gtlt_gtnn ham so
TRANSCRIPT
-
7/28/2019 Gtlt_gtnn Ham So
1/18
3: GTNN, GTLN CA HM S
1. Cc kin thc c bn
nh ngha GTNN, GTLN ca hm s
Cho hm s y=f(x) xc nh trn min D
S M gi l GTLN ca hm s y=f(x) trn D nu : 0 0
f (x) M, x D
x D,f (x ) M
=
S m gi l GTNN ca hm s y=f(x) trn D nu :0 0
f (x) m, x D
x D,f (x ) m
=
2. Cc k nng c bn
K nng tm GTNN, GTLN ca hm s y=f(x) trn mt khong, mt on
Tm GTNN, GTLN ca hm s y=f(x) lin tc trn khong (a;b)
- Tnh o hm f(x).
- Tm cc nghim 1x , 2x , , nx ca f(x) trn (a;b).
- Lp bng bin thin ca f(x) trn (a,b).
Cn c vo bng bin thin suy ra GTLN, GTNN ca f(x) trn (a;b)
Tm GTNN, GTLN ca hm s y=f(x) lin tc trn [a;b]
- Tnh o hm f(x).
- Tm cc nghim 1x , 2x , , nx ca f(x) trn [a;b].
- Tnh f(a) , f(b) , 1f (x ) , , nf (x ) .
Chn s M ln nht trong n+2 s trn x [a;b]
M max f (x)
= .
Chn s m nh nht trong n+2 s trn x [a;b]
m minf (x)
= .
3. H thng bi tp s dng phng php o hm tm GTLN,
GTNN ca hm s.Dng 1. Kho st trc tip
Nu hm s y=f(x) trn min D cho dng n gin , ta c th kho st trc tip hm s
v rt ra kt lun GTNN, GTLN ca hm s.
gii quyt tt cc bi ton dng ny, HS cn c cc k nng sau:
- Tnh f(x) chnh xc.
- Bit cch tm nghim ca phng trnh f(x)=0.
- Bit cch lp bng bin thin ca f(x) trn D rt ra kt lun GTNN, GTLN ca
hm s.Bi 1.Tm GTNN, GTLN ca hm s 2y x 4 x= +
-
7/28/2019 Gtlt_gtnn Ham So
2/18
Li gii
TX D=[-2,2]
2
xy ' 1
4 x=
; y=0 24 x x = 2 2
x 0
4 x x
= x= 2
y(-2)=-2 ; y(2)= 2 ; y( 2 )=2 2
Vy x Dmax f (x) 2 2 = ; x Dmin f (x) 2 =Bi 2.Tm GTNN, GTLN ca hm s
y =2
x+1
x +1trn on [ ]2;1 .
Li gii
Ta c :( )
,
32
-x+1y =
x +1 ,y = 0 x=1.
Do y(-1) = 0, y(1) = 2, y(2) = 53 nn
[ ]1;2max y
= y(1) = 2, [ ]y
2;1min = y(-1) = 0.
Bi 3. Tm GTNN, GTLN ca hm s2
2
x 8x 7y
x 1
+=
+ (x R)
Li gii2
2 2
8x 12x 8y '
(x 1)
=
+; y' 0= x 2= ;
1x
2=
Bng bin thint -
1
2 2 +
y + 0 - 0 +
y
9 1
1 -1
Vy x Rmin y 1 = khi x 2= ; x Rmax y 9 = khi1
x 2=
Bi 4. Tm GTNN, GTLN ca hm s
y 5cos x cos5x= vi x [- ; ]4 4
Li gii
y ' 5sin x 5sin 5x= +k
x5x x k2 2y ' 0 sin5x sin xk5x x k2
x6 3
== + = = = + = +
-
7/28/2019 Gtlt_gtnn Ham So
3/18
*)k
x2
=
Do x4 4
k
4 2 4
1 1k
2 2 k=0 x=0.
