht ca hm s

Cch 1: Tp xc nh ;

(

Cch 2: t

(;

Bi 15. ( d b TSH 2003 khi B) Tm gi tr ln nht, nh nht ca trn on

Cch 1. t . Ta c

Nhn bng bin thin ta c

Cch 2. t .

Vi th. S dng bt ng thc Csi ta c:

. Vi Bi 16. a) Lp bng bin thin v tm gi tr ln nht ca hm s

b) Cho . Chng minh rng:

Gii. a) TX: ;

.

Suy ra . Nhn BBTta c

b) Theo phn a) th ( . c bit ha bt ng thc ny ti cc gi tr ta c:

(

Cch 2. Trn mt phng ta Oxy t

.

Khi .

Do

T suy ra

Bi 17. ( 33 III.2, B thi TSH 1987 1995) Cho . Tm Max, Min ca A ( .

Gii. 1. Tm MaxA: S dng bt ng thc BunhiaCpski ta c A (. Vi th Max A (

2. Tm MinA: Xt 2 trng hp sau y Trng hp 1: Nu, xt 2 kh nng sau:

+) Nu th A>0 (

+) Nu x ( 0, y ( 0 th (A( ( =

T 2 kh nng xt suy ra vi th Min A = (1

Trng hp 2: Xt : t ( (

(

(

Ta c:

Th vo phn d ca chia cho (.

Nhn bng bin thin suy ra:

suy ra

xy ra ( ;

( x, y l nghim ca (

Kt lun: Max A (;

Bi 18. Cho tho mn iu kin: .

Tm Max, Min ca biu thc:

Gii. Do nn . V hm s nghch bin trn nn bi ton tr thnh.

1. Tm MaxS hay tm Min

.Vi th MaxS =

2. Tm MinS hay tm Max

Cch 1: Phng php tam thc bc hai:

Khng mt tnh tng qut gi s . Bin i v nh gi a v tam thc bc hai bin z

Do th hm y = f(z) l mt parabol quay b lm ln trn nn ta c: .

Vi th MinS =

Cch 2: Phng php hnh hc

Xt h ta cc vung gc Oxyz. Tp hp cc im tho mn iu kin nm trong hnh lp phng ABCDA(B(C(O cnh 1 vi A(0, 1, 1); B(1, 1, 1); C(1, 0, 1); D(0, 0, 1); A((0, 1, 0); B((1, 1, 0); C((1, 0, 0).Mt khc do nn nm trn mt phng (P):

Vy tp hp cc im tho mn iu kin gi thit nm trn thit din EIJKLN vi cc im E, I, J, K, L, N l trung im cc cnh hnh lp phng. Gi O( l hnh chiu ca O ln EIJKLN th O( l tm ca hnh lp phng v cng l tm ca lc gic u EIJKLN. Ta c O(M l hnh chiu ca OM ln EIJKLN. Do OM2 = nn OM ln nht ( O(M ln nht

( M trng vi 1 trong 6 nh E, I, J, K, L, N.

T suy ra:

Vi th MinS =

Bi 19. Cho tha mn iu kin

Tm gi tr nh nht ca

Gii. Sai lm thng gp:

( Nguyn nhn:

mu thun vi gi thit

( Phn tch v tm ti li gii :

Do S l mt biu thc i xng vi a, b, c nn d on Min S t ti

( S im ri:

( ( (

( Cch 1: Bin i v s dng bt ng thc Csi ta c

. Vi th

( Cch 2: Bin i v s dng bt ng thc BunhiaCpski ta c

(

EMBED Equation.DSMT4

. Vi th

( Cch 3: t

Do nn suy ra :

(

(

( (

( . Vi th

B. CC NG DNG GTLN, GTNN CA HM S

I. NG DNG TRONG PHNG TRNH, BT PHNG TRNH

Bi 1. Gii phng trnh:

Gii. t vi

Nhn BBT suy ra:

( Phng trnh c nghim duy nht x ( 3

Bi 2. Gii phng trnh:

Gii. PT ( . Ta c:

( ( (((x) ng bin

Mt khc (((x) lin tc v

,

( Phng trnh (((x) ( 0 c ng 1 nghim x0

Nhn bng bin thin suy ra:

Phng trnh c khng qu 2 nghim.

