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8/7/2019 Kythuat CM BDT Cosi_logo
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K thut s dngBt ng thcC-Si
8/7/2019 Kythuat CM BDT Cosi_logo
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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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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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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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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:
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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
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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:
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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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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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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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.
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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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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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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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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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 )
Bi 3: Chng minh rng:32
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
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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.
Bi 1: Chng minh rng:32
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
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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
. Tm gi tr nh nht ca
1S abc
abc
3.9.5 Cho a, b > 0. Tm gi tr nh nht ca biu thca b ab
Sa bab
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3.9.6 Cho
, , 032
a b c
a b c
. Tm gi tr nh nht ca
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
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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
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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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