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K THUT S DNG BT NG THC C-SI

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1. NHNG QUY TC CHUNG TRONG CHNG MINH BT NG THC S DNG BT NG THC C SI

Quy tc song hnh: hu ht cc BT u c tnh i xng do vic s dng cc chng minh mt cch song hnh, tun t s gip ta hnh dung ra c kt qu nhanh chng v nh hng cch gi nhanh hn.

Quy tc du bng: du bng = trong BT l rt quan trng. N gip ta kim tra tnh ng n ca chng minh. N nh hng cho ta phng php gii, da vo im ri ca BT. Chnh v vy m khi dy cho hc sinh ta rn luyn cho hc sinh c thi quen tm iu kin xy ra du bng mc d trong cc k thi hc sinh c th khng trnh by phn ny. Ta thy c u im ca du bng c bit trong phng php im ri v phng php tch nghch o trong k thut s dng BT C Si.

Quy tc v tnh ng thi ca du bng: khng ch hc sinh m ngay c mt s gio vin khi mi nghin cu v chng minh BT cng thng rt hay mc sai lm ny. p dng lin tip hoc song hnh cc BT nhng khng ch n im ri ca du bng. Mt nguyn tc khi p dng song hnh cc BT l im ri phi c ng thi xy ra, ngha l cc du = phi c cng c tha mn vi cng mt iu kin ca bin.

Quy tc bin: C s ca quy tc bin ny l cc bi ton quy hoch tuyn tnh, cc bi ton ti u, cc bi ton cc tr c iu kin rng buc, gi tr ln nht nh nht ca hm nhiu bin trn mt min ng. Ta bit rng cc gi tr ln nht, nh nht thng xy ra cc v tr bin v cc nh nm trn bin.

Quy tc i xng: cc BT thng c tnh i xng vy th vai tr ca cc bin trong BT l nh nhau do du = thng xy ra ti v tr cc bin bng nhau. Nu bi ton c gn h iu kin i xng th ta c th ch ra du = xy ra khi cc bin bng nhau v mang mt gi tr c th. Chiu ca BT : , cng s gip ta nh hng c cch chng minh: nh gi t TBC sang TBN v ngc li Trn l 5 quy tc s gip ta c nh hng chng minh BT, hc sinh s thc s hiu c cc quy tc trn qua cc v d v bnh lun phn sau. 2. BT NG THC C SI

(CAUCHY) 1. Dng tng qut (n s): x1, x2, x3 ..xn 0 ta c:

Dng 1: 1 2 1 2 ...... ...........n n n

x x x x x xn

Dng 2: 1 2 1 2 ...... ...........n n nx x x n x x x Dng 3: 1 21 2 ...........

...... n

nn x x xx x xn

Du = xy ra khi v ch khi: 1 2 ............ nx x x H qu 1: Nu: 1 2 ........ nx x x S const th: 1 2P ............

n

nSMaxn

x x x khi 1 2 ............ n

Sn

x x x H qu 2: Nu: 1 2................. nx x x P const th: 1 2 2......... nMin S n Px x x

khi 1 2 ............n

nx x x P 2. Dng c th ( 2 s, 3 s ):

n = 2: x, y 0 khi :

n = 3: x, y, z 0 khi :

2.1 2

x y xy 33

x y z xyz 2.2 2x y xy 33 x y z xyz



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2.3 2

2x y xy

3

3x y z xyz

2.4 2 4x y xy 3 27x y z xyz 2.5

1 1 4x y x y

1 1 1 9x y z x y z

2.6 21 4xy x y 3

1 4xyz x y z

Bnh lun: hc sinh d nh, ta ni: Trung bnh cng (TBC) Trung bnh nhn (TBN). Dng 2 v dng 3 khi t cnh nhau c v tm thng nhng li gip ta nhn dng khi s dng BT C Si: (3)

nh gi t TBN sang TBC khi khng c c cn thc.

3. CC K THUT S DNG

3.1 nh gi t trung bnh cng sang trung bnh nhn. nh gi t TBC sang TBN l nh gi BT theo chiu . nh gi t tng sang tch. Bi 1: Chng minh rng: 2 2 2 2 2 2 2 2 2 , ,8 a b ca b b c c a a b c

Gii Sai lm thng gp: S dng: x, y th x2 - 2xy + y2 = ( x- y)2 0 x2 + y2 2xy. Do :

2 2

2 2

2 2

222

a b abb c bcc a ca

2 2 2 2 2 2 2 2 28 , ,a b b c c a a b c a b c (Sai)

V d: 2 23 54 3

24 = 2.3.4 (-2)(-5).3 = 30 ( Sai )

Li gii ng: S dng BT C Si: x2 + y2 2 2 2x y = 2|xy| ta c:

2 2

2 2

2 2

0

0

0

2

22

a b ab

b c bcc a ca

2 2 2 2 2 2 2 2 2 2 2 2| 8 | 8 , ,a b b c c a a b c a b c a b c (ng) Bnh lun: Ch nhn cc v ca BT cng chiu ( kt qu c BT cng chiu) khi v ch khi cc v cng khng m. Cn ch rng: x2 + y2 2 2 2x y = 2|xy| v x, y khng bit m hay dng. Ni chung ta t gp bi ton s dng ngay BT C Si nh bi ton ni trn m phi qua mt v php bin i

n tnh hung thch hp ri mi s dng BT C Si. Trong bi ton trn du nh gi t TBC sang TBN. 8 = 2.2.2 gi n vic s dng bt ng thc

Csi cho 2 s, 3 cp s.



