Bai Giang Toan Kinh Te Quang 2012

8/3/2019 Bai Giang Toan Kinh Te Quang 2012

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ThS Phng Duy QuangTrng Khoa Cbn Trngi hc Ngoi Thng H ni

1

ThS PHNG DUY QUANG (ch bin)

BI GING N THI CAO HC

Mn: TON KINH T

H NI, 2011


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2

Phn 1. Ton csng dng trong kinh t

TON CAO CP 1Chuyn 1. Ma trn v nh thc

1. Ma trn v cc php ton 2. nh thc ca ma trn vung cp n

3. Ma trn nghch o

4. Hng ca ma trn

Chuyn 2. H phng trnh tuyn tnh v ng dng

1. Khi nim h phng trnh tuyn tnh

2. Phng php gii h phng trnh

TON CAO CP 2Chuyn 3. Gii hn, lin tc, vi tch phn hm mt bin s

1. Gii hn ca dy s

2. Gii hn ca hm s

3. Hm s lin tc

4 o hm, vi phn v ng dng

5. Tch phn hm mt bin s

Chuyn 4. Php tnh vi phn hm nhiu bin s v ng dng

1. Gii hn v lin tc

2. o hm ring v vi phn ca hm nhiu bin

3 Cc tr hm nhiu bin

Chuyn 5. Tng hp cc dng Ton cao cp ng dng trong phn tch kinh t


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4

TON CAO CP 1

Chuyn 1. MA TRN V NH THC

1. MA TRN

1. Cc khi nim

Cho m, n l cc s nguyn dng

nh ngha 1. Ma trn l mt bng s xp theo dng v theo ct. Mt ma trn c m

dng v n ct c gi l ma trn cp mn. Khi cho mt ma trn ta vit bng s bn

trong du ngoc trn hoc ngoc vung. Ma trn cp mn c dng tng qut nh sau:

mn2m1m

n22221

n11211

a...aa............

a...aa

a...aa

hoc

mn2m1m

n22221

n11211

a...aa............

a...aa

a...aa

Vit tt l A = (aij)n xn hoc A = [aij]n xn

V d 1. Cho ma trn

=

176

752A . A l mt ma trn cp 2 x 3 vi

a11 = 2 ; a12 = 5 ; a13 = - 7 ; a21 = 6 ; a22 = 7 ; a23 = 1

nh ngha 2.

Hai ma trn c coi l bng nhau khi v ch khi chng cng cp v cc phn t

v tr tng ng ca chng i mt bng nhau.

Ma trn chuyn v ca A l AT : AT = [aji]n xn

Ma trn i ca ma trn A l ma trn: -A = [- aij]n x n

V d 2. Cho ma trn

=

02

14

31

A . Xc nh AT, - A

Ta c

=

013

241AT ;

=

02

14

31

A


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Ma trn khng cp m x n l ma trn m mi phn tu bng 0 : nxm]0[=

Khi n = 1 ngi ta gi ma trn A l ma trn ct, cn khi m = 1 ngi ta gi ma trn

A l ma trn dng.

Ma trn vung cp n l ma trn c s dng v s ct bng nhau v bng n. Mt matrn c s dng v s ct cng bng n c gi l ma trn vung cp n. Khi cc phn

t a11, a22, , ann gi l cc phn t thuc ng cho chnh, cn cc phn t an1, n 12a ,

, a1n gi l cc phn t thuc ng cho ph.

Ma trn tam gic l ma trn vung khi c cc phn t nm v mt pha ca ng

cho chnh bng 0.

+) Ma trn A = [aij]n x nc gi l ma trn tam gic trn nu aij = 0 vi i > j:

=

nn

n1n1n1n

n21n222

n11n11211

a0...00

aa...00

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

aa...a0

aa...aa

A

+) Ma trn A = [aij]n x nc gi l ma trn tam gic di nu aij = 0 vi i < j:

=

nn1nn2n1n

1n1n21n11n

2221

11

aa...aa

0a...aa

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

00...aa

00...0a

A

V d 4. Cho mt v d v ma trn vung cp 3, ma trn tam gic trn, tam gic di cp

3.

