Kythuat CM BDT Cosi_logo

8/7/2019 Kythuat CM BDT Cosi_logo

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K thut s dngBt ng thcC-Si


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2

1. NHNG QUY TC CHUNG TRONG CHNG MINH BT NG THC SDNG 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 songhnh, 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 chngminh. N nh hng cho ta phng php gii, da vo im ri ca BT. Chnh v vy m khi dy cho hc sinh tarn 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 nghcho 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 nghincu v chng minh BT cng thng rt hay mc sai lm ny. p dng lin tip hoc song hnh cc BT nhngkhng ch n im ri ca du bng. Mt nguyn tc khi p dng song hnh cc BT l im ri phi c ngthi 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 biton cc tr c iu kin rng buc, gi tr ln nht nh nht ca hm nhiu bin trn mt min ng. Ta bit rng ccgi 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 chra 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

liTrn 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 ccv 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 21 2

.................n n n

x x xx x x

n

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

Dng 3:1 2

1 2 .................

n

nn x x x

x x x

n

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

n

SMax

nx x x

khi1 2

............ nS

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

n = 2: x, y 0 khi : n = 3: x, y, z 0 khi :

2.12

x yxy

3

3

x y zxyz

2.2 2x y xy 33x y z xyz

2.32

2x y

xy

3

3x y z

xyz


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3

2.4 2 4x y xy

3 27x y z xyz

2.51 1 4x y x y

1 1 1 9x y z x y z

2.6

2

1 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 ab

b c bc

c 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 2

3 5

4 3

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

Li gii ng:

S dng BT C Si: x2

+ y2

22 2

x y = 2|xy| ta c:2 2

2 2

2 2

0

0

0

2

2

2

a b ab

b c bc

c 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.

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 . .

C

a 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.


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4

Gii

Ta c: (1 + a + b)(a + b + ab) 3 33 1. . . 3. . . 9a b a b ab abBnh 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 ss kh c cn thc cho cc bin .

Bi 4: Chng minh rng: 3a3 + 7b3 9ab2 a, b 0Gii

Ta c: 3a3 + 7b3 3a3 + 6b3 = 3a3 + 3b3 + 3b3 3 3 3 33 3Csi a b = 9ab2Bnh 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:

, , , 01:1 1 1 1 813

1 1 1 1

a b c d

CMR abcd

a b c d

GiiT gi thit suy ra:

si

331 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

bcd

a b c d

cda

b c d a abc

a b c d a b c d dca

c d c a

abc

d a b c

1

81abcd

Bi ton tng qut 1:

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 x

CMR 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 ixng 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

CMRa b ca b c

(1)

Gii

si1 1 1(1) . . 2 2 2. . . . 8Ca b c

VTa b c

b c c a a b bc ca ab

a b c a b c

(pcm)


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Bi ton tng qut 2:

Cho: n1 2 3

1 2 31 2 3

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

........ 1

0 1 1 1 1: 1 1 1 1nnn

nx x x x

CMRx x x xx x x x

Bi 7: CMR: 1 2 3

33 3

1 1 1 1 1 8 , , 03

a b ca b c abc abc a b c

Gii

Ta c:

si33 1 1 1

1 1 1 13 3

Ca b ca b ca b c

(1)

Ta c: 1 1 1 1a b c ab bc ca a b c abc

2 2 23si

33 33 11 3C

a b c abc abc abc (2)

Ta c: 33

3 3si

2 1. 81C

abc abc abc

(3)

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

Du = (3) xy ra 3 abc =1 abc = 1Bi ton tng qut 3:Cho x1, x2, x3,., xn 0. CMR:

1 2 3

1 21 2 1 2 1 2

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

nnx 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 tamgic 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 quantrng, 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.