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Bi 2 : Chng minh rng: 8 264 ( )a b ab a b a,b 0 Gii

4 48 2 4 Si 24 2 .2 2 2 2 2 . .Ca b a b a b ab a b ab ab a b 264 ( )ab a b

Bi 3: Chng minh rng: (1 + a + b)(a + b + ab) 9ab a, b 0. Gii

Ta c: (1 + a + b)(a + b + ab) 3 33 1. . . 3. . . 9a b a b ab ab Bnh lun: 9 = 3.3 gi s dng Csi cho ba s, 2 cp. Mi bin a, b c xut hin ba ln, vy khi s dng C Si cho ba s

s kh c cn thc cho cc bin . Bi 4: Chng minh rng: 3a3 + 7b3 9ab2 a, b 0

Gii Ta c: 3a3 + 7b3 3a3 + 6b3 = 3a3 + 3b3 + 3b3 3 3 3 3 3 3

Csia b = 9ab2

Bnh lun: 9ab2 = 9.a.b.b gi n vic tch hng t 7b3 thnh hai hng t cha b3 khi p dng BT Csi ta c b2.

Khi c nh hng nh trn th vic tch cc h s khng c g kh khn.

Bi 5: Cho: , , , 0 1 : 1 1 1 1 813

1 1 1 1

a b c dCMR abcd

a b c d

Gii T gi thit suy ra:

si

3 3 1 1 1 11 1 1

1 1 1 1 1 1 1 1 1 1= - Cb c d bcd

a b c d b c d b c d

Vy:

3

3

3

3

3

3

3

3

1 01 1 1 1

1 01 1 1 1 1 d81

1 1 1 1 1 1 1 11 01 1 1 1

1 01 1 1 1

bcda b c d

cdab c d a abc

a b c d a b c ddcac d c a

abcd a b c

1 81

abcd Bi ton tng qut 1:



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Cho: 1 2 3

1 2 3

1 2 3

, , ,.............,

1

01 : ...........1 1 1 1......... 1

1 1 1 1

n

n

n

nn

x x x xCMR x x x xn

x x x x

Bnh lun: i vi nhng bi ton c iu kin l cc biu thc i xng ca bin th vic bin i iu kin mang tnh i

xng s gip ta x l cc bi ton chng minh BT d dng hn

Bi 6: Cho , , 0 1 1 1 : 1 1 1 8

1a b c CMR

a b ca b c

(1) Gii

si1 1 1(1) . . 2 2 2. . . . 8

Ca b cVTa b c

b c c a a b bc ca aba b c a b c

(pcm) Bi ton tng qut 2: Cho: n1 2 3

1 2 31 2 3

, , ,...............,........ 1

........ 10 1 1 1 1 : 1 1 1 1 n

nn

nx x x x

CMRx x x xx x x x

Bi 7: CMR: 1 2 333 3 1 1 1 1 1 8 , , 03a b c a b c abc abc a b c Gii

Ta c: si

33 1 1 1 1 1 1 13 3Ca b ca b c a b c

(1)

Ta c: 1 1 1 1a b c ab bc ca a b c abc 2 2 2 3si 33 3 3 11 3C a b c abc abc abc (2) Ta c: 333 3si 2 1. 81 Cabc abc abc (3)

Du = (1) xy ra 1+a = 1+b = 1+c a = b = c Du = (2) xy ra ab = bc = ca v a = b = c a = b= c Du = (3) xy ra 3 abc =1 abc = 1

Bi ton tng qut 3: Cho x1, x2, x3,., xn 0. CMR:

1 2 31 2 1 2 1 2 1 2 .... ...... ..... 2 ......1 1 1 1 1n n n nnnn nx x x x x x x x x x x xn

Bnh lun: Bi ton tng qut trn thng c s dng cho 3 s, p dng cho cc bi ton v BT lng gic trong tam

gic sau ny. Trong cc bi ton c iu kin rng buc vic x l cc iu kin mang tnh ng b v i xng l rt quan

trng, gip ta nh hng c hng chng minh BT ng hay sai. Trong vic nh gi t TBC sang TBN c mt k thut nh hay c s dng. l k thut tch nghch o.



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3.2 K thut tch nghch o. Bi 1: CMR: 2 . 0a b a bb a

Gii Ta c: 2 2

Csia b a bb a b a

Bi 2: CMR: 2

2 2 2 1

a a Ra

Gii

Ta c: 2 2 2

2 2 2 2

si2 2 2

1 12 1 11 11 1 1 1

Caa a aa a a a

Du = xy ra 2 2

2

11 1 1 01

a a aa

Bi 3: CMR: 1 3 0a a bb a b

Gii Ta c nhn xt: b + a b = a khng ph thuc vo bin b o hng t u a s c phn tch nh sau:

3si

.1 1 13 . 3 0C

a b a b b a b a bb a b b a b b a b

Du = xy ra 1b a b b a b a = 2 v b = 1. Bi 4: CMR: 2

4 3 01

a a ba b b

(1) Gii

V hng t u ch c a cn phi thm bt tch thnh cc hng t sau khi s dng BT s rt gn cho cc tha s di mu. Tuy nhin biu thc di mu c dng 21a b b (tha s th nht l mt a thc bc nht b, tha s 2 l mt thc bc hai ca b) do ta phi phn tch v thnh tch ca cc a thc bc nht i vi b, khi ta c th tch hng t a thnh tng cc hng t l cc tha s ca mu. Vy ta c: 21a b b = (a - b)( b + 1)( b + 1) ta phn tch a theo 2 cch sau: 2a +2 = 2(a - b) + ( b + 1) + ( b + 1) hoc a +1 = 1 12 2b ba b T ta c (1) tng ng : VT + 1 = 2

4 1 1 41 2 2 1 11b ba a b

a b b ba b b

4si

. . . .1 1 44 42 2 1 1C b ba b

a b b b PCM



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Bi 5: CMR : 3

12a 1 23 4 ( ) 1

ab a b a

b

Gii Nhn xt: Di mu s b(a-b) ta nhn thy b + ( a b ) = a. Chuyn i tt c biu thc sang bin a l 1 iu

mong mun v vic s l vi 1 bin s n gin hn. Bin tch thnh tng th y l mt mt mnh ca BT Csi. Do :