Gii:

=

611

412

521

A ;

=

600

410

521

B ;

=

611

012

001

C

Ma trn cho l ma trn vung cp n m c tt c cc phn t nm ngoi ng

cho chnh u bng 0


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Ma trn cho c tt c cc phn t thuc ng cho chnh bng 1 c gi l ma

trn n v :

=

10...00

01...00

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

00...10

00...01

E

Tp cc ma trn cp m x n trn trng s thc R, k hiu: Matm x n(R)

Tp cc ma trn vung cp n trn trng s thc R, k hiu: Mat n(R)

V d 5. Cho ma trn

=

176

752A v

=2m7

75

62

B

a) Tm AT v A

b) Tm m AT = B

Gii:

a) Ta c

=

17

75

62

AT v

=

176

752A

b) 1m1m

m7

75

62

17

75

62

BA 2

2

T ==

=

=

2. Php ton trn ma trn

a) Php cng hai ma trn v php nhn ma trn vi 1 s

nh ngha 3. Cho hai ma trn cng cp mn: nmijnmij bB;aA ==

Tng ca hai ma trn A v B l mt ma trn cp m n, k hiu A + B v c xc nh

nh sau:nmiiij

baBA

+=+

Tch ca ma trn A vi mt s l mt ma trn cp mn, k hiu A v c xc

nh nh sau:


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nmija.A

=

Hiu ca A tr B: A B = A + (-B)

Tnh ngha ta suy ra cc tnh cht cbn ca php ton tuyn tnh

Tnh cht 1. Cho A, B, C l cc ma trn bt k cp m n, ; l cc s bt k ta lun

c:

1) A + B = B + A

2) (A + B) +C = A + (B + C)

3) A + 0 = A

4) A + (-A) = 0

5) 1.A = A

6) (A + B) = A + B

7) (+ )A = A +A

8) ( )A = (B)

V d 6. Cho cc ma trn

=

=

312

212B;

110

421A . Khi

=

+

=

1116

1474

312

212).3(

110

421.2B3A2

V d 7. Cho ma trn

=

35

31B . Tm ma trn C sao cho 3B 2(B + C) = 2E

Gii:

Phng trnh cho

=

==

2/12/5

2/32/1

10

01

35

31.

2

1EB

2

1C

b) Php nhn ma trn vi ma trn

Cho hai ma trn :


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A =

mn2m1m

n22221

n11211

a...aa

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

a...aa

a...aa

; B =

np2n1n

p22221

p11211

b...bb

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

b...bb

b...bb

Trong , ma trn A c s ct bng s dng ca ma trn B.

nh ngha4.

Tch ca ma trn A vi ma trn B l mt ma trn cp mp, k hiu l AB v c xc

nh nh sau: AB =

mn2m1m

n22221

n11211

c...cc

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

c...cc

c...cc

trong ( )p,...,2,1j;m,...,2,1i;baba...babacn

1kkjiknjinj22ij11iij ===+++=

=

Ch 1.

Tch AB tn ti khi v ch khi s ct ca ma trn ng trc bng s dng ca ma

trn ng sau.

Cca ma trn AB: Ma trn AB c s dng bng s dng ca ma trn ng trc

v s ct bng s ct ca ma trn ng sau.

Cc phn t ca AB c tnh theo quy tc: Phn t ijc l tch v hng ca dng

th i ca ma trn ng trc v ct th j ca ma trn ng sau.

V d 8. Cho hai ma trn

=

13

21A v

=

231

410B . Tnh A.B v B.A

Gii :

Ta c

=

++++++=

=

1461872

2.14.33.11.31.10.32.24.13.21.11.20.1

231410.

1321B.A

Nhng s ct ca B khc s dng ca A nn khng tn ti tch BA.


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V d 9. Cho ma trn

=

023

012A ;

=

1203

0112

1321

B . Tnh A.B, BA

Gii:

Ta c

=

=

3781

1753

1203

0112

1321

.023

012B.A

Cn B.A khng tn ti

Cc tnh cht cbn ca php nhn ma trn

Tnh cht 2. Gi s php nhn cc ma trn di y u thc hin c.

1) (AB)C = A(BC)

2) A(B+C) = AB+AC; (B+C)D =BD +CD

3) (AB) = (A)B = A(B)

4) AE = A; EB =B

c bit , vi ma trn vung A: AE = EA = A

5) ( )T T TAB B A=

Ch 2. Php nhn ma trn khng c tnh cht giao hon. Nu =B.A th cha chc

=A hoc =B .

V d 10. Cho cc ma trn

=

=

01

00B;

00

10A .

Khi

=

=

10

00A.B;

00

01B.A v BAAB

V d 11. Cho

=

=

10

00B;

00

01A , ta c

=

=

00

00

10

00.