*)k
x6 3
= +
Do x4 4
k k 5 k
4 6 3 4 4 6 3 4 6 12 3 12
+
xk 15 1 6kk 04 4
x6
= = = =
y(0) 4= ; y( ) y( ) 3 36 6
= = ; y( ) y( ) 3 2
4 4
= =
Vy Miny=4 ; Maxy = 3 3
Bi 5. Tm GTNN ca 2y x 2x 1= + + ( x R)
Li gii
2
2xy ' 1
2x 1= +
+; 2
2 2
x 0 1y ' 0 2x 1 2x x
2x 1 4x 2
= + = =
+ =
Bng bin thin
x-
1
2 +
y - 0 +
y
+ +
1
2
Vy1
Miny2
= khi1
x2
=
Dng 2. Kho st gin tip
Trong nhiu bi ton tm GTNN, GTLN ca hm s nu ta kho st trc tip c th gp
nhiu kh khn , chng hn nh tm nghim ca f(x), xt du ca f(x). Do thay v kho
st trc tip f(x) ta c th kho st gin tip hm s cho bng cch sau:
- t n ph t, chuyn hm s cho v hm s mi g(t).
- Tm iu kin ca n ph t ( Bng cch kho st hm s, dng bt ng thc)
- Kho st hm s g(t) suy ra GTNN, GTLN ca hm s.
gii quyt tt dng ton ny HS cn phi c nhng k nng sau:
- K nng chn n ph t : Chn n ph t thch hp sao cho hm s ban u c th qui
ht v bin t.
-
7/28/2019 Gtlt_gtnn Ham So
4/18
- K nng tm iu kin ca n ph : tm iu kin ca t, ty theo tng bi ton c
th ta c th dng phng php o hm, dng bt ng thc, nh gi trc tip
Bi 6. Tm GTNN , GTLN ca 8 4S 2sin x cos 2x= + , xR
Li gii
Do
2 1 cos2x
sin x 2
=
nn ta qui S v cos2x
S= 4 41 cos2x
2( ) cos 2x2
+ = 4 4
1(1 cos2x) cos 2x
8 +
t t= cos2x , 1 t 1
Bi ton tr thnh tm GTNN, GTLN ca hm s 4 41
S g(t) (1 t) t8
= = +
vi 1 t 1
Ta c 3 31
g '(t) (1 t) 4t2
= + ; g(t) = 0 3 3(1 t) 8t = 1-t =2t 1
t3
=
g(1) =1 ; g(-1)=3 ; g(1
3)=
1
27
Vy MinS=1
27; MaxS= 3
Bi 7: Tm GTNN, GTLN ca hm s y= 1 sin x 1 cos x+ + + ( x R)
Li gii
Hm s xc nh vi x v y>0 vi x , do y t GTNN, GTLN ng thi vi 2y
t GTNN, GTLN.
Ta c: y2= 2 + sinx+ cosx+ 2 1+ sinx+ cosx+ sinxcosx
t t= sinx+ cosx = 2 sin x+4
= t ( )- 2 t 2
Th y2= f(t) =2t -1
2 + t+ 2 1+ t+2
=2t + 2 t+1
2 + t + 22
= 2 + t+ 2 t+1
Vy 22 t 2(t 1)
y f (t)2 t 2(t 1)
+ += =
+ + +vi
- 2 t -1
-1 t 2
1 2 , ( 2 t 1)f '(t)1 2 , ( 1 t 2 )
= + <
Bng bin thin:
-
7/28/2019 Gtlt_gtnn Ham So
5/18
t 2 1 2 +f(t) - 0 +
f(t)
4 2 2 4 2 2+
1
T bng bin thin ta c
[ 2 ; 2 ]max f( t)
= 4+2 2 ; [ 2 ; 2 ]min f (t)
= 1
x R
max y 4 2 2
= + ;x R
min y 1
=
Bi 8. Tm GTNN ca biu thc 20 20S sin (x) cos (x)= +
Li gii
Nhn xt : Ta quy S v ht 2sin x
Ta c 2 10 2 10S (sin x) (1 sin x)= +
t 2t sin x= (0 t 1) . Yu cu bi ton tr thnh tm GTNN, GTLN ca hm s
10 10S f (t) t (1 t)= = + vi t [0;1]
9 9f '(t) 10t 10(1 t)=
9 9
f '(t) 0 t (1 t)= = 1
t 2=
1 1f (0) 1; f ( ) ; f (1) 1
2 512= = = . Vy
1MinS
512= ; MaxS 1=
Bi luyn tp 1. Tm GTNN , GTLN ca biu thc sau:2012 2012S sin (x) cos (x)= +