Ta c nh gi v mu s nh sau: 2 24. 4. 4.2 4b a b ab a b a

Vy: 3 3 3si

32 2

3 si 3 32a 1 2 1 1 1 1 . .4 ( )

C Ca a a a a a ab a b a aa a

Du = xy ra 2

1 1 1

2

b a b a

a ba

Bnh lun: Trong vic x l mu s ta s dng 1 k thut l nh gi t TBN sang TBC nhm lm trit tiu bin b. i vi phn thc th vic nh gi mu s, hoc t s t TBN sang TBC hay ngc li phi ph thuc vo du

ca BT. Bi 6: Bi ton tng qut 1. Cho: 1 2 3 ............., 0 1nx x x x v k Z . CMR:

1 1 2 11 2 2 3 1

1 21...............

k kk n k n kn nn

n ka

a a a a a a a k

Gii

VT = 1 2 2 3 1 1 2 2 3 1.....1

......n n k kkn

n nn

a a a a a aa a a a a a a

a

1 11 2 1 2

1 2 2 3 1

.. ....

1...

n

n nn nk k k

n nnk k

a a a aa a a aa

k k k k a a a a a a a

1 11 2 1 2

1 2 2 3 1

1 2 .. .. ...

. 11 2 ..n

n nn nk kk

n nn

n k

k k

a a a aa a a an k ak k k k a a a a a a a

1 2 1

1 2n k n k

n k

k

Tm li: Trong k thut tch nghch o k thut cn tch phn nguyn theo mu s khi chuyn sang TBN th cc phn cha bin s b trit tiu ch cn li hng s.

Tuy nhin trong k thut tch nghch o i vi bi ton c iu kin rng buc ca n th vic tch nghch o hc sinh thng b mc sai lm. Mt k thut thng c s dng trong k thut tch nghch o, nh gi t TBN sang TBC l k thut chn im ri.



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3.3 K thut chn im ri Trong k thut chn im ri, vic s dng du = trong BT Csi v cc quy tc v tnh ng thi ca du

= , quy tc bin v quy tc i xng s c s dng tm im ri ca bin. Bi 1: Cho a 2 . Tm gi tr nh nht (GTNN) ca 1S a a

Gii Sai lm thng gp ca hc sinh: 1S a a 2

1a a =2

Du = xy ra 1aa

a = 1 v l v gi thit l a 2. Cch lm ng: Ta chn im ri: ta phi tch hng t a hoc hng t 1

a sao cho khi p dng BT Csi du = xy ra khi a = 2.

C cc hnh thc tch sau: 1 1; (1)

1; (2)1,

1; (3)

; (4)

aa

aa

aa

aa

aa

Vy ta c: 514 4 2

1 3 1 3 3.224 4 4a a a aS a a . Du = xy ra a = 2.

Bnh lun: Ta s dng iu kin du = v im ri l a = 2 da trn quy tc bin tm ra = 4. y ta thy tnh ng thi ca du = trong vic p dng BT Csi cho 2 s ,

41aa v

34a t gi tr ln

nht khi a = 2, tc l chng c cng im ri l a = 2.

Bi 2: Cho a 2. Tm gi tr nh nht ca biu thc: 21S a a

Gii

S chn im ri: a = 2 2

2

1 14

a

a

2 14 = 8.

Sai lm thng gp:

2 2 2.1 1 7 1 7 2 7 2 7.2 2 7 928 8 8 8 8 8 4 4 48 8.2

a a a a aS a a a a a

MinS = 94

Nguyn nhn sai lm:

Chng hn ta chn s im ri (1): (s im ri (2), (3), (4) hc sinh t lm)

1 2

1 12

a

a

2 12 = 4.



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Mc d chn im ri a = 2 v MinS = 94

l p s ng nhng cch gii trn mc sai lm trong vic nh gi mu

s: Nu a 2 th 2 2 248 8.2a l nh gi sai.

thc hin li gii ng ta cn phi kt hp vi k thut tch nghch o, phi bin i S sao cho sau khi s dng BT Csi s kh ht bin s a mu s. Li gii ng: 32 2 2

si . .1 1 6 1 6 3 6 3 6.2 938 8 8 8 8 8 4 8 4 8 4Ca a a a a a aS a a a a

Vi a = 2 th Min S = 94

Bi 3: Cho , , 0

32

a b c

a b c

. Tm gi tr nh nht ca

1 1 1S a b c a b c Gii

Sai lm thng gp: 6 . .1 1 1 1 1 16 . . . 6S a b c a b ca b c a b c Min S = 6

Nguyn nhn sai lm : Min S = 6 31

21 1 1 3a b c a b ca cb tri vi gii thit.

Phn tch v tm ti li gii:

Do S l mt biu thc i xng vi a, b, c nn d on MinS t ti im ri 12a b c

S im ri: 12a b c 12

1 1 1 2

a b c

a b c

2 412

Hoc ta c s im ri sau:

12a b c

2 2 421 1 1 2

a b c

a b c

2 412

Vy ta c cch gii theo s 2 nh sau:

6 . .1 1 1 1 1 14 4 4 3 6 4 .4 .4 . 3S a b c a b c a b c a b ca b c a b c 3 1512 3.2 2

. Vi 12a b c th MinS = 152

Bi 4: Cho, , 0

32

a b c

a b c

. Tm GTNN ca

2 2 22 2 2

1 1 1S a b cb c a

Gii Sai lm thng gp:



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2 2 2 2 2 22 2 2 2 2 2

3 6. . . .1 1 1 1 1 13 3a b c a b cb c a b c aS

2 2 2 62 2 26 . . . . .

