[Www.vietmaths.com]Ung Dung Bat Dang Thuc Tim Gtln Va Gtnn

Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh

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PHN M U

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I. L DO CHN TI C th ni trong chng trnh ton bc trung hc ph thng th phn kin thc v bt ng thc l kh kh. Ni v bt ng thc th c rt nhiu bt ng thc c cc nh Ton hc ni ting tm ra v chng minh. i vi phn kin thc ny th c hai dng bi tp l chng minh bt ng thc v vn dng bt ng thc gii cc bi ton c lin quan. L mt sinh vin ngnh ton ti khng ph nhn ci kh ca bt ng thc v mun tm hiu thm v cc ng dng ca bt ng thc phc v cho vic ging dy ton sau ny. Do ti chn ti Vn dng bt ng thc t m gi tr ln nht, gi tr nh nht v gii phng trnh tm hiu thm. Khi vn dng bt ng thc gii cc bi ton dng ny th c rt nhiu bt ng thc chng ta vn dng. y ti ch gii hn trong ba bt ng thc l bt ng thc Csi, Bunhiacopski v bt ng thc vect. Trong ti ny ti trnh by cch v n dng ba bt ng thc trn tm gi tr ln nht, gi tr nh nht v gii phng trnh rn luyn kh nng vn dng bt ng thc gii ton v qua c th tch ly c kinh nghim trong gii ton ging dy sau n y.

III. I TNG NGHIN CU

i tng ca ti l ba bt ng thc Csi, Bunhiacopski v bt ng thc vect cng vi cc bi ton tm gi tr ln nht, nh nht v cc phng trnh. ti ny ch yu xoay quanh ba i t ng trn bn cnh ti cng gii thiu v chng minh mt s bt ng thc thng d ng khc.

Phm vi ca ti ny ch xoay ch yu vo ba bt ng thc nu trn gii cc bi ton tm gi tr ln nht, nh nht v gii phng trnh. VI. PHNG PHP NGHIN C U

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Tm v tham kho ti liu, su tm phn tch v bi tp gii minh ha, tham kho kin ca cn b hng dn

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IV. PHM VI NGHIN CU

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Mc tiu chnh ca ti ny l tng hp cc bi ton tm gi tr ln nht, nh nht v gii phng trnh bng bt ng thc ch yu vn dng ba bt ng thc ni trn. Qua y ti hi vng s a ra y cc dng vn ca cc bt ng thc ni trn.

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II. MC CH NGHIN CU

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PHN NI DUNG

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Phn 1: S LC V BT NG THC

1.2. Tnh cht c bn ca bt ng thc

1.3.1. Bt ng thc cha tr tuyt i a b a b du = xy ra

ie t

1.3. Mt s bt ng thc c bn

m

a b a c b c a c b c a b a b c a c b a b c d a c e b e f a b v m 0 ma a b v m 0 ma a b 0 c d 0 ac bd an bn a b 0 a b

dmb mb

f

n

a

tab 0

a

b

a b

a1 a2 ... an n n a1.a2 ...an n Du = xy ra a1 a2 ... an M rng: Cho n s dng a1 , a2 ,..., an n c: 1 ... n 1 . Th: 2

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a1 a2 ... an a1 a2 ... an 1.3.2. Bt ng thc Csi Cho hai s dng a, b ta c: a b 2 ab Du = xy ra a b Tng qut: cho n s khng m a1 , a2 ,..., an n

w .v

h2 , ta lun c: 2 v n s

s

.c o1

Cho hai s thc a, b bt k, ta nh ngha: a b a b 0

m,2

1.1. nh ngha bt ng thc

,....,

n

dng

a1 1 .a2 2 ...an n 1a1 Du = xy ra a1 a2

a ... ... an2 2

n n

a

Trang 4


1.3.3. Bt ng thc Bunhiacopski Bt ng thc Bunhiacopski: Cho hai b s a, b v c, d ta c: 2 ac bd a2 b2 c2 d 2 a b Du = xy ra c d Tng qut: Cho n s a1 , a2 ,..., an v b1, b2 ,.., bn ty ta c:

a1b1 a2b2 ... anbnDu = xy ra

2

a12a2 b2 ...

2 2 2 2 a2 ... an b12 b2 ... bn

a1 b1

an bn

a1a2 ...am a1m

b1b2 ...bm

c1c 2 ...cm

m

m

Cho a 1 v r N : n Nu n 1 th 1 a

1 na du = xy ran

a

Du = xy ra a1 : b1 : ... : c1 a2 : b2 : ... : c2 1.3.4. Bt ng thc Bernuolli

t

m b1m ... c1m a2

m m m b2 ... c2 ... am

h

M rng: Cho m b s, mi b gm n s khng m: ai , b i ,...c i Khi ta c:

sw

.c oa

m m bm ... cm

... an : bn : ... : cn

u.v

u .v

u vu v

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ie t

Nu a n 1 th 1 a 1.3.5. Bt ng thc vect

1 na

u

v

u

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u v

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u v

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Trang 5

mi 1, 2,..., m 0 hoc n 1


Phn 2: TM GI TR LN NHT V GI TR NH NHT CA HM S HOC BIU THC

2.1 KIN THC CN NH

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i vi vic tm gi tr ln nht, gi tr nh nht ca biu thc (hm s) th c th k n cc phng php sau: phng php kh o st, phng php nh gi thng thng v phng php s dng bt ng thc. Trong cc ph ng php nu trn th phng php s dng cc bt ng thc c th coi l mt trong nhng phng php thng dng v hiu qu nht tm gi tr ln nht v nh nht ca biu thc v hm s. i vi phng php ny, ta s dng cc bt ng thc thng dng nh: bt ng thc Csi, Bunhiacopski, Schwartz, Bernouli, bt ng thc vect nh gi biu thc P (h oc hm s f ( x1, x2 ,..., x n ) ), t suy ra gi tr ln nht, gi tr nh nht cn t m. Phng php ny, nh tn g i ca n, da trc tip v o nh ngha ca gi tr ln nht v nh nht ca biu thc v hm s. Lc chung ca phng php ny c th miu t nh sau: - Trc ht chng minh mt bt ng thc c dng P ( x1 , x2 ,..., xn ) D vi bi ton tm gi tr nh nht (hoc P ( x1 , x2 ,..., xn ) D i vi bi ton tm gi tr ln nht), y P l biu thc hoc hm s xc nh trn D. - Sau ch ra mt phn t ( x01 , x02 ,..., x0 n ) D sao cho P( x01 , x02 ,..., x0 n ) . Ty theo dng ca bi ton c th m ta chn mt bt ng thc thch hp p dng vo vic tm gi tr nh nht v ln nht. Do phm vi ca ti, y ch gii thiu phng php s dng ba bt ng thc l: Csi, Bunhiacopski v phng php b t ng thc vect.

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2.1.2. Tm gi tr nh nht, gi tr ln nht ca biu thc (h m s) bng phng php vn dng bt ng thc

t1 x3

Cho biu thc P( x1, x2 ,..., xn ) ( hm s f ( x1, x2 ,..., x n ) ), xc nh trn D ( x1, x2 ,..., xn ) D v - Nu P( x1, x2 ,..., x n ) M (hoc f ( x1, x2 ,..., xn ) M ) ( x1, x2 ,..., x n ) D sao cho: P( x1, x2 ,..., xn ) M th M gi l gi tr ln nht ca P( x1, x2 ,..., xn ) (hoc f ( x1, x2 ,..., x n ) ). K hiu l maxP hoc Pmax ( max f ( x1, x2 ,..., x n) hoc f ( x1, x2 ,..., x n )max ). - Nu P( x1, x2 ,..., x n ) m ( hoc f ( x1, x2 ,..., x n ) m ) th m gi l gi tr nh nht ca P( x1 , x2 ,..., xn ) ( hm s f ( x1, x2 ,..., x n ) ). K hiu l minP hoc Pmin (min f ( x1, x2 ,..., x n ) hoc f ( x1 , x2 ,..., x n )min ).

h55

s1 2 x 33

V d : Tm gi tr nh nht ca hm s f ( x) x 2 Gii: Ta c: f ( x)1 2 x 3 1 2 x 3 1 2 x 3 1 x3

Trang 6

.c o1 x65

2.1.1. nh ngha

2 ( x 0) x35 ( BT Csi) 27

m


Du = xy ra Vy Min f x =5

1 2 x 35

1 x3

x5

3

x

5

3

5 ti x 27

3

2.2. BI TP 2.2.1. S dng bt ng thc Csi

p dng bt ng thc Csi ta c:

1

a b

2

a b1

1

b c

m

a

Gii:

2

b c8 abc abc 8

t

Bi 1: Cho ba s thc dng a, b, c . Tm gi tr nh nht ca biu thc sau: a b c P 1 1 1 b c a

h12

s1111 1 a

Lu : bit c bi ton no s dng bt ng Csi ta cn ch n cc thnh phn ca hm s hoc biu thc. Nu n c dng tch hoc l tng ca hai phn khng m v c bit sau khi vn dng bt ng thc Csi th xut hin biu thc ca gi thit ban u v a c v hng s th ta c th s dng bt ng thc Csi nh gi tm gi tr ln nht, nh nht.