3.2 K thut tch nghch o.

Bi 1: CMR: 2 . 0a b a bb a

Gii

Ta c: 2 2Csia b a b

b a b a

Bi 2: CMR:2

2 22

1

aa R

a

Gii

Ta c: 2 2 2

2 2 2 2

si2

2 21 12 1 1

1 11 1 1 1

Caaa a

a a a a

Du = xy ra 2 22

11 1 1 01

a a aa

Bi 3: CMR:

1 3 0a a bb a b


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6

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:

3

si.

1 1 13 . 3 0

Ca b a b b a b a b

b 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)

GiiV 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 2

1a 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 th nh tch ca cc a thc bc nht i vi b, khi tac th tch hng t a thnh tng cc hng t l cc tha s ca mu.

Vy ta c: 2

1a 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 1

2 2

b ba b

T ta c (1) tng ng :

VT + 1 =

24 1 1 4

12 2 1 11

b ba a b

a b b ba b b

4

si. . . .

1 1 44 42 2 1 1

C b ba b

a b b b

PCM

Bi 5: CMR :3

12a 1 23

4 ( )1

a

b 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 iumong 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 4

b a b ab a b a

Vy:3 3 3si

3

2 2

3 si3 3

2a 1 2 1 1 1 1. .4 ( )

C Ca a aa a a a

b a b a aa a

Du = xy ra 2

11 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.


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Cho:1 2 3

............., 0 1nx x x x v k Z . CMR:

1 1 2 1

1 2 2 3 1

1 21

...............k kk n k n k

n 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 a

a a a a a a a

a

1 11 2 1 2

1 2 2 3 1

.. ....

1...

n

n nn n

k 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 .. .. ...

.1

1 2 ..

n

n nn n

k kk

n nn

n k

k k

a a a aa a a an k a

k 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 ccphn 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 TBNsang TBC l k thut chn im ri.

3.3 K thut chn im riTrong 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) ca1

S aa

Gii

Sai lm thng gp ca hc sinh:1

S a a 21

a a =2

Du = xy ra 1

aa

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 t1a

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

a aa

aa

a

aa

Vy ta c:5

14 4 2

1 3 1 3 3.224 4 4

a a a aS

a a . Du = xy ra a = 2.

Bnh lun:

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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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 ,4

1aa

v34a

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:2

1S 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 92

8 8 8 8 8 8 4 4 48 8.2a a a a a

S aa a a a

MinS = 94

Nguyn nhn sai lm:

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 th2 2 2

48 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 dngBT Csi s kh ht bin s a mu s.

Li gii ng: 32 2 2si

. .1 1 6 1 6 3 6 3 6.2 93

8 8 8 8 8 8 4 8 4 8 4Ca a a a a a a

S aa a a

Vi a = 2 th Min S =9

4

Bi 3: Cho , , 0 32

a b c

a b c

. Tm gi tr nh nht ca1 1 1

S a b ca b c

GiiSai 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 3

12

1 1 13a b c a b c

a 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 ri12

a b c

S im ri:12

a b c

12

1 1 1 2

a b c

a b c

2 412

Hoc ta c s im ri sau:


9/26

9

12

a 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

12

a b c th MinS = 152

Bi 4: Cho

, , 032

a b c

a b c

. Tm GTNN ca 2 2 22 2 2

1 1 1S a b c

b c a

GiiSai lm thng gp:

2 2 2 2 2 22 2 2 2 2 2

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

b c a b c a

S

2 2 2 62 2 2

6 . . . . .1 1 1

3 2 2 2 3 8 3 2a b cb c a

MinS = 3 2 .