1 1 13 2 2 2 3 8 3 2a b cb c a

MinS = 3 2 . Nguyn nhn sai lm: MinS = 3 2 31

21 1 1 3a b c a b ca cb tri 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 MinS t ti 12a b c

2 2 2

2 2 2

11 44 16441 1 1

a b c

a b c

Li gii 2 2 2

2 2 2 2 2 2

16 16 16

..... ..... .....1 1 1 1 1 1 16 16 16 16 16 16

S a b cb b c c a a

2 2 22 2 2 2 2 2

16 16 16

17 17 1717 . ..... 17 . ..... 17 . .....1 1 1 1 1 1

16 16 16 16 16 16a b c

b b c c a a

2 2 217 17 17 1717 17

16 32 16 32 16 32 8 16 8 16 8 1617 17 17 17 16 16 16 16 16 16a b c a b c

b c a b c a

3 1717 17 17

8 16 8 16 8 16 8 5 5 5 517

. . 3. 17.

3 1717 316 16 16 16 2 2 2 2

a b c ab c a a b c a b c

15

172 2 2.

3

3 17 3 172

2 a b c

. Du = xy ra khi 12a b c Min S = 3 17

2

Bnh lun: Vic chn im ri cho bi ton trn gii quyt mt cch ng n vmt ton hc nhng cch lm trn tng

i cng knh. Nu chng ta p dng vic chn im ri cho BT Bunhiacpski th bi ton s nhanh gn hn p hn.

Trong bi ton trn chng ta dng mt k thut nh gi t TBN sang TBC, chiu ca du ca BT khng ch ph thuc vo chiu nh gi m n cn ph thuc vo biu thc nh gi nm mu s hay t s

Bi 5: Cho a, b, c, d > 0. Tm gi tr nh nht ca biu thc: a b c d b c d c d a a b d a b cS

b c d c d a a b d a b c a b c d

Gii Sai lm 1 thng gp:



11

.

.

.

.

2 2

2 2

2 2

2 2

a b c d a b c db c d a b c d a

b c d a b c d ac d a b c d a b

c a b d c a b da b d c a b d c

d a b c d a b ca b c d a b c d

S 2 + 2 + 2 + 2 = 8

Sai lm 2 thng gp: S dng BT Csi cho 8 s:

8 . . . . . . .8 8a b c d b c d c d a a b d a b cSb c d c d a a b d a b c a b c d

Nguyn nhn sai lm:

Min S = 8 a b c db c d ac d a bd a b c

a + b + c + d = 3(a + b + c + d) 1 = 3 V l.

Phn tch v tm ti li gii tm Min S ta cn ch S l mt biu thc i xng vi a, b, c, d do Min S nu c thng t ti im ri t do l : a = b = c = d > 0.(ni l im ri t do v a, b, c, d khng mang mt gi tr c th). Vy ta cho trc a = b = c = d d on 4 40 12

3 3Min S . T suy ra cc nh gi ca cc BT b phn phi c iu kin du bng

xy ra l tp con ca iu kin d on: a = b = c = d > 0. Ta c s im ri: Cho a = b = c = d > 0 ta c:

11 33 933

a b c db c d c d a a b d a b cb c d c d a a b d a b c

a b c d

Cch 1: S dng BT Csi ta c:

8

, , ,, , ,

. . . . . . .

8 .9 9 9

89 9 9 9

a b c da b c d

a b c d b c db c d a a

a b c d b c d c d a a b d a b cb c d c d a a b d a b c a b c d

S

89

b c c d a b a ba a b b c c d d

d a d ca b c d

12.12. . . . . . . . . . . . .83

8 8 8 40129 3 9 3

b c d c d a a b d a b ca a a b b b c c c d d d

Vi a = b = c = d > 0 th Min S = 40/3. 3.4 K thut nh gi t trung bnh nhn (TBN) sang trung bnh cng (TBC)



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Nu nh nh gi t TBC sang TBN l nh gi vi du , nh gi t tng sang tch, hiu nm na l thay du + bng du . th ngc li nh gi t TBN sang trung bnh cng l thay du . bng du + . V cng cn phi ch lm sao khi bin tch thnh tng, th tng cng phi trit tiu ht bin, ch cn li hng s. Bi 1 : CMR , , , 0ab cd a c b d a b c d (1)

Gii (1) 1

ab cda c b d a c b d Theo BT Csi ta c:

1 1 1 1 1 1 12 2 2 2a b c b a c b dVT a c b c a c b d a c b c (pcm)

Bnh lun: Nu gi nguyn v tri th khi bin tch thnh tng ta khng th trit tiu n s ta c php bin i tng

ng (1) sau bin tch thnh tng ta s c cc phn thc c cng mu s. Du gi cho ta nu s dng BT Csi th ta phi nh gi t TBN sang TBC

Bi 2: CMR 0 0a cc a c c b c ab b c (1)

Gii

Ta c (1) tng ng vi : 1 c b cc a cab ab

Theo BT Csi ta c:

1 1 1 12 2 2

c b c b cc a c a cc c a bab ab b a a b a b

(pcm)

Bi 3: CMR 3 3 1 1 1 1 , , 0 abc a b c a b c (1) Gii

Ta c bin i sau, (1) tng ng: 33 3 33

1.1.1 1.1.1 1 1 1 11 1 1 1 1 1

abcabc a b ca b c a b c

Theo BT Csi ta c:

1 1 1 1 1 1 1 1 1 1.3 13 1 1 1 3 1 1 1 3 1 1 1 3

a b c a b cVTa b c a b c a b c

Du = xy ra a = b = c > 0. Ta c bi ton tng qut 1: CMR: 1 2 1 2 1 1 2 2 ....... ....... ........ , 0 1,nn nn n n n i ia a a bb b a b a b a b a b i n Bi 4 : Chng minh rng: 2 4 16 ( ) ( ) , 0ab a b a b a b

Gii Ta c:

2 22 22 2 4

2 24 ( ) ( )16 ( ) 4.(4 )( ) 4 4 ( )ab a b a bab a b ab a b a b



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Bi 5: Cho , , 0

1a b ca b c

Chng minh rng 8 729abc a b b c c a

Gii S im ri:

Ta nhn thy biu thc c tnh i xng do du = ca BT s xy ra khi 13

a b c . Nhng thc t ta ch cn quan tm l sau khi s dng BT Csi ta cn suy ra c iu kin xy ra du = l: a = b = c. Do ta c li gii sau:

33 3 3si 1 2 8

3 3 3 3 729C a b b c c aa b cabc a b b c c a

Trong k thut nh gi t TBN sang TBC ta thy thng nhn thm cc hng s sao cho sau bin tch thnh tng cc tng trit tiu cc bin. c bit l i vi nhng bi ton c thm iu kin rng buc ca n s th vic nhn thm hng s cc em hc sinh d mc sai lm. Sau y ta li nghin cu thm 2 phng php na l phng php nhn thm hng s, v chn im ri trong vic nh gi t TBN sang TBC. Do trnh by phng php im ri trn nn trong mc ny ta trnh by gp c 2 phn .



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3.5 K thut nhn thm hng s trong nh gi t TBN sang TBC

Bi 1: Chng minh rng: 1 1 , 1a b b a ab a b Gii

Bi ny chng ta hon ton c th chia c 2 v cho ab sau p dng phng php nh gi t TBN sang TBC nh phn trc trnh by, tuy nhin y ta p dng mt phng php mi: phng php nhn thm hng s

Ta c :

si

si

.

.

1 11 1 1 2

1 11 1 1 .2

2

2

C

C

b aba b a b a

a abb a b a b

1 1 +2 2ab aba b b a ab Du = xy ra 1 1 2

1 1 2b ba a

Bnh lun: Ta thy vic nhn thm hng s 1 vo biu thc khng hon ton t nhin, ti sao li nhn thm 1 m khng phi

l 2. Thc cht ca vn l chng ta chn im ri ca BT theo quy tc bin l a = b = 1/2. Nu khng nhn thc c r vn trn hc sinh s mc sai lm nh trong VD sau. Bi 2: Cho

, , 01

a b ca b c

Tm gi tr ln nht: S a b b c c a

Gii Sai lm thng gp:

si

si

si

2

2

2

1.1

1.1

1.1

C

C

C

a ba b a b

b cb c b c

c ac a c a

2 3 52 2

a b ca b b c c a

Nguyn nhn sai lm Du = xy ra a + b = b + c = c + a = 1 a + b + c = 2 tri vi gi thit. Phn tch v tm ti li gii: Do vai tr ca a, b, c trong cc biu thc l nh nhau do im ri ca BT s l 1

3a b c t ta d on

Max S = 6 . a + b = b + c = c + a = 23

hng s cn nhn thm l 23

. Vy li gii ng l:



15

si

si

si

. .

. .

. .

23 2 3 3. 2 3 2 2

23 2 3 3. 2 3 2 2

23 2 3 3. 2 3 2 2

C

C

C

a ba b a b

b cb c b c

c ac a c a

.

22 3.3 33 .2 62 2 2

a b ca b b c c a

Bi ton trn nu cho u bi theo yu cu sau th hc sinh s c nh hng tt hn: Cho , , 01

a b ca b c

Chng

minh rng: 6S a b b c c a . Tuy nhin nu nm c k thut im ri th vic vit u bi theo hng no cng c th gii quyt c. Bi 3: Cho

0 30 4

xy

Tm Max A = (3 x )(12 3y)(2x + 3y)

Gii

A = 3

si

6 2x 12 3 2x+3y1 6 2 12 3 2 3 366 3

C yx y x y

Du = xy ra 6 -2x = 12 - 3y = 2x + 3y = 6 02

xy

Bnh lun: Vic chn im ri trong bi ton ny i vi hc sinh thng b lng tng. Tuy nhin cn c vo yu cu khi

nh gi t TBN sang TBC cn phi trit tiu ht bin cho nn cn c vo cc h s ca tch ta nhn thm 2 vo tha s th nht l mt iu hp l.

Bi 4: Cho x, y > 0. Tm Min f(x, y) = 3

2

x yxy

Gii

Ta c: 2 3 3 31 1 4x+2y+2y 1 4 44x 2 216 16 3 16 3 27xy y y x y x y f(x,y) =

3 3

2 3

4 4 f( , ) 4 27 2727

=x y x y Min x yxy x y

Du = xy ra 4x = 2y = 2y y = 2x > 0. l tp hp tt c cc im thuc ng thng y = 2x vi x dng. Thc ra vic h s nh trn c th ty c min l sao cho khi sau khi p dng BT Csi ta bin tch thnh tng ca x + y. ( C th nhn thm h s nh sau: 2x.y.y). Bnh lun:



16

Trong bi ton trn yu cu l tm Min nn ta c th s dng k thut nh gi t TBN sang TBC cho phn

di mu s v nh gi t TNB sang TBC l nh gi vi du nn nghch o ca n s l . Ta cng c th nh gi t s t TBC sang TBN c chiu Bi ton tng qut 1:

Cho

2 3

1 2 3 ...

1 2 32 31 2 3 4

1

..........., , ........... 0. . . ...........

nn

nn

x x x xx x x x Tm Min f x x x x

Bi 5: Chng minh rng: 21 (1) ( 1)n n n N nn

Gii

Vi n = 1, 2 ta nhn thy (1) ng. Vi n 3 ta c:

22

1 1....... 1 2 2 2 2.1.1......1 1n nnn

n n n n n nn n nn n n n

Bi ton tng qut 2: Chng minh rng: 1 11 1

m n

m n Nm n

(1)

Gii

Ta bin i (1) v bt ng thc tng ng sau: 1111 m

nnm

Ta c: . ....... .1 1 1 11 1 1 1 1.1.........1n m

m

mn

nm m m m

si.......