.c oc a 22.1b

Suy ra Hay

a b c 1 1 b c a

P 8

Du = xy ra khi v ch khi a b c 1 Vy Pmin

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Bi 2: Cho ba s thc a, b, c 0 tha

1

1 a 1 b 1 cabc

Tm gi tr ln nht ca biu thc M

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Gii:

Ta c:

11 1 a

1

11 1 b

1 a 1 b 1 c1

21

1 1 a1 1 c

1 b 1 cc 1 b 1 c

Trang 7

mc a



b

c

1 b 1 c

2

bc 1 b 1 c

1 1 a

2

bc 1 b 1 c

(1)

1 1 b 1 1 c

2

2

ab 1 a 1 b

.c o s8 a 2b 2 c 2 2 2 1 a 1 b 1 c2

ac 1 a 1 c

m1 1 c

Tng t, ta c:

(2)

(3)

T (1) , (2) v (3) nhn v vi v ta c:

1 1 a

1 1 b

1 a 1 b 1 cSuy ra:

M

abc

1 8

Du = xy ra khi v ch khi

11 8

1

m1

aa b c 1 2(tha iu kin ban u)

1 a 1 b 1 cVy M max Cch khc: ti a b c

ie t

1 2

T gi thit ta c:

1 b 1 c

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1 a 1 c

t8

1

1 a 1 b

h

abc 1 a 1 b 1 c

21 a 1 b 1 c

2 a b c

3 ab bc ac 2 1 a 1 b 1 c(1)


w

1 2abc ab bc ac

w

2abc ab bc ac

4 4 2a 3b 3c 3

(2)

T (1) v (2) ta c: 1 4 4 2a3b3c3

1 8abc hay M

abc

1 8

Du = xy ra khi v ch khi 2abc ab bc ac

a b c

1 2

Trang 8


Vy M max =

1 8

ti a b c

1 2

Bi ton tng qut:i

Tm gi tr ln nht ca biu thc M Lp lun nh trn ta c M max

.c oa2

a1.a2 ....an

m... an 1 n 1(1) (2) (3) (4) (5) (6)

Cho a1 , a2 ,..., an

0 tha mn :

n

1 1 1 ai

n 1

2 n ti a1

Gii: p dng bt thc Csi ta c:4

a1 x 2

1 x2

4

1 x .4 1 x

4

1 x 1 x

4

1 x .1 1 x .1

1 x 1 2 1 x 1 2

4

4

T (1), (2), (3) cng v theo v ta c:

Nhn thy (4) xy ra khi v ch khi (1), (2) v (3) ng thi xy ra khi v ch khi x 0 .

Li p dng bt ng thc Csi, ta c:

w .v

f ( x) 1

1 x

ie t1 x

m1 x 2

w

1 x

1 x .1

1

w

1 x

1 x .1

1

1 x 2

T (5), (6) a n:

1 x

1 x 2

t1 xx D

h1

xc nh trn D

x

R : 1 x 1 . Tm gi tr ln nht ca f ( x) trn D.

s

Bi 3: Cho hm s f ( x)

4

1 x2

4

1 x

4

1 x

1 x

1 x 3

(7)

Du = (7) xy ra khi v ch khi (5) v (6) ng thi xy ra khi v ch khi x 0 . T (4) v (7) suy ra f ( x) 3

x D.

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Ta li c f (0) 3, v 0 D . Do : max f ( x) = 3.

Bi 4: Tm gi tr nh nht ca hm s thc sau:

f ( x)

vi

0 x 1

Ta c: f ( x)

1 1 x 1 x

1 x x x 1 x

1

1 x 1 x

1 x x 2 x 1 xp dng bt ng thc Csi ta c:

f ( x)

Bi 5: Cho ba s thc dng a, b, c .

m

Vy min f ( x) 4 ti x

1 2

a

Du = xy ra khi v ch khi

1 x x

tx 1 x a

1 x x 1 x x 2 2 . x 1 x x 1 x

h

s2 4x 1 2 b c a c a b b c x y z 2

Tm gi tr nh nht ca biu thc P Gii: t:

x b c,

ie t

y c a,

z

a b

a b cV

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1

x y z

2

a

y z x , 2

b

z x y , 2

.c ox

Gii:

m1 1 x x x

1 1 x 1 x

c

(*)

w

T ta c:

P

y z x 2x

z x y 2y

x y z 2z

1 y z 2 x

z x y

x y z

3

w

1 2

y x

x y

z x

x z

z y

y z

3

Trang 10


1 2 2 2 3 2

3 2

( Bt ng thc Csi)

Du = xy ra

T (*) ta c a b c Vy Pmin

Bi 6: Cho ba s thc dng a, b, c tha: a b c 1 . Tm gi tr ln nht ca biu thc S abc a b b c c a Gii:

p dng BT Csi cho ba s dng, ta c:

a b c 33 abcV

m

1 3 3 abc

a

t(1)

ie t

a b

b c

c a

33 a b b c c a(2)

2 33 a b b c c aT (1) v (2) nhn v vi v ta c:

w .v

2 9 3 abc a b b c c a8 93 S

S

8 729 1 3

Du = xy ra khi v ch khi a b c

w

Vy

Smax

8 729

w

Bi 7: Tm gi tr ln nht ca hm s f ( x ) trn min D

x

R: 1 x

1 . 2

Trang 11

h93 Sx 2

s

3 vi mi s thc dng a, b, c tha a b c . 2

1 x 2 x2

.c o

y x z x z y

x y x z y z

x

y

z

m


Gii: Nhn thy D l min xc nh ca f ( x) . p dng bt ng thc Csi ta c:

1 x 2x2 x 1 2

1. 1 x 2 x 2 1 x 2x2 2

m

1

1 x 2 x2 2

x D

Do :

f ( x)

f ( x) 1 x 2T suy ra:

f ( x) 1

x D

Mt khc du = xy ra th

1 1 x 2x2 1 x2 1 xTa li c: f (0) 1 Vy max f ( x) 1x D

t1 1 22 1 . x0

hx 0 D1 1 x2

Bi 8: Cho hm s f ( x)

Tm gi tr nh nht ca f ( x) vi x Gii:

w .v

Ta c: f ( x )

1 x

ie t

1 x

2

2

1 x2

m1 x2

2 1 x

p dng bt ng thc Csi ta c:2

w

w

f ( x)

1 2 x .2 x

16

Du = xy ra

x 1 > 0.

Vy min f ( x ) 16 ti x 1x 0

Bi 9: Cho ba s thc dng a, b, c . Tm gi tr nh nht ca biu thc sau: Trang 12

a

1 x

s

.c o


A

abc 1

1 aGii:

1 b

1 c

a b

b c

c a

a b c

A

ab


ab

a b

2a ,

bc

b c

2b , 1 aa 1 c 1 a 6

ac 1 b 1 c

c a

T suy ra: A

2a 2b 2c1 a 2 b.a b

A

a b c 1 a

a

A 2 a.

1 bb

2 c.

Vy MinA = 6 ti

a

c 1

Bi ton tng qut:Cho P

ie t

a1.a2 ...an 1

1 a1

m

Du = xy ra

c 1

1 a2

...