M nn phng trnh (1) c ng 2 nghim v

Bi 3. Tm m BPT: c nghim ng

Gii. ( (

Ta c: ( 0 (

;

Nhn BBT ta c ,

Bi 4. Tm m PT: (1) c nghim

Gii. Do ( nn t

( ; . Khi (1) (

( (2)

Ta c: ( Bng bin thin

Nhn bng bin thin suy ra:

(2) c nghim

th

( . Vy (1) c nghim th .

Bi 5. Tm m h BPT: (1) c nghim.Gii. (1) ( (2).

Ta c: ;

(((x) ( 0 ( . Nhn BBTsuy ra:

(2) c nghim th ( ( (3 ( m ( 7 Bi 6. Tm m ( 0 h: (1) c nghim.

Gii

(1) ( ( (2)

Xt . Ta c:

Nhn BBT suy ra: ((m) ( ((2) ( 1,(m ( 0

kt hp vi suy ra h (2) c nghim th m ( 2, khi h (2) tr thnh:

c nghim. Vy (1) c nghim m ( 2.

II. NG DNG GTLN, GTNN CHNG MINH BT NG THC

Bi 1. Chng minh rng: ,

BT (

Ta c:

( Bng bin thin.

Nhn bng bin thin suy ra:

( (pcm)

Bi 2. Cho CMR: T (

Ta c: T ( .

Xt hm s vi x > 0

Ta c .

Nhn bng bin thin ( .

Khi :

ng thc xy ra .Bi 3. Cho 3 ( n l. Chng minh rng: (x ( 0 ta c:

t .

Ta cn chng minh < 1

Ta c:

(

(

Do 3 ( n l nn (((x) cng du vi ((2x)

Nhn bng bin thin suy ra:

( (pcm)

Bi 4. Chng minh rng: (a, b > 0.

Xt f(t) = vi

f((t) =

EMBED Equation.DSMT4 f((t) = 0 ( t = 1 ( Bng bin thin ca f(t)

T BBT ( ( f(t) < 1 (t > 0 ( ( .

Du bng xy ra ( a = b > 0.

III. BI TP V NH

Bi 1. Cho (ABC c . Tm gi tr nh nht ca hm s:

Bi 2. Tm Max, Min ca: y (

Bi 3. Cho ab (0. Tm Min ca

Bi 4. Cho . Tm Max, Min ca

Bi 5. Gi s phng trnh c nghim x1, x2.

Tm p ( 0 sao cho nh nht.

Bi 6. Tm Min ca

Bi 7. Cho x, y ( 0 v . Tm Max, Min ca .

Bi 8. Cho . Tm Max, Min ca .

Bi 9. Tm m PT: c nghim.

Bi 10 Tm m PT: c nghim.

Bi 11 Tm m PT: c 4 nghim phn bit.

Bi 12 Tm m PT: c nghim duy nht.

Bi 13 Tm m PT: c nghim .

Bi 14 Tm m PT: c ng 2 nghim .

Bi 15 Tm m h BPT: c nghim.

Bi 16 a. Tm m : c 2 nghim phn bit.

b. Cho . CMR:

Bi 17 Chng minh: ,

x234(( (0((2

O(

1

3/ 2

N

L

I

x

z

M

J

3/ 2

K

1

1

E

O

3/ 2

y

t(1t1t21(((0(0((1 EMBED Equation.DSMT4 EMBED Equation.DSMT4 1

y

1

x

t01+(f((0+f1

EMBED Equation.3 1

x((0((f ((0(f1

x(( EMBED Equation.DSMT4 ((f ((0(f EMBED Equation.DSMT4

x((0((f ( (0(f

0

x0 EMBED Equation.DSMT4 23f ((0(( f0CT821

m02(((( (0((171((

t(1 EMBED Equation.DSMT4 1(((t) (0(((t)4

04

x(((66((f ((0+0(

(

EMBED Equation.DSMT4

EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4

x(( 0x0 1((f ( (0(f

((x0)

O

3

2

1

B

A

C

a+b+c

a+b

a

x EMBED Equation.DSMT4 1/3 EMBED Equation.DSMT4 y ( +0(0y(1 EMBED Equation.DSMT4 1

x( 2 EMBED Equation.DSMT4 2y ( +0(0y(2 EMBED Equation.DSMT4 2

x0 EMBED Equation.DSMT4 1y (0 (0(0y4

EMBED Equation.DSMT4 1

P2

P1

C

B

A

P1

P2

C

B

A

P1

P2

C

B

A

1413

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