Nguyn nhn sai lm:

MinS = 3 2 312

1 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 ti12

a b c

2 2 2

2 2 2

11 44 16441 1 1

a b c

a b c

Li gii

2 2 22 2 2 2 2 2

16 16 16

..... ..... .....1 1 1 1 1 1

16 16 16 16 16 16S a b c

b 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 1 116 16 16 16 16 16

a b cb b c c a a

2 2 2

17 17 17 1717 1716 32 16 32 16 32 8 16 8 16 8 16

17 17 17 1716 16 16 16 16 16

a 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 a

b c a a b c a b c


10/26

10

15

172 2 2

.3

3 17 3 172

2 a b c

. Du = xy ra khi12

a b c Min S = 3 172

Bnh lun:

Vic chn im ri cho bi ton trn gii quyt mt cch ng n vmt ton hc nhng cch lm trn tngi cng knh. Nu chng ta p dng vic chn im ri cho BT Bunhiacpski th bi ton s nhanh gn hnp hn.

Trong bi ton trn chng ta dng mt k thut nh gi t TBN sang TBC, chiu ca du ca BT khng chph 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:

.

.

.

.

2 2

2 2

2 2

2 2

a b c d a b c d

b c d a b c d a

b c d a b c d a

c d a b c d a b

c a b d c a b d

a b d c a b d c

d a b c d a b c

a 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 d

b c d a

c d a b

d 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 tdo 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 on4 40

123 3

Min 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 d

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

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

a b c d

Cch 1: S dng BT Csi ta c:


11/26

11

8

, , ,, , ,

. . . . . . .

8.9 9 9

89 9 9 9

a b c da b c d

a b c d b c d

b c d a a

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

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

S

89

b c c d a b a b

a a b b c c d d

d a d c

a 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)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 cnphi 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)

1ab cd a c b d a c b d

Theo BT Csi ta c:

1 1 1 1 1 1 12 2 2 2

a b c b a c b d VT

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 tngng (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

0

a c

c a c c b c ab b c

(1)Gii

Ta c (1) tng ng vi :

1c b cc a c

ab ab

Theo BT Csi ta c:

1 1 1 12 2 2

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

ab ab b a a b a b

(pcm)

Bi 3: CMR

3 31 1 1 1 , , 0abc a b c a b c (1)

GiiTa c bin i sau, (1) tng ng:

33 3 331.1.11.1.1 1 1 1 1

1 1 1 1 1 1abc

abc 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 cVT

a b c a b c a b c

Du = xy ra a = b = c > 0.


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12

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: 2 416 ( ) ( ) , 0ab a b a b a b

Gii

Ta c:

2 22 22 2 4

2 2

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

Bi 5: Cho, , 0

1a b c

a b c

Chng minh rng 8

729abc a b b c c a

GiiS im ri:

Ta nhn thy biu thc c tnh i xng do du = ca BT s xy ra khi1

3a 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 ligii sau:

3

3 3 3si 1 2 8

3 3 3 3 729

C a b b c c aa b c

abc 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 tngcc tng trit tiu cc bin. c bit l i vi nhng bi ton c thm iu kin rng buc ca n s th vic nhnthm hng s cc em hc sinh d mc sai lm. Sau y ta li nghin cu thm 2 phng php na l phng phpnhn 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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13

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 nhphn 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 21 1

1 1 1 .2

2

2

C

C

b aba b a b a

a abb a b a b

1 1 +2 2

ab aba b b a ab

Du = xy ra 1 1 21 1 2

b b

a 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, , 0

1a b c

a b c

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

GiiSai 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 5

2 2

a b ca b b c c a

Nguyn nhn sai lmDu = 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 l1

3a b c t ta d on

Max S = 6 . a + b = b + c = c + a =2

3 hng s cn nhn thm l

2

3. Vy li gii ng l:

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


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14

.

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, , 0

1a b c

a 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 bitheo hng no cng c th gii quyt c.

Bi 3: Cho 0 30 4

x

y

Tm Max A = (3x )(123y)(2x + 3y)

Gii

A =

3si 6 2x 12 3 2x+3y1 6 2 12 3 2 3 36

6 3

C yx y x y

Du = xy ra 6 -2x = 12 - 3y = 2x + 3y = 6 0

2

x

y

Bnh lun:

Vic chn im ri trong bi ton ny i vi hc sinh thng b lng tng. Tuy nhi n cn c vo yu cu khinh 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 votha s th nht l mt iu hp l.