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a1 a2 .a3 ...an

a2 a1.a3 ...an

...

t1 anc3 b3

1 b

1 c

hb

s1 b1

a b cc 1 c

(BT Csi)

an a1.a2 ...an

vi ai 0

i 1, n Th MinP = 2n ti a1

a2

... an 11 ab 2 1 bc 2 1 ca 2 c3 a3 b3 0, b 0, c 0 v abc 1

Tm gi tr nh nht ca P vi a

w

w

Bi 10: Cho biu thc sau: P

a 3 b3 c 3

Gii:

Ta c: P

3

a3 b3

a3 c3

b3 c3

b3 a3

c3 a3

Trang 13

.c o2c

a b

bc

b c

ac

c a

1 a

1 b

1 c

a b c

a1 a2 ... an

m

Ta vit biu thc A li di dng sau:


a 4b 2 c3

ab 5 c3

b 4c 2 a3

bc 5 a3

a 5c b3

a 2c 4 b3

ab 2 bc 2 ca 2

(1)


m6

a3 b3

a3 c3

b3 c3

b3 a3

c3 a3

c3 b3

66

a3 a3 b3 b3 c3 c3 . . . . . b3 c3 c3 a 3 a 3 b3

(2)

66

a 4b 2 c3

ab5 c3

b 4c 2 a3

bc5 a3

a 5c b3

a 2c 4 b33abc 3

.c o6n

a 4b 2 c3

ab 5 c3

b 4c 2 a3

bc 5 a3

a 5c b3

a 2c 4 b3

a 4b 2 ab 5 b 4c 2 bc 5 a 5c a 2c 4 . . . . . c3 c3 a3 a3 b3 b3

6abc

(3) (4)

ab 2 bc 2 ca 2

3 3 ab 2 .bc 2 .ca 2

T (1), (2), (3) v (4) ta c:P 3 6 6 3 18

Du = xy ra

a

b

c 1

Vy Pmin = 18 ti a b c 1

Bi 11: Cho n s dng x1 , x2 , x3 ,..., xn

m

an 2 tha mn x1

tx2 ... xn

h

s

1

Gii: t a

ie t

a a Tm gi tr ln nht ca biu thc S x1a1 .x2 2 ...xn n , Trong : a1 , a2 , a3 ,..., an l n s dng cho trc.

w .v

a1 a2 ... an ,b1 b2

bibn

ai a

i 1, 2, .., n th bi

0

V b1 b2 ... bnx1 a1 x . 2 a2

1 . p dng bt ng thc Csi m rng ta c:x ... n an b1 x1 a11 aa a a1a1 .a2 2 ...an n

b b2 x2 ... n xn a2 anx1 x 2 .. x 1 a

wS

a a x1a1 .x2 2 ...xn n

Du = xy ra

1 aa x1 a1 1 a

w

xn x1 x2 ... xn x1 x2 x2 ... a2 an a1 a2 ... an a1 a2 xn ai x1 x2 ... xi i 1, 2,..., n a1 a2 an a

...

xn an

Vy Smax

1 a1 a2 an a1 .a2 ...an aa

Trang 14


P

Gii: p dng bt ng thc Bunhiacopski ta c:2

3 4 a b c

h

1. 4a 3 1. 4b 3 1. 4c 3

1 1 1 4a 3 ab 3 ac 39

tP 4a 3 4b 3 4c 3Du = xy ra

3 4.3 93 7

m

4a 3 4b 3 1 1 a b c 1

a

s4c 3 12

.c o63

Bi 1: Cho a, b, c

3 v a b c 3 . Tm gi tr ln nht ca biu thc sau: 4 4a 3 4b 3 4c 3 .

a

Vy MinP = 3 7 ti a

w

Bi 2: Cho cc hng s dng a, b, c v cc s dng x, y , z thay i sao cho a b c 1 . Tm gi tr nh nht ca biu thc A x y z . x y z Gii: a b c Ta c: a b c x y z x y z p dng bt ng thc Bunhiacopski ta c:

w .v

ie t

a , b, c

3 4

b

c 1.

2

w

a

b

c2

a x xx y z

b y y

c z z

a x

mb

2.2.2. S dng bt ng thc Bunhiacopski Lu : p dng c bt ng thc Bunhiacopski th hm s hoc biu thc hoc cc biu thc gi thit phi c dng tch ca hai biu thc hoc tng ca cc biu thc m chng l tch ca hai tha s. V sau khi p dng bt ng thc Bunhiacopski th phi c phn a v biu thc gi thit ban u v a c v hng s.

c 1

b y

c z

x

y z

a

b

c

Du = xy ra

a x x

b y y

c z z

a x

b y

c z

(1)

Trang 15


Mt khc:

a x

b y

c z

1a b a a b b c

(2)

T (1) v (2) suy ra: x

z2

c

a

b

c

Vy maxA =

a

b

c

Bi 3: Tm gi tr nh nht ca hm s f ( x, y , z ) x 4 y 4 z 4 , trn min D x, y , z : x , y , z 0 v xy yz zx 1 Gii:

p dng bt ng thc Bunhiacopski ta d c:

t

1.x 2 1. y 2 1.z 2 xyV xy

2

3 x4 y2z2

y4 z22

z4 x2

h

s(1)

a

Li p dng bt ng thc Bunhiacopski ta c:

.c ox2 y2 z22

yz

zx

2

x2x2 y2

y2

z2

m

yz

zx 1 nn:

m(2)yz zx 1

y

c

1

ie ty z x

T (1) v (2) ta c: 3 x 4

y4

z4

1

w .vy z1 3

1 3 Du = xy ra khi v ch khi (1) v (2) ng thi xy ra x2 y 2 z 2 x y z kt hp vi iu kin xy f ( x, y , z )

Ta c: x

3 3

Vy Max f ( x, y , z )( x, y,z ) D

w

Bi 4: Cho cc s dng a, b, c tha a 2 b 2 c 2 a3 b3 c3 thc P a 2b 3c b 2c 3a c 2a 3b Gii: Ta c: Trang 16

w

1 . Tm gi tr nh nht ca biu


P

a2

a4 2ab 3ac

b2

b4 2bc 3ba

c2

c4 2ca 3cb

(1)

p dng bt ng thc Bunhiacopski cho hai d y s sau:

a2 a2 2ab 3ac2

,2 2

b2 b2 2bc 3ba

,

c2 c2 2ca 3cb

ta c:

a

b

2

c

a2 . a2

a4 2ab 3ac

b2

b4 2bc 3ba

.c oc21

2ab 3ac b 2

2bc 3ba c 2

M a 2 b 2

c2P

1 , t (2) suy ra1 1 5 ab bc ca

t

h

a2 b2

c2

2

P a 2 b 2 c 2 5 ab bc ca

s

mc4 . 2ca 3cb 2ca 3cb(2) (3)

a2

2 ab 3ac , b 2

2bc 3ba , c 2

2ca 3cb v

Mt khc theo bt ng thc C si ta c:

a2 b2 b c2 2

2ab 2bc 2ca

a

2

ie t1

c

2

m3 3

a

ab bc ca

a2 b2 c2

T (3) ta c: P

1 5 ab bc caa b c

1 1 5.1

1 6

Du = xy ra Vy MinP =

1 6

w

Bi 5: Cho hai s dng a, b tha 0 a2 b2 1 a b thc M 1 a 1 b a b Gii:

w

w .v

a 1,0 b 1 . Tm gi tr nh nht ca biu

Ta c:

Trang 17


M

p dng bt ng thc Bunhiacopski ta c:1 1 a 1 a 1 1 b 1 b 1 1 1 a 1 b 1 a b a b 1 a b 1 a2

1 b

1 a 1 b

a b1 1 a 1 1 b

ta b1

1 a b 1 a b

w .v

5 2 Bi ton tng qut: 2 a12 a2 Cho P 1 a1 1 a2 2n 1 Th minP nVy minM =

...

ie t

2 an 1 an

m

a

Du = xy ra

h1 3vi 0 ai

1

1

1

9 2

M

9 2 2

5 2

s1 i 1, n

a1 a2 ... an

Bi 6: Cho hm s thc f ( x)

x 2007

2009 x 2 . Tm gi tr ln nht v nh

nht ca f ( x ) trn min xc nh ca n. Gii:

Ta c min xc nh ca f ( x) : D Mt khc: f ( x)

w

2009; 2009

x 2007

2009 x 2

w

V f ( x) 0,x D

x Dx D

0; 2009x D

Do : max f ( x) Vi x D , ta c:

max f ( x) v min f ( x)2007. 2007 1. 2009 x 2

f ( x)

x

Theo bt ng thc Bunhiacopski th : Trang 18

.c oa b

f ( x)

max f ( x)x D

mf ( x) l hm l

a2 b2 1 1 a 1 b 2 1 a 1 b a b a2 1 a2 b2 1 b2 1 2 1 a 1 b a b 1 1 1 2 1 a 1 b a b


2007. 2007 1. 2009

x2

2008 2007 2009 2008 4016 x 2

x2

x D

x D

Bi 7: Cho x, y, z thc T

0 tha mny2

xy

yz

t

x

y

y

z

z

x

Gii:

ie t

p dng bt ng thc Bunhiacopski ta c: 1 x y y z z x x y z y

m

a

x

2

z

2

hzx 1 . Tm gi tr nh nht ca biu zz z x

min f ( x)

2008 2008 ti x

2008

sxz z x

Vy max f ( x)

2008 2008 ti x

2008

.c ox y2T x

p dng bt ng thc Cosi ta c: x 2 4016 x 2 f ( x) 2008. 2008.2008 2 2007 1 2009 x 2 Du = xy ra x 2007 2 2 x 4016 x

2008

mz2

Suy ra: f ( x)

x 2008 4016 x 2

2008. x 2 4016

x2

x

y

z x2 x

2

x x y2

y

x

y

y y z y y z

y

z y

x z

w .vy y zy z x

z2

T

1 x 2

z x 1 2y z

x

Du = xy ra

w

1 3

Vy minT =

1 ti x 2

y

z

1 . 3

Bi 8: Cho ba s dng a, b, c tha mn: a b c 1 . Tm gi tr nh nht ca biu 1 1 1 1 thc P 2 2 2 a b c ab bc ca Gii:

w

Trang 19


p dng bt ng thc Bunhiacopski ta c:

100

1 a2 b22

c2 1 ab

a

2

b

2

c

2

1 3 ab ab

1 3 bc bc

1 3 ca ca

2

a

1 b2

c

2 2

1 bc

1 ca

a2 b2

c 2 9ab 9bc 9ca

P a b c

7 ab bc ca

P 1 7 ab bc ca

a 2 b2 c2Do :

ab bc ca (suy ra t bt ng thc Cosi)2

P 30Du = xy ra Vy minP = 30 ti a

a

b

c

Bi ton tng qut: Cho n s dng a1 , a2 ,..., an n

ie t

b

c

1 3

w .v

t P =

1

a1 a2 ... a n

1 a1a2

m

1 3

a

100 P 1

7 a b c 3

10 P 3

2 v a1 a2 ... an 1 .1 a 2 a3 ... 1 an 1an 1 ana1

Th min P

n n

3

n

2

2

ta2 ... an

h

M ta li c: 1 2 a b c ab bc ca 3 Tht vy, t trn ta c: 2 a b c 3 ab bc ca

s1 n2

2

khi a1

2.3. S dng bt ng thc vect Lu : s dng bt ng thc vect th biu thc gi thit hoc biu thc cn tm gi tr ln nht, nh nht c dn g tng bnh phng ca cc s hng hoc cn bc hai ca tng bnh phng hoc l tng ca cc tch ca cc tha s . Bi 1: Cho hai s thc x, y tha mn 2 x 3 y 1 . Tm gi tr nh nht ca tng S 3x 2 2 y 2 Gii: Ta c S 3 x 2

w

w

2 y2

3x

2

2y

Trang 20

.c o

m


Trong mt phng ta Oxy chn 2 3 4 9 35 u , u 3 2 6 3 2

v u.v

3 x, 2 y

v

3x 2

2 y2 35 . 3x 2 6 2 y2 3x 2 2 y2

2 x 3 y 1 u .v2 3x

Du = xy ra

4 y 9x

Kt hp vi iu kin ban u ta c: x Vy minS =

4 ,y 35

h

sx2 y2

6 ti x 35y

4 ,y 35

9 35

.c o9 35z2 a 1 b b 1 a2 a2 b2 2

3 2y

Bi 2: Cho x

z 1 . Tm gi tr nh nht ca biu thc P

Gii:

Trong mt phng ta Oxy chn: u x, y , z u x2 y2 z 2

m

vTa c: u.v

z , x, yu .v2 x2 3 x2 3 x2 x2

vxz xy

x2

a

y2

y2 y2 y2 y2

ie t

yz

x2z2 z2

t

z2y2 z2

z2

2 xz 2 xy 2 yz y2 y 2 xz 2 xy 2 yz 1 x

z2 z2

x2

w

1 3 x y z 1 Du = xy ra x y z z x y 3 1 1 Vy minP = khi x y z 3 3 2 2 Bi 3: Cho a b 1. Tm gi tr nh nht ca biu thc A z2

w

Trong mt phng ta Oxy chn: u a, b u a 2 b2 1

v

1 b, 1 a

w .v

Gii:

v

a b 2

Theo bt ng thc Bunhiacopski ta c: 1.a 1.b Trang 21

m6 35


Do : v

2

2 u .v x y 2 2 2

A u.v

a 1 b b 1 a

Gii: Trong mt phng ta Oxy ta chn: 1 1 u x, u x2 x x2

v

y,

1 y

v

y2

1 y2

wu

z,

ww x

ie t

1 x

z2

1 z2

v

y

z,

1 x

p dng bt ng thc u

w .v

m

1 y

a

1 z

t

vz2

w1 z2

ux

h

Tm gi tr nh nht ca biu thc sau P

x2

1 x2

s

Bi 4: Cho ba s dng x, y, z v x

y

z 1.

.c oy2 1 y2 z2 1 z2

a b 1 b 1 a Kt hp vi iu kin ban u a 2 b 2 2 Suy ra: a b 2 2 Vy A max 2 2 khi a b 2Du = xy ra

1

vy

w ta c:z2

m1 x 1 y 1 z2

x

2

1 x2

y

2

1 y2

(1)

Nhn thy: x

y

z

2

w

1 x

1 y

1 z

2

81 x

y 1 x

z 1 y 1 y

2

80 x 1 z 1 z2

y

z

2

(2)

w

p dng bt ng thc Cosi ta c:

81 x

y

z

2

1 x

1 y

1 z

2

2 .9 x

y

z

1 x1 xyz

2.9.33 xyz .33T (2) v (3) ta c: Trang 22

2.81

(3)


x

y

z

2

1 x

1 y

1 z

2

2.81 80

82

V do (1) nn:

P

x2

1 x2

y2

1 y2y

z2z

Du = xy ra khi v ch khi x Vy Pmin

82 khi x

y

z

1 . 3

Gii: Trong mt phng ta Oxy ta chn:

u v w

4a, ax 4b, by 4c, cz

u v w

16a 2 16b 2 16c 2

ax by

2

u v wTa c: u

4 a b c , ax by cz

ie t

m

2

cz

2

a

t8,6 u v w 10

v

w

u v w

w

c. Khng c vect no bng vect 0 a kb

w .v

P 16a 2 ax 16b 2 by 16c 2 Gi tr nh nht ca P: P min = 10 Du = xy ra khi v ch khi: a. C hai trong ba vect bng vect 0 b. C mt trong ba vect bng vect 0 Gi s u 0 th w k v k 0