Bi 9. Tm GTNN, GTLN ca biu thc
S x 4 4 x 4 (x 4)(4 x) 5= + + + +
Li gii
iu kin 4 x 4
t t x 4 4 x= + + 2t x 4 4 x 2 (x 4)(4 x) = + + + +
2t 8
(x 4)(4 x)2
+ =
Ta c2
2t 8S t 4( ) 5 2t t 212= + = + +
Tm iu ca t:
-
7/28/2019 Gtlt_gtnn Ham So
6/18
Xt hm s g(x) x 4 4 x= + + vi x [ 4;4]
1 1
g'(x)2 x 4 2 4 x
= +
; g '(x) 0= x=0
g( 4) 2 2; g(0) 4; g(4) 2 2 = = =
x [ 4 ;4 ]min g(x) 2 2
=; x [ 4 ;4 ]
max g(x) 4
= t [2 2;4]
S' 4t 1 0 t [2 2;4]= + < S l hm nghch bin trn [2 2;4]
MinS S(4) 7 ; MaxS S(2 2) 5 2 2= = = = +
Bi luyn tp: Tm GTNN, GTLN ca biu thc sau:
S x 1 8 x 4 (x 1)(8 x) 5= + + + + vi x [ 1;8]
Bi 10. Tm GTNN, GTLN ca2010 2011
S sin (x).cos (x)= vi x [0; ]2
Li gii
Nhn xt:
i, S 0 vi mi x [0; ]2
ii, tm GTNN, GTLN ca S ta tm GTNN, GTLN ca 2S (v khi 2S c th quy
ht v2
sin x hoc2
cos x ).Ta c 2 4020 4022S sin (x).cos (x)=
= 2 2010 2 2011(sin x) .(1 sin x)
t 2t sin x= (0 t 1). Khi 2 2010 2011S f (t) t .(1 t)= =
2009 2011 2010 2010f '(t) 2010t (1 t) 2011.t (1 t)=
2009 2010f '(t) t (1 t) [2010 4021t]=
t 0
f '(t) 0 t 1
2010t
4021
=
= =
=
f (0) 0 ;f (1) 0= = ;2010 2011
4021
2010 (2010) .(2011)f ( )
4021 (4021)=
Vy Min S =0 ;2010 2011
4021
(2010) .(2011)MaxS
(4021)=
Bi luyn tp. Tm GTNN, GTLN ca biu thc
-
7/28/2019 Gtlt_gtnn Ham So
7/18
15 20S sin x.cos x= vi x [0; ]2
Bi 11. Tm GTNN, GTLN ca hm s :6 6y sin x cos x 2cos4x sin 2x 5= + + + , vi x R
Li gii
Nhn xt : 6 6 23sin x cos x 1 sin 2x4
+ =
2cos4x 1 2sin 2x=
Do ta a y v ht sin2x
Do y = 23
1 sin 2x4
+2( 21 2sin 2x )+sin2x-5
219
y sin 2x sin 2x 24= +
t t sin 2x= ( 1 t 1) . Yu cu bi ton tr thnh tm GTNN, GTLN ca hm s
219y t t 24
= + vi 1 t 1
19 2y ' t 1 ;y ' 0 t
2 19= + = =
Ta c 31 23 2 37y( 1) ;y(1) ; y( )4 4 19 19
= = =
Do x R
31miny
4= ;
x R
37maxy
19=
Bi luyn tp: Tm GTNN, GTLN ca hm s :6 6 4 4y sin x cos x 4(sin x cos x) 2cos2x 2= + + + , vi x R
Bi 12. Tm GTNN, GTLN ca hm s
1y 2(1 sin 2x.cos4x) (cos 4x cos8x)
2= +
Li gii
Ta c y 2 2sin 2x.cos4x sin 6x.sin 2x= +
2 32 sin 2x.(1 2sin 2x) (3sin 2x 4sin 2x).sin 2x= +
4 3 24sin 2x 4sin 2x 3sin 2x 2sin 2x 2= + +
t t sin 2x= ( 1 t 1)
Yu cu bi ton tr thnh tm GTNN, GTLN ca hm s4 3 2y 4t 4t 3t 2t 2= + + vi t [ 1;1]
-
7/28/2019 Gtlt_gtnn Ham So
8/18
3 2y ' 16t 12t 6t 2= + ;1 1
y' 0 t 1; t ; t2 4
= = = =
y( 1) 5 = ;1
y( ) 12
= ;1 145
y( )4 64
= ; y(1) 1=
Vy min y 1= ; max y 5=
Bi 13. Tm GTNN ca hm s y x(x 2)(x 4)(x 6) 5= + + + + vi x 4 .