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

3

2

x y

xy

Gii

Ta c: 23 3

31 1 4x+2y+2y 1 4 44x 2 216 16 3 16 3 27

xy y y x y x y

f(x,y) =

3 3

2 3 4 4f( , )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 tngca x + y. ( C th nhn thm h s nh sau: 2x.y.y).Bnh lun:

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.

. . ...........

n

n

nn

x x x xx x x x Tm Min f

x x x x

Bi 5: Chng minh rng:2

1 (1) ( 1)n n n N nn

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


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15

22

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

n nn n n n

n n nn n n n

Bi ton tng qut 2:

Chng minh rng:1 1

1 1

m n

m n Nm n

(1)

Gii

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

111

m

n

nm

Ta c: . ....... .1 1 1 11 1 1 1 1.1.........1

n m

m

m

nnm m m m

si.......

1 1 1 11 1 1 1 1 ......... 1 1

11

mn m

Cm n m

m m m mn 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 rakhi 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 thdu


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16

Do S lmt biu thc i xng vi a, b, c nn Max S thng xy ra ti iu kin:

, , 01

a b c

a b c

1

3a b c

2

3

2

3

2

3

a b

b c

c a

Vy hng s cn nhn thm l2

3.2

3

Ta c li gii:

3

3

3

33

3

3

3

3

9.

4

9.

4

9.

4

2 23 3. .3

2 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

Vy Max S = 3 18 . Du = xy ra

2

3

2

3

2

3

a b

b c

c a

1

3a b c

3.6 K thut ghp i xngTrong 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 x

x 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 ca b c

Giip dng BT Csi ta c:

.

.

.

1

212

12

bc ca bc cac

a b a bca ab ca ab

ab c b c

bc ab bc abc

a 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 20a b c b c a abc

b c a a b c

Gii


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17

p dng BT Csi ta 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

1

2

a b a b a ac c c cb b

b c b c b bc a c a a a

a c a c c c

a ab b b b

2 2 2

2 22

a b c b c a b c ac 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) 1

8p 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:

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 22

2

1 1 1 1 1 22

21 1 1 1 1 22

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 12

p 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 Gii

p dng BT Csi ta c:


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19

3.7 K thut ghp cp nghch o cho 3 s, n sNi dung cn nm ccc thao tc sau:

1. 1 1 1 9 , , 0x y z x y zx y z

2. 1 22

1 21 2 , ,........, 0

1 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 ca b c

(1)

Gii

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

9a b c b c a c a b

a b c

1 1 1 9a b ca 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 )


c a b

a b b c c a

, , 0a b c (BT Nesbit)

Gii

Ta c bin i tng ng sau: 33 91 1 1 2 2c a ba b b c c a

92

a b c a b c a b c

a b b c c a

21 1 1 9

a b ca 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

2

a b cc a bc a b

a b b c c a

3

1 1 12

a b cc a bc a b

a b b c c a

3

2

a b cc a ba b c

a b b c c a


20/26

20

3

2

c a b

a b b c c a

91 1 12

c a b

a 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 sC nhng bi ton v mt biu thc ton hc tng i cng knh hoc kh gii, kh nhn bit c phng hnggii,ta c th chuyn bi ton t tnh th kh bin i v trng thi d bin i hn. Phng php trn gi l phngphp i bin.


c 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 z

c a y a b c

a b z

.