2

2

h

P

16a 2

ax

2

16b 2

by

2

16c 2

scz2

Bi 5: Cho a b c

2 v ax by cz

6 . Tm gi tr nh nht ca biu thc

a b b c

ax by by cz

k

0

m 0

ax kby by mcz k, m 0 a b c 2 ax by cz 0

w

Trang 23

.c ocz2

1 z2 1 3

82

10

x y z 3 a b c 2 a , b, c 0

m


Bi 6: Cho cc s dng x, y, z tha xy thc F x 4 y 4 z 4 Gii: Trong khng gian Oxyz chn: u x2 , y2 , z 2 u x4 y4

yz

zx

4 . Tm gi tr b nht ca biu

z4

v

1,1,12

v

3

Ta c: u.v M: u.v

x22

y22

z23 x4 y4 z4 x2 y2

u .v

y2 z2

z2

2 yz 2 xy yz

t

x 2 2 zx 2 x2 y2 z 2

hx2 y2 z 2 xy 16 3

s

Mt khc ta c: x 2 y 2 2 xy

.c oz22

mzx yz zx = 4

T ta c: 3 x 4

y4

z4

4 2 16

a2 1344a 82

x4

y4

z4

Vy: minF = 16 khi x

y

ie tu v w a2 5,5

Bi 7: Tm gi tr nh nht ca biu thc A a 2 4a 8 a 2 2ab b 2 Gii:

mz

b 2 6b 10

Trong mt phng ta Oxy chn:

u v

w .va 2,2 a b, 2

a b

4

w

b 3,1

b 2 6b 10 u v w 5 2

w

Ta c: u v w

w

u v w

u

v

w5 2

4a 8 a 2 2ab b 2 4 b 2 6b 10 a 2 2 b 3 Du = xy ra a 0, b 2 a 2 1 a bTrang 24

a2


Vy A min

5 2 ti a

0, b

2

Bi 8: Cho a R . Tm gi tr nh nht ca biu thc M a 2 4a 13 a 2 2a 5 Gii:

Ta c: M a 2 9 a 1 Trong mt phng t Oxy chn:

2

2

4

u v

a 2,3 a 1,2 u v 3,5 v

u

a 2 a 12

2

9

4

u v

34

Bi 9: Cho ba s dng a, b, c tha: ab bc ca biu thc B

m

Vy: M min

34 khi a

1 5

a

Du = xy ra khi v ch khi a

1 5

tca

M: u

v

u v

a 2

2

9

ha 12

s4 34

abc . Tm gi tr nh nht ca

b2 ab

2a 2

ie t

c2

2b 2

a2

2c 2

bc

Gii: Ta c: B

1 2 1 2 1 2 2 2 2 a b b c c2 Trong mt phng ta Oxy chn: 1 2 1 2 u , u 2 a b a b2

w .v

2 a2

w

v

1 2 , b c 1 2 , c a

v w

1 b2 1 c2

2 c2 2 a2

w

w

1 1 1 1 , 2 c a b c 1 1 1 Mt khc: ab bc ca abc 1 a b cV u v

w

1 a

1 b

Trang 25

.c o

m

Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh Do : u M: u

v w v w

1, 2 u v w

u v w

3

2.4. BI TP NGH

h1 x

Bi 1: Tm GTNN ca biu thc sau: a3 b3 c3 M 1 b 1 c 1 a 1 c 1 a 1 b Vi b 0, b 0, c 0 v abc 1 Bi 2: Tm GTLN ca hm s f ( x, y , z ) trn min D

s2 1 y 2 1 z

2

x, y , z : x

0, y

0, z

t2 x2

0 v x

w .v

M 5 x 2 2 xy 5y2 Bi 6: Tm gi tr nh nht ca hm s 2 2 x3 x12 x2 f ( x1 , x2 , x3 , x4 , x5 ) x2 x3 x4 x3 x4 x5 x4 x5 2 2 x5 x4 x5 x3 x2 x1 x2 x3Trn min D Bi 7: Cho x, y

ie t

Bai 3: Tm gi tr ln nht ca biu thc A ab bc 2ca vi a, b, c l cc s thc tha a 2 b 2 c 2 1 Bi 4: Tm gi tr ln nht ca biu thc P abc Trong a, b, c l cc s thc tha a 2 2b 2 2a 2c 2 b 2c 2 3a 2b 2c 2 9 Bi 5: Cho x 2 y 2 1 . Tm gi tr ln nht ca biu thc sau:

m

a

x1 , x2 , x3 , x4 , x5 : x12x2 9y2 4x2

2 x3

R . Tm gi tr nh nht ca biu thc sau:

w

Bi 8: Cho bit x 2 y 2 z 2 27 . Hy tm gi tr nh nht v ln nht ca hm s f ( x, y, z ) x y z xy yz zx Bi 9: Tm gi tr nh nht ca biu thc P a100 10a10 10 Bi 10: Cho x, y , z tha mn h sau:

w

A

9x

2

y2

y2

x2

x2

xy

y2

y 2 yz z 2 16 Tm gi tr ln nht ca biu thc sau: P xy yz

Trang 26

.c oy z 1

1 2 1 2 1 2 3 2 2 2 2 a b b c c2 a2 Du = xy ra khi v ch khi a b c 3 Vy B min 3 khi a b c 3 B

2 x4

2 x5 1

4y2

3zx

mx1


Bi 11: Cho

a , b, c 0 . a b c 12 2 2

P 1 x 1 y 1 x 1 y Bi 14: Cho ba1 s thc a, b, c bt k. Tm gi tr nh nht ca biu thc2 2 2

p dng bt ng thc vect: u

v

u v

w

w

w .v

ie t

Bi 15: Cho x > 0, y > 0, z > 0 v x + y + z = 1. x y z Tm gi tr ln nht ca biu thc M = x 1 y 1 z 1 Hng dn: p dng bt ng thc Bunhiacopski cho hai d y: 1 1 1 v 1 x , 1 y , 1 z , , 1 x 1 y 1 z

m

Trang 27

a

t

h

P= b 1 c a b 1 c a Hng dn: Trong mt phng Oxy chn u b 1, c a , v b 1, a c

s

.c o2

F Bi 13: Cho a, b

a b b c b c 0,1 . Tm gi tr ln nht ca biu thc:

m

1 1 1 Tm gi tr nh nht ca biu thc P a b c a b c Bi 12: Cho a b c 1 v a, b, c 0 . Tm gi tr ln nht ca biu thc


Phn 3: GII PHNG TRNH BNG PHNG PHP S DNG BT NG THC

3.1. Vn dng bt ng thc Csi

w .v

p dng bt ng thc Csi cho hai s khng m, ta c: 1 1 1 (1) 2 x8 x4 2x8 2 2x8 . 8 8 8 V

x4

1 4

ie t

Lu : p dng c bt ng thc Csi gii th: mt trong hai v ca phng trnh sau khi p dng bt ng thc Csi phi ln h n hoc bng (nh hn hoc bng) v cn li, hoc sau khi p dng bt ng thc th c mt ng thc c lng c nh hn (ln hn) hoc bng v cn li p dng c iu kin xy ra ca bt ng thc Csi . 3 Bi 1: Gii phng trnh: x 2 2 x 8 8 Gii:

2 x4.

1 4

m

x4

a

1 4

tx2

hx2x2 2

Ni v phng trnh th c rt nhiu loi phng trnh nh phng rnh bc hai, bc ba,phng trnh v t, phng trnh m, phng trnh logarit.Mi phng trnh c th c nhiu phng php gii khc nhau mu mc hay khng mu mc. Trong s cc phng php gii ca cc phng trnh th phng php s dng bt ng thc c th coi l phng php c o v sng to i hi ngi gii ton phi linh hot. S dng phng php ny ta c th s dng nhiu bt ng thc khc nhau, c th vn dng ring l hoc kt hp nhiu bt ng thc. Sau y l mt s bi ton gii phng trnh bng phng php vn dng bt ng thc m bt ng thc c s dng ch yu l bt ng thc Csi, Bunhiacopski v bt ng thc vect.

s2 x82 2

T (1) v (2) ta c: 2 x 8

x4

3 82x8 x4

x41 8 1 4

Ta c du = xy ra, do

w

Vy nghim ca phng trnh l x

w

Bi 2: Gii phng trnh: Gii:

x 3

5

x

x2

8 x 18

Trang 28

.c o3 8 x2

m(2)


iu kin: 3 x 5 p dng bt ng thc Csi cho hai s khng m, ta c: x 3 5 x x 3 .1 5 x .1

x 42

2

22

x 4 0 x 4 Vy phng trnh cho c nghim l xBi 3: Gii cc phng trnh sau: a. x 2 4 x 5 2 2 x 3 x 1 3 b. 1 x 1 3 Gii:

2

4

a. x 2

4x 5

2 2x 3x

ie t

3 2 p dng bt ng thc Csi cho hai s khng m: 2 x 3 v 1, ta c: 2x 3 1 2 2x 3 x2 4x 5

2x 42

x2

4x 5

m

iu kin: 2 x 3 0

Du ng thc xy ra trong (1) khi v ch khi:

w

w

w .v

x 1 0 x 1 Th li x 1 l nghim ca phng trnh cho. Vy phng trnh cho c nghim duy nht l x x 1 3 (k: x b. 1) 1 x 1 3 p dng bt ng thc Csi , ta c: x 1 x 1 3 3 (1) 2 . 2 x 1 3 x 1 3

a

x2

t

2x 1 0

h1

sx x

.c o04

Do :

x 3

5

x

x

2

8 x 18

x 4

m22 22 4

x 3 1 5 x 1 2 2 2 Mt khc: x 2 8 x 18 x 2 8 x 16 2

x 1 x2 2x 1 9 x2 2x 8 x 1 3 Th li x 2 v x 4 l nghim ca phng trnh 3Vy phng trnh cho c nghim l x2 v x

Trang 29


Bi 4: Gii phng trnh sau: 8 x 2

1 x

5 2

1 1 4 x

1 1 4 x

Gii:

w .v

0 1 x 0 x iu kin: x 1 1 1 0 x x 0 p dng bt ng thc Csi cho hai s khng m, ta c: 1 1 1 1 x 1 x .1 x 1. x x x x

x

ie t

Du = xy ra, ta c:

w

1 1 1 1 x 1 x 1 2 x 2 x 1 x 1 x x2 1 x 1 x 1 x 2 x x 1 0

m

a

Bi 5: Gii phng trnh sau: x

x

1 x

t1 1 x

h1 2

1 1 1 1 1 23 1 5 . . . . 5.5 8 x 2 . 2 4 4 x x x x 2 x 2 Du ng thc xy ra khi v ch khi: x 0 x 0 x 0 x 1 1 1 1 32 2.x 4 x5 8x 2 5 x 4 4 x Th li: x 4 tha mn Vy nghim ca phng trnh l x 4 55 8 x 2 .

s5

w

x

1 2

5

Kt hp vi iu kin ban u ta c: x Trang 30

.c o1 4

mx

Gii: iu kin: x 0 p dng bt ng thc Csi ta c: 1 1 1 1 1 8x 2 8x 2 x 4 x 4 x


Vy nghim ca phng trnh l x Bi 6: Gii phng trnh sau: Gii: iu kin:

1 2

5

x2

4x 9

x2

4x 9

6

x2 2

4x 9 x2 9

x2 4xx4 2

4x 92

2 x4

x2 2x2

.c o4 x 9. x 2 2 811 3 3 2 3

x 2 x2 4x 9 0 x 2 5 0 p dng bt ng thc Csi cho hai s d ng, ta c:

x2

4x 9

0

x 2

2

5 0

R

9 x2iu kin: 3 x

0 0 0

m

Gii:

3 x

p dng bt ng thc Csi cho hai s khng m, ta c:

ie t

3

x

a

Bi 7: Gii phng trnh sau:

9 x2

3 x

t

h

Du = xy ra, do :

x 0 2x 0 Vy nghim ca phng trnh cho l: x

0

s0

2

3 x

3 2 3

3

3 x 3 x 2Do vy:

w .v

9 x2

3 x

3 x

3 x 3 x 1 3 x 3 . 2 3

3 x .3

m4x 9 2.3 6 811 3 3 x .3

1 3 x 3 . 2 3

3 x3 x 3 2 3x 0

3 x 3 3x 0

w

9 x

2

3 x

3 x 3 x

Vy phng trnh cho c nghim l:

w

Bi 8: Gii phng trnh sau: Gii:0

3

25 x 2 x 2

9

4x

3 x

iu kin: x

Trang 31


Ta c:

3

25 x 2 x 23

9

4x9 9

25 x 4 2 x 2

3 x 4x25x 2

3 5x 2 2x2 9

3.3 25 x 4 2 x 2

p dng bt ng thc Csi cho ba s d ng: 5 x 2 ; 5 x 2 ; 2 x 2

m0

9 c:

Bi 9: Gii phng trnh 2 7 x 3 11x 2 Gii: Ta c: 2 7 x 3 11x 2

25 x 12

25 x 122

x2x

2 7x 4 x

x 3

a2

iu kin: 7 x 4 x 27x 4

x 3

m

0

t

6x 16x 1

h2

sx2x 1 22

0 (v x

x 3

4 7 p dng bt ng thc Csi cho hai s khng m: 7 x 4; x 2 x

ie t

.c o6x 12

(*) 5 x 2 5 x 2 2 x 2 9 3.3 25 x 4 2 x 2 9 Du = ng thc (*) xy ra khi v ch khi: 5x 2 2x 2 9 3x 2 9 x2 3 x 3 Th li: x 3 l nghim ca phng trnh cho 3 Vy phng trnh cho c nghim l x

11 4

x 3 c:

w .v

7x 4

x

2

x 3

2 7x 4 x

x 3

x2

6 x 1 2 7 x 3 11x 2

25 x 12

3.2. Vn dng bt ng thc BunhiacopskiLu : p dng c bt ng thc Bunhiacopski t h phng trnh phi c dng tch ca hai biu thc hoc tng ca cc biu thc m chng l tch ca hai tha s. V sau khi p dng bt ng thc Bunhiacopski th phi c phn a v

w

Du = xy ra khi v ch khi: 7 x 4 x 2 x 3 x 2 8x 7 0 x 1 (tha iu kin) x 7 Th li: x 1; x 7 l nghim Vy phng trnh cho c nghim l x 1; x 7

w

Trang 32


biu thc gi thit ban u v a c v hng s. Sau vn dng iu kin bng nhau ca bt ng thc Bunhiacopski a ra nghim ca phng trnh. Bi 1: Gii phng trnh: Gii: iu kin: x 0 p dng bt ng thc Bunhiacopski cho hai cp s: x 1 c: 2 2 ; x 1 v ; x 1 x 1

ie t

(1) x 9 x 1 x 1 Du = trong (1) xy ra khi v ch khi: 1 2 2 2 2 1 x 1 2 2 1 x 1 . x 1 x x 1 x 1 x x 1 x x 1 8 1 1 (tha iu kin) 8x x 1 7 x 1 x x 1 x 7 1 Vy phng trnh c nghim l x 7

8

x 1 .

1

x

m

a

w

p dng bt ng thc Bunhiacopski cho cc cp s sau: 2; 3 v x 2 3 x 6; x 2 2 x 7 ta c:

w .v

Bi 2: Gii phng trnh: 13 x 2 3 x 6

2

t

h

2 2 x 1

x

1 2 2. x 1

s2

2

x 1.

.c o2

x

x 1

mx2 2x 7 5 x 2 12 x 332

2 2 x 1

x

x 9

Gii:

2 2 32 x 2 3 x 6

2

x2

2x 7

2 2 2

w

2 x 2 3x 6

3 x2

2x 7

5 x 2 12 x 33

Du = xy ra khi v ch khi: 3 x 2 3x 6 2 x 2 2 x 7

Trang 33


x2 5x 4 0 x 1 x 4 Vy phng trnh cho c nghim l x 1; xBi 3: Gii phng trnh sau trn tp s N: 2 x 2 4 y 2 28 17 x 4 y 4 14 y 2 Gii: p dng bt ng thc Bunhiacopski ta c: 2 2 x 2 4 y 2 28 1. x 2 4 y 2 72 2