1 1 1 11 1 1 1 1 ......... 1 111

mn m

Cm n mm m m m

n n n

Bnh lun Cn phi bnh lun v du = : trong bi ton trn ta coi 1/m = a th th khi du bng trong BT Csi xy ra

khi v ch khi 1+ a = 1 a = 0. Nhng thc t th iu trn tng ng vi m tin ti +, khi m l hu hn th du



17

Bi 6: Cho , , 0

1a b ca b c

Tm Max

3 3 3S a b b c c a Gii

Sai lm thng gp:

3

3

3

3

3

3

1 1.1.1

31 1

.1.13

1 1.1.1

3

a ba b a b

b cb c b c

c ac a c a

3 3 3 2 6 83 3

a b cS a b b c c a

Max S = 83

Nguyn nhn sai lm:

Max S = 83

11 2 3 2 3 1

a bb c a b c V lc a

Phn tch v tm ti li gii: Do S lmt biu thc i xng vi a, b, c nn Max S thng xy ra ti iu kin:

, , 0

1a b ca b c

13

a b c

232323

a b

b c

c a

Vy hng s cn nhn thm l 23

.23

Ta c li gii:

3

3

3

33

3

3

3

3

9 .4

9 .4

9 .4

2 23 3. .

32 23 3. .

32 23 3. .

3

2 23 3

2 23 3

2 23 3

a ba b a b

b cb c b c

c ac a c a

3 3 3 33 39 9. .4 4

2 4 6 183 3

a b cS a b b c c a



18

Vy Max S = 3 18 . Du = xy ra

232323

a b

b c

c a

13

a b c

3.6 K thut ghp i xng Trong k thut ghp i xng chng ta cn nm c mt s kiu thao tc sau:

Php cng: 2

2 2 2

x y z x y y z z xx y y z z xx y z

Php nhn: 2 2 2 x ; xyz= xy x x, y, z 0x y z xy yz z yz z Bi 1: Chng minh rng: , , 0bc ca ab a b c a b c

a b c

Gii p dng BT Csi ta c:

.

.

.

12121 2

bc ca bc ca ca b a bca ab ca ab ab c b cbc ab bc ab ca c a c

bc ca ab a b ca b c . Du = xy ra a = b = c.

Bi 2: Chng minh rng: 2 2 2

2 2 2 0 a b c b c a abcb c a a b c


2 2 2 2

2 22 2

2 2 2 2

2 2 2 2

2 2 2 2

2 22 2

.

.

.

12

12

12

a b a b a ac c c cb b

b c b c b bc a c a a aa c a c c c

a ab b b b

2 2 2

2 22a b c b c a b c a

c a a c a cb b b

Bi 3: Cho tam gic ABC, a,b,c l s o ba cnh ca tam gic. CMR: a) 18p a p b p c abc ; b) 1 1 1 1 1 12p a p c a cp b b

Gii a) p dng BT Csi ta c:



19

2

12 8

2

2

2

2

p a p bp a p b

p b p cp b p c p a p b p c abc

p a p cp a p c

c

a

b

b) p dng BT Csi ta c:

1 1 1 1 1 2 2

21 1 1 1 1 2 2

21 1 1 1 1 2 2

2

p a p b cp a p bp a p b

p b p c ap b p cp b p c

p a p c bp a p cp a p c

1 1 1 1 1 12p a p c a cp b b

Du = xy ra cho c a) v b) khi vo ch khi ABC u: a = b = c

( p l na chu vi ca tam gic ABC: 2

a b cp ) Bi 4: Cho ABC, a, b, c l s o ba cnh ca tam gic. Chng minh rng: b c a c a b a b c abc


2

2

2

0

0

0

b c a c a bb c a c a b c

c a b a b cc a b a b c a

b c a a b cb c a a b c b

0 b c a c a b a b c abc Du = xy ra ABC u: a = b = c.



20

3.7 K thut ghp cp nghch o cho 3 s, n s Ni dung cn nm ccc thao tc sau: 1. 1 1 1 9 , , 0x y z x y zx y z

2. 1 221 21 2

, ,........, 01 1 1........ ......... nnn

x x xnx x x x x x

Bi 1: Chng minh rng : 6 , , 0b c c a a b a b c

a b c (1)

Gii Ta bin i (1) tng ng: 1 1 1 9b c c a a ba b c

9 a b c b c a c a ba b c 1 1 1 9 a b c a b c (pcm )

Bi 2: Chng minh rng: 2 2 2 9 , , 0a b ca b b c c a a b c Gii

Ta bin i tng ng BT nh sau: 1 1 12 9a b c a b b c c a 1 1 1 9a b b c a c a b b c c a (pcm )

Bi 3: Chng minh rng: 32

c a ba b b c c a , , 0a b c (BT Nesbit)

Gii Ta c bin i tng ng sau: 33 91 1 1 2 2

c a ba b b c c a

92a b c a b c a b c

a b b c c a

21 1 1 9a b c a b b c c a 1 1 1 9a b b c a c a b b c c a (pcm)

Bi 4: Chng minh rng: 2 2 2

, , 02c a b a b c a b ca b b c c a

Gii

Ta bin i BT nh sau: 2 2 2 3 2a b cc a bc a ba b b c c a



21

31 1 1 2a b cc a bc a ba b b c c a

3 2a b cc a ba b c a b b c c a 3

2c a b

a b b c c a 91 1 1 2

c a ba b b c c a

1 1 1 9a b b c a c a b b c c a 3.8 K thut i bin s C nhng bi ton v mt biu thc ton hc tng i cng knh hoc kh gii, kh nhn bit c phng hng gii,ta c th chuyn bi ton t tnh th kh bin i v trng thi d bin i hn. Phng php trn gi l phng php i bin. Bi 1: Chng minh rng: 3

2c a b

a b b c c a , , 0a b c (BT Nesbit) Gii

t: 00 ; ;

2 2 20

b c xy z x z x y x y zc a y a b c

a b z

.