00

01B.A


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c) Lu tha ca ma trn vung: Cho A l ma trn vung cp n. Ta xc nh

A0 = E; An = An -1. A ( n l s nguyn dng)

V d 12. Cho

=

dc

baA . Chng minh rng, ma trn A tho mn phng trnh

=++ )bcad(X)da(X 2

Gii:

Ta c

+

+

=++

10

01).bcad(

dc

ba).da(

dc

ba.

dc

baE)bcad(A)da(A 2

= =

=

+

++

++

++

++

00

00

bcad0

0bcad

)da(d)da(c

)da(b)da(a

dbcc)da(

b)da(bca2

2

. (pcm)

V d 13. Cho ma trn

=

10

11A . Tnh A2, A3, ..., An (n l s t nhin)

Gii:

Ta c

=

=

10

21

10

11

10

11A 2 ;

=

=

10

31

10

11

10

21A3 ; .... ; tng t ta c th d

on

=

10

n1A n . D dng chng minh c bng quy np cng thc An.

nh ngha 5. Php bin i scp trn ma trn A = [aij]m x n l cc php bin i c

dng

i) i ch 2 dng (ct) cho nhau: )cc(dd jiji

ii) nhn mt dng (ct) vi mt s khc 0: )kc(kd ii

iii) nhn mt dng (ct) vi mt s ri cng vo dng (ct) khc: )chc(dhd jiji ++

V d 15. Cho ma trn

=

4211

5212

6421

A . Thc hin cc php bin i scp sau: (1)

nhn dng 2 vi 2

(2) hon v dng 1 cho dng 2


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2. NH THC CA MA TRN VUNG

1. Khi nim nh thc

Cho ma trnA =

nn2n1n

n22221

n11211

a...aa

............a...aa

a...aa

. Xt phn t aij ca A, bi dng i v ct j ca A

ta c ma trn vung cp n -1, k hiu Mij: gi l ma trn con con ng vi phn t aij.

V d 1.

=

333231

232221

131211

aaa

aaa

aaa

A . Tm cc ma trn con ng vi cc phn t ca A

nh ngha 1. Cho mt ma trn A vung cp n: A =

nn2n1n

n22221

n11211

a...aa

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

a...aaa...aa

.

nh thc ca A, k hiu det(A) hoc A c nh ngha nh sau:

* nh thc cp 1: A = [a11] th det(A) = a11

* nh thc cp 2:

= 2221

1211

aa

aa

A th 211222112221

1211

aaaaaa

aa

)Adet( ==

V d 2. Tnh nh thc 22.614.1142

61==

V d 3. Gii phng trnh: 049

25x 2=

Gii: Tnh nh thc ta c: VT = 4x2 25.9

215x

49.25xPT 2 ==

* nh thc cp 3:


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322311332112312213322113312312332211

333231

232221

131211

a.a.aa.a.aa.a.aa.a.aa.a.aa.a.a

aaa

aaa

aaa

Adet ++==

Quy tc Sariut: nh thc cp 3 c 6 s hng, m mi s hng l tch ca 3 phn t m

mi dng, mi ct ch c mt i biu duy nht.* Cc s hng mang du cng: cc s hng m cc phn t nm trn ng cho chnh

hoc cc phn t nm trn cc nh ca tam gic c 3 nh c mt cnh song song vi

ng cho chnh.

* Cc s hng mang du tr: cc s hng m cc phn t nm trn ng cho ph hoc

cc phn t nm trn cc nh ca tam gic c 3 nh c mt cnh song song vi ng

cho ph. nh quy tc tnh nh thc cp 3, ngi ta thng dng quy tc Sarrus

sau:

T quy tc Sarrus trn, chng ta cn mt quy tc khc tnh nhanh nh thc cp

3: ghp thm ct th nht v ct th hai vo bn phi nh thc hoc ghp thm dng th

nht v dng th hai xung bn di nh thc ri nhn cc phn t trn cc ng chonh quy tc th hin trn hnh:

V d 4.Tnh nh thc122

102

321

3

=

Gii: Ta c = 3 1.0.1 + 2.(-2).1 + 3.2.(-2) 3.0.2 1.(-2).2) 1.1.(-2) = -10

Du + Du -

1 1 1 1 1

2 2 2 2 2

3 3 3 3 3

a b c a b

a b c a b

a b c a b

Du - Du +

1 1 1

2 2 2

3 3 3

1 1 1

2 2 2

a b c

a b c

a b c

a b c

a b c

Du -

Du +


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V d 5. Gii phng trnh 0124

111

1xx 2

=

Gii:

Ta c

=

==+=

2x

1x02x3x

124

111

1xx2

2

nh thc cp n (n 3 ):

det(A) = )Mdet()1(a ijji

n

1jij

+

=

(vi i bt k)

hoc det(A) = )Mdet()1(a ijji

n

1iij

+

=

(vi j bt k)

V d 6. Gii phng trnh : 0

1242008

1112009

1xx2010

00020112

=

Gii :t

1242008

1112009

1xx2010

00020112

4 = . S dng cng thc khai trin nh thc theo dng 1 ta

c )2x3x.(2011

124

111

1xx

)1.(2011 2

2

114 +==

+ .