4

Bi 4: Gii phng trnh sau: Gii:

x2

m

1 42 x 2 y2 7 17 x 4 y 4 14 y 2 49 Du ng thc xy ra khi v ch khi : 4x2 y 2 7 2x y 2x y 7 V x, y N nn 2 x y 0 2x y 7 x 2 Ta c: 2x y 1 y 3 Vy phng trnh cho c nghim duy nht l x, y 2;3

a

t2x 1

h2

2x

s3x 23x 2

x2

2x

ie t

0

3x 2 4 x 1 0 p dng bt ng thc Bunhiacopski ta c:

w .v

iu kin: 2 x 1 0

x

1 2

x . x 2 1. 2 x 1

x2

12

x 2

.c o4x 1

49

2x 1

m2

w

Du ng thc trong (1) xy ra khi v ch khi: x . x 2 1. 2 x 1 2x2 x x 2

w

x 1 x 2 2x 1

x 1 3x 1

4x 1

(1)

x2

x 1 0

x x

1 2 1 2

5 5

Trang 34


Kt hp iu kin ban u ta c nghim l x

1 25

5

Vy nghim ca phng trnh cho l x Bi 5: Gii phng trnh Gii: iu kin: 5 x 2 4 x 0 1 x 5 p dng bt ng thc Bunhiacopski ta c: 1 2 x 2 1. 5 x 2 4 x

1 2

5 x2

4x

2x 3 5

4

Bi 6: Gii phng trnh sau: 13 x 1 9 x 1 16 x Gii:

iu kin: x 1 p dng bt ng thc Bunhiacopski ta c: 13 x 1 9 x 1 13. 13 x 13 27 . 3 x 3

13 27 13 x 13 3 x 3 40 16 x 10 2 10 16 x 10 10 16 x 10 16 x (Bt ng thc Csi) ng thc xy ra khi v ch khi: 27 . 13 x 13 13. 3 x 3 5 (tha iu kin) x 4 10 16 x 10

w

w .v

ie t

m

a

5 x2 4x 2x 3 5 4 Du ng thc xy ra trong (2) 2 5 x2 x 2 0 6 5 x 2 2 5 5 x 20 x 16 0

t

h5 4

1

4 1. x 2

4x 4

5 x2

s4x4x1 x

Bi 7: Gii phng trnh sau: 4 x Gii:

w

Vy phng trnh cho c nghim l x4

1 x

x

.c o1 3 5 (2)x 224

Trang 35

m8


x 0 0 x 1 1 x 0 p dng bt ng thc Bunhiacopski v Bunhiacopski m rng ta c: 1. x 1. 1 x 1 1 x 1 x 2iu kin:

1.4 x 1.4 1 x4

4

1 1 1 1 1 1 x 1 x

4

Vy nghim ca phng trnh cho l x Bi 8: Gii phng trnh sau: Gii: iu kin:

s

1 2

h

3x 2 1x 1 x

x2

x

x x2 1

.c o1 2 2 7x2 x 4(*)2 5x 2 x

x 41 x x 1 x 2 48 Du = trong ng thc xy ra khi v ch khi: x 1 x 1 (tha iu kin) x 4 4 2 x 1 x

3x 2 1 0 x2 x 0

3

x 2 1 1 3x 2 1 x 23x 2 1 x2 x Du = xy ra khi v ch

m

3 p dng bt ng thc Bunhiacopski m rng ta c: 1. 3 x 2 1 1. x 2 x x x 2 1

ie t

x x2 1

a

x

tx2 1x2

x x x

1 1 2 1 x 1(1)

3x 2 1 x2 1 x

x2

x

w .v

3x

2

1

x 1

Do (*) nn 5 x 2 x 0 p dng bt thc Csi ta c: 1 1 7x2 x 4 5x 2 x 2 x 2 2 2 2 2 2 1 .2 . 5 x 2 x .2 x 2 2 2 2 1 7x2 x 4 5 x 2 x .2 x 2 2 2 2 Du = xy ra khi v ch khi 5x 2 x 2 x 2 2 3x 2 x 4 0

w

w

Trang 36

m8


Bi 1: Gii phng trnh sau: Gii: iu kin: xR

x2

2x 5

x2

s2

Lu : p dng c bt ng thc vect vo vic gii phng trnh i hi phng trnh c cha cn bc hai ca hai tng b nh phng ta phn tch thnh ln vect, hoc cha tng ca hai tch cho thy c s phn tch ca tch v hng ca hai vect. T ta p dng cc bt ng thc vect bit c lng v vn dng iu kin xy ra ca du = t m nghim ca phng trnh.

6 x 10

w u

u v v

2,1 x 12

w 4

5

ie t

x 3

m

Ta vit li phng trnh: x 1 4 x 3 1 5 Trong mt phng ta Oxy chn cc vect c ta sau: u x 1,2 ; v x 3,1

a

2

t

h

.c ou

3.3. Vn dng bt ng thc vect

2

1k v vi k >0

Do (*) nn: u v Nn:

u

v , du = xy ra

Vy nghim ca phng trnh l x Bi 2: Gii phng trnh sau:

w .v

x 1 (iu kin: x 0 ) 2 x 3 x 1 2x 6 x 5 (tha iu kin)5

x2

2x

10

x2

6x

m5(*)

1 4 x 3 T (1) v (2) ta c nghim ca phng trnh l: x Vy nghim ca phng trnh cho l x 1

x

(2)

1

13

41

w

Gii:R2 2

w

iu kin: x

Ta vit li phng trnh: x 1 9 3 x 4 41 Trong mt phng ta Oxy chn cc vect c ta sau:

(*)

u v

x 1,3 3 x, 2

u v

x 1

2

92

3 x

4

Trang 37


u v

4,5

u v

41 u v ux

Kt hp vi (*) nn: u v Do :

kv

Vy nghim ca phng trnh l x Bi 3: Gii phng trnh: 3 x x 1 Gii: 5 iu kin: 1 x 2

5 2x

40 34 x 10 x 2

.c ox3x32003

7 5

m(1) (2)

x 1 3 x

3 2

2x 2

9 3x

7 5

(1)

3 x

x 1

5 2x

3 x

Trong mt phng ta Oxy chn cc vect c ta sau:

u vu.v

3 x,1

u

3 x v

1

3 x

x 12

5 2x

V (2) nn: Do :

u.v

ie tu .vu

u .v

3 x

1. 4 x

mkv

x 1, 5 2 x

4 x

40 34 x 10 x 2(k>0)

w .v

5 3 x 1 (iu kin: x ; x 1) 2 x 1 5 2x 2 x 3 17 x 2 49 x 46 0 x 2 (tha iu kin) Vy nghim ca phng trnh l x = 2Bi 4: Gii phng trnh sau:

x2

ax 2 10 x 267

2

8 x 816

w

Gii: iu kin: x R Trong mt phng ta Oxy chn

w

u v

4 x,20 2

u v u v

x2

8 x 816

5 x,11 2 9,31 2

x 2 10 x 267 81 2.312u v

u v

t

Theo bi ta c: u v

Trang 38

h2

s4 4 x2003


Du = xy ra

4 x 5 x

20 2 11 2

44 11x 100 20 x

31x

56

x

Vy nghim ca phng trnh l: x Bi 5: Gii phng trnh sau: Gii: iu kin: 2 x 4 Trong mt phng ta Oxy chn:

x 2

4 x

x2

v

x 2, 4 x

v

2

u .vx 2

2, u.v4 x

x 2

4 x

a

t1 x 22

h1 4 x2

u

1,1

u

2

2 du = xy ra

Nhn thy: x 2

6 x 11

x 2 Vy nghim ca phng trnh l: xBi 6: Gii phng trnh sau:

ie t

Do du = xy ra

x 3 x 3

m

2

s4 x 3 x 31 2x 1 2x

Gii: 1 1 Diu kin: x 2 2 Trong mt phng ta Oxy chn:

v

w

u

1,1

1 2x , 1 2x1 2x 2

w .v

1 2x

1 2x

1 2x 1 2x

u

2

v

2

w

u.v u .v

1 2x

M: u.v

u .v

1 2x

1 2x

2

Nhn thy:

1 2x 1 2x

1 2x 1 2x

2

1 2x 1 2x . 1 2x 1 2x

Trang 39

.c o6 x 11

2 (BT Csi)

m

56 31 56 31


1 2xDu = xy ra

1 2x 1 2x 1 2x x 0

1 2x 1 2x 1 2x 1 2x Vy nghim ca phng trnh l: x = 0Bi 7: Gii phng trnh sau: Gii: iu kin: x 4 Ta vit li phng trnh di dng sau:x 2 x 4 3

x 4 x 4

u v 2

x 4 1,0 x 4 ,0 1,0v

u v 1 2

x 4 1 x 4

2

u vM: u

u vu v

Theo (*) du = xy ra

m

ie t

4 x 1 2 4 x4 x 1 2

a

t

2

h0R

1

w .v

x 4 1 2 1 2 x 4 1 1 2 1 1 2 12

x 4 x 4

w

Vy nghim ca phng trnh l: x

4

1 2 1

sx 42

x 4 1 + 2 x 4 = 1 (*) Trong mt phng ta Oxy chn:

2

2

.c o2

vi

w

Bi 8: Gii phng trnh sau: Gii:

x 1 x 3

2x 3

2

2x 1

iu kin: x 1 Trong mt phng ta Oxy chn: Trang 40

m1

R


u v u.v u .v

x 1, x 3 1,1 v x 1 2 x 32

u

x 1

x 3

2

2x 3

2x 1

Du = xy ra

x 1 1 x 3

x 3 1

x x 32

3 x 3

x2

7 x 10

0

2)

x 3 4 x 1

ie t

Gii cc phng trnh sau: 3 1) 4 x 2x 8 3) x 4 4) x 2 5)

x 8 6 x 1

m

3.4. BI TP NGH

4

2 x4

4

2 x4

4

2x 44x 9

3 x3

4x

w .v

x2

x2

4x 9

x3 2

6) x 3 3 x 2 8 x 40 2x 3 7) x 1 8)

84 4 x 4 50 3 x 12

x22

2x 2 2x 5

x22

2x 2

2 2 29

w

9) x x 1 10) x

3x

2 xx

2

1

2 x 10

w

Trang 41

a

x 5 Vy nghim ca phng trnh l x = 5

6

t1

hx x

x

3

s2 5

x 1

x 1

.c ox2 6x 9

Ta c: u.v

u .v

m


KT LUNCc dng ton lin quan n bt ng thc thng khng d nn cc dng ton ny thng ch s dng tuyn chn cc hc sinh gii. Ban u, n ch bit di dng chng minh cc bt ng thc tr n c s cc bt ng thc thng dng, nhng sau cc dng ton ra i trn c s cc bt ng thc thng dng bit nh: tm gi tr ln nht, gi tr nh nht, gii ph ng trnh, h phng trnh, bt phng trnh v h bt phng trnh.

Trong ti ny ti ch nghin cu hai dng ton l tm gi tr ln nht, gi

- p dng c cc bt ng thc gii ton i hi k nng nhn xt ca

- Mc d cc dng ton v bt ng thc rt kh, kh nht l a v ng dng bt

tun th cc nguyn tc bin i ng thc nhn xt nhy bn a v dng ca bt ng thc cn ng dng th bi ton s tr nn khng kh. Qua ti ti hc c rt nhiu kinh nghim trong gii ton bt ng thc v thy c mi lin h ca cc bt ng thc vi nhau.

w

w

w .v

ie t

m

ng thc cn vn dng nhng khi ta bit s dng thnh tho cc bt ng thc v

Trang 42

a

hot a v ng dng ca bt ng thc cn p dng.

t

ngi gii phi nhy bn, v k nng bin i tng ng cc biu thc phi linh

h

v bt ng thc vect. Qua qu trnh thc hin ti rt ra c cc iu sau:

s

tr nh nht v gii phng trnh da trn ba bt ng thc l: Csi, Bunhiacopski

.c o

m


TI LIU THAM KHO1) Phan Huy Khi, Chuyn bi dng hc sinh gii Ton THCS: Gi tr ln nht v nh nht ca hm s, NXB Gio Dc, nm 2008.

2) T sch Ton hc & tui tr, Cc b i thi Olympic Ton THPT (1990

3) V Giang Giai, Chuyn Bt ng Thc, NXB HQG H Ni, nm 2002. 4) Nguyn Th Hng, Bt ng thc v bt phng trnh i s, NXB HQG T.P H Ch Minh, nm 2003.

5) H Vn Chng, Tuyn tp 700 bi ton bt ng thc luyn thi v o cc

6) Nguyn c Tun, Nguyn Anh Ho ng, Trn Vn Hnh, Nguyn o n V, Gii phng trnh bt phng trnh h phng trnh h bt phng

7) Trn nh Th, Dng hnh hc gii tch gii phng trnh bt phng trnh h phng trnh bt ng thc....,NXB HQG H Ni, nm 2008. 8) Trn Vn K, Chn lc 394 bi ton bt ng thc gi tr ln nht gi tr nh nht, NXB T.P H Ch Minh, nm 2002.

w

w

w .v

ie t

m

Trang 43

a

trnh bng bt ng thc, NXB HQG T.P H Ch Minh, nm 2006.

t

h

trng H C bi dng hc sinh gii PTTH, NXB Tr, nm 1993.

s

.c o

2000).

m


PH LCCc bi ton tm gi tr ln nht, gi tr nh nht trong cc thi i hc gn y. 1. (Khi A nm 2006) Cho hai s thc x 0, y 0 thay i v tho mn iu kin: 1 1 x y xy x 2 y 2 xy . Tm gi tr ln nht ca biu thc A 3 x y3 2. (Khi B nm 2006) Cho x, y l cc s thc thay i. Tm gi tr nh nht ca biu thc

w

w

w .v

x 1 y 1 z 1 y z 2 yz 2 zx 2 xy 5. (Khi B nm 2008) Cho hai s thc x, y thay i tho mn h thc x 2 y 2 1 . Tm gi tr ln 2 x 2 6 xy nht v gi tr nh nht ca biu thc P 1 2 xy 2 y 2 6. (Khi D nm 2008) Cho x, y l hai s thc khng m thay i. T m gi tr ln nht v gi tr nh x y 1 xy nht ca biu thc P 2 2 1 x 1 yP

x

ie t

m

Trang 44

a

t

x 1 y2 x 1 y2 y 2 A 3. (Khi A nm 2007) Cho x, y, z l cc s thc dng thay i v tho mn iu kin: xyz 1 . Tm x2 y z y2 z x z2 x y gi tr nh nht ca biu thc P = y y 2z z z z 2x x x x 2 y y 4. (Khi B nm 2007) Cho x, y , z l cc s thc dng thay i. Tm gi tr nh nht ca biu thc

2

2

h

s

.c o

m


MC LCPHN M U ................................................................................................................... 1 I. L DO CHN TI ............................................................................................. 2 II. MC CH NGHIN CU...................................................................................... 2 III. I TNG NGHIN CU ..................................................................................... 2 IV. PHM VI NGHIN CU .......................................................................................... 2 VI. PHNG PHP NGHIN C U ......................................................................... 2 PHN NI DUNG ................................................................................................................ 3 Phn 1: S LC V BT NG THC ..................................................................... 4 1.1. nh ngha bt ng thc ................................................................................... 4 1.2. Tnh cht c bn ca bt ng thc ....................................................................... 4 1.3. Mt s bt ng thc c bn................................................................................... 4 Phn 2: TM GI TR LN NHT V GI TR NH NHT ...................................... 6 CA HM S HOC BIU THC ................................................................................ 6 2.1 KIN THC CN NH ......................................................................................... 6 2.1.1. nh ngha........................................................................................................ 6 2.1.2. Tm gi tr nh nht, gi tr ln nht ca biu thc (h m s) bng phng php vn dng bt ng thc ..................................................................................... 6 2.2. BI TP ................................................................................................................. 7 2.2.1. S dng bt ng thc Csi ............................................................................. 7 2.2.2. S dng bt ng thc Bunhi acopski ............................................................ 15 2.3. S dng bt ng thc vect ................................................................................. 20 2.4. BI TP NGH.............................................................................................. 26 Phn 3: GII PHNG TRNH BNG PHNG PHP ........................................... 28 S DNG BT NG THC ...................................................................................... 28 3.1. Vn dng bt ng thc Csi ................................................................................ 28 3.2. Vn dng bt ng thc Bunhiacopski .................................................................32 3.3. Vn dng bt ng thc vect .............................................................................. 37 3.4. BI TP NGH.............................................................................................. 41 KT LUN...................................................................................................................... 42 TI LIU THAM KHO ............................................................................................... 43 PH LC ........................................................................................................................ 44 MC LC ....................................................................................................................... 45

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ie t

m

Trang 45

a

t

h

s

.c o

m

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