Khi bt ng thc cho tng ng vi bt ng thc sau: 6

2 2 2y z x z x y x y z y x z x y z

x y z x y x z z y

Bt ng thc trn hin nhin ng, Tht vy p dng BT Csi ta c:

VT . . .2 2 2 2 2 2 6y x z x y zx y x z z y Du = xy ra x = y = z a = b = c

Bi 2: Cho ABC. Chng minh rng: 2 2 2a b c a b cb c a c a b a b c

Gii

t: 00 ; ;

2 2 20

b c a xy z z x x yc a b y a b c

a b c z

.

Khi bt ng thc cho tng ng vi bt ng thc sau:

2 2 2

4 4 4

y z z x x y x y zx y z (2)

Ta c: VT (2) 1 1 1 2 2 2

yz zx xy yz zx zx xy yz xyx y z x y y z x z



22

si

. . .C yz zx zx xy yz xy x y zx y y z x z

Bi 3:Cho ABC. CMR : ( b + c a ).( c + a b ).( a + b c ) abc (1) Gii

t: 00 ; ;

2 2 20

b c a xy z z x x yc a b y a b c

a b c z

.

Khi ta c BT (1) tng ng vi bt ng thc sau: . .2 2 2

x y y z z xxyz p dng BT Csi ta c: . . . . x

2 2 2x y y z z x xy yz z xyz (pcm)

Bi 4: Cho ABC. CMR: 2 2 2 1 1 1 pp a p b p cp a p cp b (1) Gii

t: 000

p a xp b yp c z

th (1) 2 2 21 1 1 x y z

xyzx y z (2)

Ta c:

VT (2) = 2 2 2 2 2 2 2 2 2 2 2 2. . . 1 1 12 2 2

1 1 1 1 1 1 1 1 1 1 1 1 x y y z x z x y y z x z

1 1 1

x x y z

xy yz z xyz

Du = xy ra x = y = z a = b = c ABC u. Bi 5: Chng minh rng nu a, b, c > 0 va abc = 1 th : 1 1 1 1

2 2 2a b c

Gii Bt ng thc cho tng ng vi:

1 1 1 11 1 12 2 2a b c

12 2 2a b c

a b c

t ; ; ;x y za b cy z x

tha iu kin . . 1. . x y za b cy z x

. Bt ng thc cho tng ng vi: 1

2 2 2x y z

x y y z z x

p dng bt ng thc Bunhiacopski ta c: 22 2 2 2 2 2x y zx x y y y z z z x x y zx y y z z x

2 2

2 12 2 2 2 2 2x y z x y zx y z

x y y z z x x x y y y z z z x x y z

3.9. MT S BI TP VN DNG

K thut chn im ri v nh gi t TBC sang TBN:



23

3.9.1 Cho a 6. Tm gi tr nh nht ca biu thc 2 18S aa

3.9.2 Cho 0 < a 1

2. Tm gi tr nh nht ca biu thc 2

12S aa

3.9.3 Cho

, 01

a ba b


1S abab

3.9.4 Cho , , 0

1a b ca b c


1S abcabc

3.9.5 Cho a, b > 0. Tm gi tr nh nht ca biu thc a b abSa bab

3.9.6 Cho , , 0

32

a b c

a b c


1 1 1S a b ca b c

3.9.7 Cho , , 0

32

a b c

a b c


2 2 2 1 1 1S a b ca b c

3.9.8 Cho a, b, c, d > 0. Tm gi tr nh nht ca biu thc: 3.9.9

2 2 2 21 1 1 13 3 3 3a b c dS b c d a

3.9.10 Cho , , 0

1a b ca b c

Chng minh rng: 2 2 2

1 1 1 2 2 2 81Sa b c ab bc ca

3.9.11 Cho , , 0

1a b ca b c

Chng minh rng:

2 2 2 1 1 1 28a b cSb c a a b c

K thut chn im ri v nh gi t TBN sang TBC: 3.9.12

2 2

11 1R: -2 21 1

a b abCM

a b

3.9.13 Cho , , 0

1a b ca b c

Chng minh rng

8 27

ab bc ca abc

3.9.14 Cho , , 0

1a b ca b c

Chng minh rng 16abc a b

K thut chn im ri v nhn thm hng s trong nh gi t TBN sang TBC

3.9.15 Cho 34 2

2 3 42 2

ab Tm Max Sc

ab c bc a ca b

3.9.16 Cho x, y, z >0. Tm Min f(x, y, z) = 6

2 3

x y zxy z

3.9.17 Chng minh rng: 11 (1) 1n n n Nn



24

3.9.18 Chng minh rng: 3 ...........2 1 3 1 11 12 3

n nS nn

3.9.19 ( Gi y: CMR 2

1 11n nn k )

3.9.20 Cho , , , 0

1a b c da b c d

Tm Max a b c b c d c d a d a bS

3.9.21 Cho , , , 0

1a b c da b c d

Tm Max

3 3 3 32 2 2 2S a b b c c d d a

3.9.22 Cho a 2, b 6; c 12. Tm Min 3 42 6 12bc a ca b ab cabc

S K thut ghp cp nghch o cho 3 s, n s

3.9.23 Cho , , 0

1a b ca b c

CMR :

1 1 1 92a b b c c a

3.9.24 Cho , , 0

1a b ca b c

CMR: 2 2 2

1 1 1 92 2 2a bc b ca c ab 3.9.25 Cho tam gic ABC, M thuc min trong tam gic. Gi MA, MB, MC th t giao vi BC, AC, AB ti

D, E, F. Chng minh: a) 1MD ME MF

DA EB FC ; b) 2MA MB MC

DA EB FC ; c) 6

DMA MB MCM ME MF

; d) . . 8

DMA MB MCM ME MF

e ) 9 / 2DA EB FCMA MB MC

; f) D 3/ 2M ME MFMA MB MC

5. MT S NG DNG KHC CA BT NG THC

p dng BT gii phng trnh v h phng trnh Bi 1: Gii phng trnh 11 2 ( )

2x y z x y z

Gii iu kin : x 0, y 1, z 2. p dng bt ng thc Csi cho hai s khng m ta c:

1.12

( 1) 11 ( 1).12

( 2) 1 12 2 .12 2

xx x

yy y

z zz z

Suy ra : 11 2 2x y z x y z Du = xy ra khi v ch khi

321

1211

1

zyx

zyx

.