=

==+

2x

1x02x3xPT 2

2. Tnh cht ca nh thc

A =[aij]n x n vi )Adet(n =

Dng i ca nh thc c gi l tng ca 2 dng nu:

( ) ( ) ( )i1 i2 ij in i1 i2 ij in i1 i2 ij in ij ij ija a ....a ....a b b ....b ....b c c ....c ....c ;a b c ( j 1,n)= + = + =

Dng i l t hp hp tuyn tnh ca cc dng khc nu


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)n,1j(aa kjn

k1k

kij == =

. K hiu =

=n

ik1k

kki dd ; dk = (ak1 ak2 ... akn)

Tnh cht 1. (Tnh cht chuyn v)

nh thc ca ma trn vung bng nh thc ca ma trn chuyn v ca n: det(AT)

= det(A)

V d 1. Cho

=

dc

baA . CMR det(AT) =det(A)

Bn c tgii

Ch 1. T tnh cht chuyn v, mi tnh cht ca nh thc ng cho dng th cng

ng cho ct v ngc li. Do , trong cc tnh cht ca nh thc, ch pht biu cho cc

dng, cc tnh cht vn gi nguyn gi tr khi thay ch "dng" bng ch "ct".

Tnh cht 2. (Tnh phn xng).

i ch hai dng cho nhau v gi nguyn v tr cc dng cn li th nh thc i du.

V d 2. Xtdc

bav

ba

dc

Bn c tgii

H qu 1. Mt nh thc c hai dng ging nhau th bng khng.

Chng minhGi nh thc c hai hng nh nhau l n . i ch hai hng ta c, theo tnh cht 2

ta c

n = - n 002 nn ==

Tnh cht 3. (Tnh thun nht).Nu nhn cc phn t mt dng no vi cng mt s

k th c nh thc mi bng k ln nh thc c

nn2n1n

in2i1i

n11211

nn2n1n

in2i1i

n11211

a

...

...

...

a

...

a

...a...aa............

a...aa

.k

a

...

...

...

a

...

a

...ka...kaka............

a...aa

=

nh l ny c th pht biu: Nu mt nh thc c mt dng c nhn t chung th a

nhn t chung ra ngoi du nh thc


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H qu 2. Mt nh thc c hai dng t l vi nhau th bng khng.

Chng minh: Tht vy, nu a h s t l ra ngoi du nh thc th c mt nh thc

c hai dng ging nhau nn n bng khng.

V d 2.19. Chng minh nh thc sau chia ht cho 17:

91176

4112204356817

76212

4

=

Gii:

Ta c D.17

91176

4112

12241

76212

.17

91176

4112

)12.(172.17)4.(171.17

76212

4 =

=

= .

V D l nh thc to bi cc s nguyn nn D cng l s nguyn. Do 174 M

Tnh cht 3. (Tnh cng tnh).Nu nh thc c mt dng l tng hai dng th nh thc

bng tng ca hai nh thc.

11 12 1n 11 12 1n 11 12 1n

i1 i1 i2 i2 in in i1 i2 in i1 i2 in

n1 n2 nn n1 n2 nn n1 n2 nn

a a a a a a a a a

b c b c b c b b b c c c

a a a a a a a a a

= ++ + +

L L L

L L L L L L L L L L L L

L L L

L L L L L L L L L L L L

L L L

H qu 3. Nu nh thc c mt dng l t hp tuyn tnh ca cc dng khc th nh

thc y bng khng.

l h qu ca tnh cht cng tnh v tnh thun nht.

H qu 4. Nu cng vo mt dng mt t hp tuyn tnh ca cc dng khc th nh thc

khng i.

Tcc tnh cht ca nh thc, ta thng sdng cc php bin i scp trn ma trn

trong qu trnh tnh nh thc cp n:

* i ch 2 dng (ct) cho nhau: )cc(dd jiji , php bin i ny nh thc i du

* Nhn mt dng (ct) vi mt s khc 0: )kc(kd ii , php bin i ny nh thc tng ln

k ln.


