1.3.gia tri lon nhat - nho nhat cua ham so
DESCRIPTION
Tìm giá trị lớn nhât, nhỏ nhất của hàm sốTRANSCRIPT
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
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