Li gii
Ta c 2 2y (x 6x)(x 6x 8) 5= + + + +
t 2t x 6x= +
Khi 2y t 8t 5= + +
Xt hm s 2g(x) x 6x= + vi x 4
g '(x) 2x 6;g '(x) 0 x 3= + = =
x - -4 -3 +g(x) - 0 +
g(x)
-8 +
-9
Suy ra t [ 9; ) +Yu cu bi ton tr thnh tm GTNN ca hm s 2y t 8t 5= + + vi t [ 9; ) + .
Ta c y' 2t 8 ; y ' 0 t 4= + = =
Bng bin thin
t - -9 -4 +
y - 0 +
y
14 +
-11Vy Miny=-11.
Trong nhiu bi ton tm GTNN, GTLN ca hm s khi bi c nhiu hn hai bin ta phi
tm cch qui v mt bin , sau tm GTNN, GTLN ca hm s theo bin s mi.
Sau y l cc bi ton minh ha
Bi 14. Tm GTNN, GTLN ca2 2
2 2
2 2
x xy yS (x y 0)
2x y
+ += + >
+Li gii
-
7/28/2019 Gtlt_gtnn Ham So
9/18
V t s v mu s ca S l cc biu thc ng cp bc hai i x, y nn ta xt TH y=0 v y
0 chia t s v mu s ca S cho 2y , sau chuyn v bin sx
ty
= .
TH1: y= 0 2
2
x 1S
2x 2= =
TH2: y 0 . Chia c t s v mu s ca S cho 2y ta c :
2
2
2
2
x x1
y ySx
2 1y
+ +=
+
tx
ty
= . Khi 2
2
t t 1S
2t 1
+ +=
+2
2 2
2t 2t 1S'
(2t 1)
+=
+; 2
1 3S' 0 2t 2t 1 0 t
2
= + = =
Bng bin thint - 1 3
2
1 3
2
++
S - 0 + 0 -
S1
2
3
2 3 2
3
2 3 2+
1
2
Kt hp TH1 v TH2 ta c : 32 3 2+
S 32 3 2
Vy MinS =3
2 3 2+khi
x 1 3
y 2
= ; Max S =
3
2 3 2khi
x 1 3
y 2
+=
Bi luyn tp : Tm GTNN, GTLN ca cc biu thc sau:
a,2 2
2 2
2 2
x xy yM (x y 0)
3x y
+ = + >
+
b,2 2
2 2
2 2
x xy 2yN (x y 0)
4x y
= + >
+
Bi 15. Cho a.b 0 . Tm GTNN ca4 4 2 2
4 4 2 2
a b a b a by ( )
b a b a b a= + + + +
Li gii
ta b
t
b a
= + . Ta ca b a b
| t | | | | | | | 2
b a b a
= + = + ( Theo C Si )
2 2
2
2 2
a bt 2
b a+ =
4 44 2
4 4
a bt 4t 2
b a+ = +
-
7/28/2019 Gtlt_gtnn Ham So
10/18
4 2 2y t 4t 2 (t 2) t= + + = 4 2t 5t t 4 + +
3y'(t) 4t 10t 1= + 2y ''(t) 12t 10 0= > vi mi t 2
Bng bin thin ca y(t)t - -2 2 +
y(t) + +
y(t)
-11
-
+
13
Suy ra y '(t) 0< vi t 2 ; y '(t) 0> vi t 2
Bng bin thin ca f(t)
t - -2 2 +
y(t) - +
y
-
-2
+
2
Vy Miny=-2 ; Maxy=2 .