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

62 2 2

y 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 y

c 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 yx y z

x y z

(2)

Ta c: VT (2) 1 1 12 2 2

yz zx xy yz zx zx xy yz xy

x y z x y y z x z

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 + ab ).( a + bc ) abc (1)Gii

t:

00 ; ;

2 2 20

b c a xy z z x x y

c 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


21/26

21

p dng BT Csi ta c: . . . . x2 2 2

x y y z z xxy 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 x

p b y

p c z

th (1)2 2 2

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

1 1 1x

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

12 2 2a b c

GiiBt ng thc cho tng ng vi:

11 1 1

1 1 12 2 2a b c

12 2 2

a b c

a b c

t ; ; ;x y z

a b cy z x

tha iu kin . . 1. .x y z

a b cy z x

. Bt ng thc cho tng ng vi:

12 2 2

x y z

x y y z z x

p dng bt ng thc Bunhiacopski ta c:

2

2 2 2

2 2 2

x y zx x y y y z z z x x y z

x y y z z x

2 2

21

2 2 2 2 2 2

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

3.9.1 Cho a 6. Tm gi tr nh nht ca biu thc 218

S aa

3.9.2 Cho 0 < a 1

2. Tm gi tr nh nht ca biu thc

2

12S a

a

3.9.3 Cho, 0

1a b

a b

. Tm gi tr nh nht ca

1S ab

ab

3.9.4 Cho, , 0

1a b c

a b c


1S abc

abc

3.9.5 Cho a, b > 0. Tm gi tr nh nht ca biu thca b ab

Sa bab


22/26

22

3.9.6 Cho

, , 032

a b c

a b c


1 1 1S a b c

a b c

3.9.7 Cho

, , 032

a b c

a b c

. Tm gi tr nh nht ca 2 2 2

1 1 1S a b c

a b c

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

3.9.92 2 2 21 1 1 13 3 3 3a b c d

Sb c d a

3.9.10 Cho, , 0

1

a b c

a b c

Chng minh rng:

2 2 2

1 1 1 2 2 281S

a b c ab bc ca

3.9.11 Cho, , 0

1

a b c

a b c

Chng minh rng:

2 2 2 1 1 128

a b cS

b 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 c

a b c

Chng minh rng

8

27ab bc ca abc

3.9.14 Cho, , 0

1a b c

a 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

3

4

2

2 3 42 2

a

b Tm Max S

c

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 z

xy z

3.9.17 Chng minh rng:1

1 (1) 1n n n Nn

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

2 3n

nS n

n

3.9.19 ( Gi y: CMR 21 11n n

n k

)

3.9.20 Cho

, , , 0

1

a b c d

a b c d

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

3.9.21 Cho, , , 0

1

a b c d

a 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 Min3 42 6 12bc a ca b ab cabc

S

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


23/26

23

3.9.23 Cho, , 0

1a b c

a b c

CMR :

1 1 1 92a b b c c a

3.9.24 Cho, , 0

1a b c

a b c

CMR:

2 2 2

1 1 19

2 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) 6DMA MB MC

M ME MF ;

d) . . 8D

MA MB MC

M ME MF e ) 9 / 2

DA EB FC

MA MB MC ; f)

D3/ 2

M ME MF

MA 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 trnh1

1 2 ( )2

x 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 : 1

1 22

x y z x y z

Du = xy ra khi v ch khi

32

1

1211

1

zy

x

zy

x

.

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

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

Giiiu 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)

21 1

1 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 2

1 1 1 1 32 2

x xx x


24/26


25/26

25

V xi 1 nni

ix

x1

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

1

21

xy

x

yz

y

zx

z

GiiR 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, tac:

222

2

2 21 2

1 2

xxx x y x

x x

Tng t:22

2 2

2 2x .1 1

y zz y v z

y z

Vy : y x z y, suy ra x = y = z.

Thay y = x vo phng trnh th nht ta c:22

2

22 1 1 ( v x 0)

1

xx x x x

x

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

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 2

1 2

1 1 1.. 4nn

a a aa a a

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

i

i

aa

vi i = 1, 2, , n

Suy ra 4 2n hay n 2:

Vi n = 1: h1

1

2

1 2

a

a

v nghim; Vi n = 2: h

21

1 2

2

1 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 thcBI 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


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