Vy phng trnh c nghim (x, y, z) = (1; 2; 3)



25

Bi 2: Gii phng trnh: 4 4 42 21 1 1x x x = 3 Gii

iu kin: -1 x 1. p dng bt ng thc C-si ta c:

24

4

4

1 11 1 . 1 (1)2

1 11 1 .1 (2)2

1 11 1 .1 (3)2

x xx x x

xx x

xx x

Cng (1), (2), (3) ta c: 4 2 4 41 1 1 1 1 1x x x x x Mt khc, li theo bt ng thc Csi ta c:

(1 ) 1 21 (1 ).12 2

(1 ) 1 21 (1 ).12 2

x xx x

x xx x

2 21 1 1 1 32 2

x xx x

T (4) v (5) suy ra: 4 4 421 1 1 3x x x Du = xy ra khi v ch khi:

1 11 1 01 1

x xx xx

Vy phng trnh c nghim duy nht x = 0 Bi3: Gii phng trnh: 2 2 21 1 2x x x x x x (1)

Gii

p dng bt ng thc C-si, ta c:

2 22

2 22

( 1) 112 2

( 1) 1 212 2

x x x xx x

x x x xx x

2 21 1 1x x x x x (2) Kt hp (1) v (2) ta c: .10)1(12 22 xxxxx Th li ta c x = 1 l nghim duy nht ca phng trnh

Bi 4: Gii h phng trnh: ( 1) ( 1) 21 1

x y y x xyx y y x xy

Gii iu kin: x 1, y 1. p dng bt ng thc C-si, ta c:

1 ( 1)1 1.( 1) 12 2 2 x x xyx x y x (1)

Tng t: -1 12 2y xyy x y (2)

Cng (1), (2) ta c 1 1x y y x xy .



26

Du = xy ra khi v ch khi 1 1 21 1

x x yy

.

Th li thy: x = y = 2 cng tha mn phng trnh th nht ca h Vy h c nghim duy nht (2;2)

Bi 5: Cho s nguyn n >1. Gii h phng trnh:

1 22

2 33

11

1 12

1 12

............................1 12n

x xx

x xx

x xx

Gii T h cho suy ra x1, x2, , xn l cng du. Gi s xi > 0 vi mi i. p dng bt ng thc C-si, ta c:

1 22

1 1 12

x xx

.Tng t: xi 1 vi mi i.

Cng n phng trnh ca h theo tng v ta c: 1 21 2

1 1 1... ...nn

x x xx x x

V xi 1 nn i

i xx

1 vi mi i, suy ra: 1 21 2

1 1 1... ...nn

x x xx x x

Du = xy ra khi v ch khi x1 = x2 = = xn = 1

Bi 6: Gii h phng trnh:

2

2

2

2

2

2

212

12

1

x yx

y zy

z xz

Gii R rng h c nghim x = y = z = 0. Vi x,y,z 0, t h cho suy ra x>0, y>0, z>0. p dng bt ng thc C-si, ta c:

222

22 21 2

1 2xxx x y x

x x

Tng t: 22

2 22 2 x .

1 1y zz y v zy z

Vy : y x z y, suy ra x = y = z. Thay y = x vo phng trnh th nht ta c:

22

22 2 1 1 ( v x 0)

1x x x x xx

Vy h c hai nghim (x, y, z) = {(0; 0; 0) ; (1; 1; 1)} Bi 7: Tm s nguyn dng n v cc s dng a1 = a2 = = an tha cc iu kin



27

n1 2

n1 2

a a ..... a 2 (1)1 1 1.... 2 (2)a a a

Gii:

Ly (1) cng (2) v theo v, ta c: 1 21 2

1 1 1.. 4nn

a a aa a a

p dng bt ng thc C-si, ta c: 1 2 ii

aa

vi i = 1, 2, , n Suy ra 4 2n hay n 2:

Vi n = 1: h 1

1

21 2

a

a

v nghim; Vi n = 2: h

21

1 2

21 1 2

aa

a a

c nghim a1 = a2 = 1

Vy: n = 2 v a1 = a2 = 1

Sau y s l mt s bi tp tng t gip hc sinh n luyn kin thc BI TP HC SINH VN DNG

1. Gii cc phng trnh sau: 2 2 2) ( 1)( 2)( 8) 32 ( , , 0)a x y z xyz x y z

2 2) x 2-x 4 4 3b y y 16 4 1225) 82 3 1 665x-3 1 665

c x y zy z

3

23

4( 1) 1 4x 1) 10x ( 1)

y ydy

2. Gii phng trnh:

2) -1 3 2( 3) 2 2.a x x x x 3) .

2 2 2 4x y zb

x y z y z x z x y

3. Gii h phng trnh:

2

2

3

2 4

4

2 4 6

21

31

41

x yx

y zy y

z xz z z

4. Xc nh s nguyn dng n v cc s dng x1, x2 , , xn tha:

n1 2

n1 2

x x ... x 9 1 1 1... 1 x x x



28

5. Gii h phng trnh: 4 4 41x y z

x y z xyz

6. Gii h phng trnh: 1 2

1 2

1 1 .... 1

1 1 .... 1

n

n

n kx x x nn

n kx x x nn

[]-K ĩ Thuật Sử Dụng Bất Đẳng Thức Côsi

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Transcript of []-K ĩ Thuật Sử Dụng Bất Đẳng Thức Côsi