Nhn xt
i, ta b
tb a
= + gip ta chuyn y v ht bin t.
ii, xt du ca y ta tnh y , lp bng bin thin ca y, sau suy ra du ca y
trn cc khong ( ; 2] v [2; )+ .
Bi 16. Cho x, y, z > 0 v x +y+z 1. Tm GTNN ca biu thc
2 2 2
3 3 3
3 3 3S x y z
x y z= + + + + +
Li gii
Nhn xt: Ta quy S v x+ y +z
p dng bt ng thc Bunhiacopxki ta c:
2 2 2 2 2 2 2(x y z )(1 1 1 ) (x y z)+ + + + + + 2 2 2 21
x y z (x y z)3
+ + + +
p dng bt ng thc C Si ta c:
33 3 3 3 3 3 3
3
3 3 3 3 3 3 9 9 81
3 . . x y zx y z x y z xyz (x y z)( )3
+ + = =+ + + +
-
7/28/2019 Gtlt_gtnn Ham So
11/18
Vy2
3
(x y z) 81S
3 (x y z)
+ + +
+ +
t t= x+y+z (0 t 1)<
Khi 2
3
t 81S f (t)
3 t = + ;
5
4 4
2t 243 2t 729f '(t) 0 t (0;1]
3 t 3t
= = <
f(t) nghch bin trn (0;1] t (0;1]
244minS min f (t) f (1)
3= = =
Bi 17. Cho A, B, C l 3 gc ca mt tam gic. Tm GTNN ca biu thc
2 2 2A B C A B CP sin sin sin cot cot cot2 2 2 2 2 2
= + + + + +
Li gii
Ta c 2 2 2A B C 1 1 1
P sin sin sin 3A B C2 2 2
sin sin sin2 2 2
= + + + + +
Trong tam gic ABC ta cA B C 3
sin sin sin2 2 2 2
+ +
Ta nh gi cc biu thc theoA B C
t sin sin sin2 2 2
= + + vi3
t (0; ]2
p dng bt ng thc C Si ta c:
2 2 2 22 2 23
1 1 1 3 27A B C A B CA B Csin sin sin (sin sin sin )sin .sin .sin2 2 2 2 2 22 2 2
+ + + +
Vy 227
P t 3 f (t)t
+ =
3
54f '(t) 1 0
t= < vi mi
3t (0; ]
2
Bng bin thint
03
2f(t) -
P=f(t)
+
21
2
Vy
21
MinP 2=
3
t 2 =
Bi 18: Cho 0,0 yx v x + y = 1. Tm GTNN, GTLN ca biu thc
P = 2 x y3 + 3 .
-
7/28/2019 Gtlt_gtnn Ham So
12/18
Li gii
Do x 0, y 0 v x + y = 1 y = 1- x v 0 x 1 Ta c P = 2 x y3 + 3 = 2 x x3
3 +3
.
t t = 3x 1 t 3 . Khi 23
P f (t) tt
= = + vi [ ]t 1;3 .
2
3f '(t) 2t 0
t
= = 33
t =
2
Bng bin thin
t 1 3
3
23 +
f(t) - 0 +
f(t)
4 10
39
34
T bng bin thin ta c maxP = 10 3x = 3 x = 1
y = 0
minP = 34
93 3x = 3
2
3
33
3
3
3x = log
2
3
y =1- log 2
Bi 19: Cho x 0, y 0 v x + y = 1. Tm GTNN, GTLN ca biu thc
P =x y
+y+1 x+1
.
Li gii
Ta c: P =x y
+
y+1 x+1
=( )
2x+ y - 2xy+1
2x+xy
=2-2xy
2+xy
.
t xy = t , v x+ y =1, x 0, y 0 xy 0
1 2 xy
10 xy
4
10 t
4
Khi P = f(t) =2 - 2 t
2 + tvi
10 t
4 .
Do( )
2
-6f '(t) < 0
2 + t= vi
1t 0;
4
nn hm s f(t) lun nghch bin trong on
4
1;0 maxP = f(0) = 1 khi t = xy = 0
x = 0 , y = 1
y = 0 , x = 1
-
7/28/2019 Gtlt_gtnn Ham So
13/18
minP = f(4
1) =3
2khi t =
4
1 x = y =
2
1.
Trong cc k thi chn HS gii thng c bi ton tm GTLN, GTNN ca hm s c nhiu
bin ph thuc ln nhau. gii nhng bi ton dng ny ta c th dng phng php kho
st ln lt tng bin, ngha l : tm GTNN ( hoc GTLN ) ca hm s vi bin th nht v
cc bin cn li coi l tham s , ri tm GTLN (GTNN) ca hm s vi bin th hai v ng
vi gi tr xc nh ca bin th nht m cc bin cn li coi l tham s
Bi 20. Cho min {D (x, y) |0 x 1;0 y 2}=
Tm GTNN ca hm s f (x, y) (1 x)(2 y)(4x 2y)=
Li gii
f(x,y) (1 x)(2 y)[2(2-y)-4(1-x)]=
t u=1-x ; v=1-y u [0;1] ; v [0;2]
f ( x, y ) g(u, v) uv(2v 4u ) = = 2 22uv 4u v=
Coi u l n , v l tham s . Ta c
2g '(u,v) 8uv 2v= +
vg '(u,v) 0 u
4= =
Bng bin thin
u- 0
v
41 +
g'(u,v) + 0 -
g(u,v)
3v
4
0 22v 4v
V 22v 4v 2v(v 2) 0 v [0; 2] = nn 2 2g(u,v) 2v 4v 2(v 1) 2 2 =
Vy Ming(u,v) =-2u 1
v 1
= =
Suy ra Minf(x,y)=-2x 0
y 1
= =
Bi 21. Cho hm s f (x, y,z) xy yz zx 2xyz= + + trn min
{ }D (x, y,z) |0 x, y, zv x y z 1= + + =
Tm GTNN, GTLN ca f(x,y,z)
-
7/28/2019 Gtlt_gtnn Ham So
14/18
Li gii
*) Tm GTNN ca f(x,y,z)
Gi s1
z min{x, y,z} z [0; ]3
=
f (x, y,z) xy (x y)z 2xyz xy(1 2z) z(1 z)= + + = +
23 2(1 z) 1(1 2z) z(1 z) (2z z 1)
4 4
+ =
Xt hm s 3 21
F(z) (2z z 1)4
= vi1
z [0; ]3
21 z(1 3z) 1F'(z) (6z 2z) 0 z [0; ]4 2 3
= = >
z - 0 1 +F(z) +
F(z)
7
27
1
4
Vy Max f(x,y,z) =7
27
1x y z
3 = = =
*) Tm Min f(x,y,z)
f (x, y,z) (1 y z)y yz z(1 y z) 2(1 y z)yz= + +
2 2 2 2y y z zy z 2yz 2y z 2z y= + + +
Xt 2 2 2G(z) z (2y 1) z(1 3y 2y ) y y= + + +
2G '(z) 2(2y 1)z 1 3y 2y
2(2y 1)z (2y 1)(y 1)
= + += +
Gi s1
y min{x, y,z} y [0; ]3
=
1 yG '(z) 0 z
2
= =
Bng bin thin
z-
01 y
2
1 +
G(z) + 0 -
G(z)
1 yG( )
2
-
7/28/2019 Gtlt_gtnn Ham So
15/18
2y y 2y
V 2 2 2y (y y ) 2y y y(2y 1) 0 = =
2G(z) y 0 ;y 0
G(z) 0z 1
== =
Vy Min f(x,y,z)=0x y 0
z 1= = =
Bi luyn tp 1: Cho x, y,z 0 v x+y+z=1. Tm GTLN ca biu thc:
1 1 1 1 1 1S xyz[x( ) y( ) z( )]
y z x z y x= + + + + +
Bi luyn tp 2: Cho x, y, z tha mn 0 x, y,z 1
Tm GTLN ca biu thc3 3 3 2 2 2
P 2(x y z ) (x y y z z x)= + + + +
Bi 22. Cho x, y, z >0 tha mnx y z 4
xyz 2
+ + = =
Tm GTNN, GTLN ca biu thc 4 4 4S x y z= + +
( thi chn HSG QG nm 2004 )
Li gii
t1
2
3
S x y z 4
S xy yz zx
S xyz 2
= + + = = + + = =
Ta biu din S theo 2f ( S ) . Cn c vo bi, tm min bin thin ca 2S . Sau ta kho st
2f ( S ) tm GTNN, GTLN ca S.
*) Ta biu din 2S theo z
2
2
2 2S xy yz zx z(4 z) 4z z
z z= + + = + = +
*) Ta tm min bin thin ca z
Dox y 4 z
2xy
z
+ =
= 2
8(4 z)
z 2(z 2)(z 6z 4) 0 +
-
7/28/2019 Gtlt_gtnn Ham So
16/18
z - 3 5 2 3 5+ +
z-2 - - 0 + +
2z 6z 4 + + 0 - - 0 +
VT - + - +
Vy z [3 5;2] [3 5; ) + +
T gi thit suy ra z (0;4)
Do z [3 5;2]
Xt 222
S 4z zz
= + vi z [3 5;2]
2'
2 2 2
2 (1 z)(2z 2z 2)S 4 2z
z z
= =
Bng bin thin
z- 1 5
2
3 5 1
1 5
2
+2 +
'2S 0 - 0 + 0 -
2S 5 5 1
2
5 5 1
2
5 5
Suy ra 2S5 5 1
[5; ]2
Khi 4 4 4 2 2 2 2 2 2 2 2 2 2x y z (x y z ) 2(x y y z z x )+ + = + + + +
2 2 2
1 2[S 2S ] 2[(xy yz zx) 2xyz(x y z)]= + + + +
= 2 2 21 2 2 1 3[S 2S ] 2[S -2S .S ]
= 2 22 2[16 2S ] 2[S 16]
= 2 22 2 2256 64S 4S 2S 32 + +
= 22 22S 64S 288 + = 2f (S )
2 2f '(S ) 4S 64 0= < 2S5 5 1
[5; ]2
. Bng bin thin
-
7/28/2019 Gtlt_gtnn Ham So
17/18
2S 5
5 5 1
2
f( 2S ) -
f( 2S )
18
383 165 5 Vy MinS=383 165 5 ; MaxS=18.
rn luyn k nng gii cc bi ton dng trn, ta c bi ton sau:
Bi luyn tp: Cho x, y, z >0 tha mnx y z 4
xyz 2
+ + = =
Tm GTNN, GTLN ca cc biu thc sau:
a, 2 2 2M x y z= + +
b, 3 3 3N x y z= + +
c, 3 3 3 3 3 3P x y xy x z xz y z yz= + + + + +
BI TP NGH
Bi 1. Tm GTNN ca hm s:
y x 2 2 x 4 (x 2)(2 x) 5= + + + + ( x [ 2;2] )Bi 2. Tm GTNN, GTLN ca hm s:
2
2
x x 3y
2x 1
=
+
Bi 3. Tm GTNN, GTLN ca hm s:2 2
2 2
x xy yy
2x y
+=
+( 2 2x y 0+ > )
Bi 4. Tm GTNN, GTLN ca hm s:
y (x 2)(x 4)(x 6)(x 8) 5= + + + + + ( x [ 4;2] )
Bi 5. Tm GTNN, GTLN ca hm s:6 6y sin x cos x 2cos4x sin 2x 5= + + + +
Bi 6. Tm GTNN, GTLN ca hm s:x yy 9 3= + vi x, y 0 v x+y=1
Bi 7. Tm GTNN, GTLN ca hm s:
y 4cos x cos4x= vi x [ ; ]2 2
-
7/28/2019 Gtlt_gtnn Ham So
18/18
Bi 8. Tm GTNN, GTLN ca hm s:4 2
4 2
3cos x 4sin xy
3sin x 2cos x
+=
+
Bi 9. Tm GTNN, GTLN ca hm s:
x x
x x
1 1S 9 4(3 ) 5
3 9
= + + + vi x [ 1;1]