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* Nhn mt dng (ct) vi mt s cng vo dng (ct) khc: )chc(dhd jiji ++ , php bin

i ny khng lm thay i gi tr ca nh thc.

V d 4. Tnh nh thcy'ccxy'bbxy'aax

'c'b'a

cba

3

+++

=

Gii:

Nhn dng 1 vi (-x), dng 2 vi (-y) cng vo dng 3 ta c: 0000

'c'b'a

cba321 dydxd

3 ==+

V d 5. Tnh nh thc

2222

2222

2222

2222

4

)3d()3c()3b()3a(

)2d()2c()2b()2a(

)1d()1c()1b()1a(dcba

++++

++++

++++=

Gii:

Nhn dng 1 vi (-1), ri cng ln lt vo dng 2, dng 3, dng 4 c:

9d69c69b69a64d44c44b44a4

1d21c21b21a2

dcba 2222

dd

4,3,2i4

i1

++++++++

++++=

+

=

Sau nhn dng 2 vi (- 2) cng vo dng 3, nhn dng 2 vi (-3) cng vo dng 4

c:

6666

2222

1d21c21b21a2

dcba 2222

dd2

dd34

32

42

++++=

+

+= 0 (v c 2 dng t l nhau)

V d 6. Tnh nh thc

12

ac

2

cb

2

ba1bac

1acb

1cba

4

+++

=

Gii:


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Cng cc ct vo ct 1 ta c:

12

ac

2

cb1cba

1ba1cba

1ac1cba

1cb1cba

4

+++++

+++

+++

+++

=

t nhn t chung ca ct 1 ra ngoi:

0

12

ac

2

cb1

1ba1

1ac1

1cb1

).1cba(4 =++

+++=

3.Cc phng php tnh nh thc

Cho nh thc cp n:

nmnj1n

inij1i

n1j111

n

a...a...a

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

a...a...a

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

a...a...a

=

a) Sdng nh ngha bng cng thc khai trin:

Phn b i s ca ija

Xa i dng th i v ct th j (dng v ct cha phn t ija ) ca A ta c mt ma

trn con (n - 1), k hiu l ijM . nh thc ca ijM c gi l nh thc con cp n -1

tng ng vi phn t aij ca A v )Mdet()1(A ijji

ij+= c gi l phn b i s ca

phn t ija ca nh thc d. Cho nh thc cp n l n . Khi n c th tnh theo hai

cch sau:

i) Cng thc khai trin theo dng th i :

==

+ ==n

1jijij

n

1jij

jiijn Aa)Mdet(.)1(a (1)

ii) Cng thc khai trin theo ct th j:


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==

+ ==n

1iijij

n

1iij

jiijn Aa)Mdet(.)1(a (2)

H qu.i vi nh thc cp n l n , ta c

i)

==

= kikhi0kikhiAa n

n

1jkjij (3)

ii)

==

= kjkhi0

kjkhiAa n

n

1iikij (4)

Nhn xt: Mc ch ca cng thc (1) hoc (2) l chuyn vic tnh nh thc cp n v

tnh nh thc cp n -1, ri t cp n -1 chuyn v cp n -2, , cho n nh thc cp 3, 2.

Khi p dng cng thc (1) hoc (2), ta nn chn dng hoc ct c cha nhiu phn t 0

nht khai trin. Nu khng c dng hoc ct nh vy ta bin i nh thc a vnhthc mi bng nh thc ban u nhng c dng hoc ct nh vy.

V d 1. Tnh nh thc a)054

213

112

3 = b)

421

213

121

3

=

Gii: a) Khai trin nh thc theo dng 3 ta c:

7512023

12

.)1.(521

11

.)1.(42313

3 ==++=++

b) Khai trin nh thc theo ct 1 ta c:

35530021

12.)1)(1(

42

12.)1.(3

42

21.)1.(1 1312113 ==

+

+= +++

V d 2. Tnh nh thc a)

1253

3142

3131

5011

4

= b)

11432

4100

3010

2001

4

=

Gii:

a) Nhn ct 1 vi (-1) cng vo ct 2, nhn ct 1 vi (-5) cng vo ct 4; ri khai trin

nh thc theo ct 1, ta c


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1428

1316

814

1428

1316

814

.)1.(1

14283

13162

8141

0001

11cc

cc54

21

41

=

=

= +

+

+

Cng dng 1 vo dng 2, nhn dng 1 vi (-2) cng vo dng 2, ri khai trin nh thctheo ct 2 ta c:

203016

52.)1.(1

30016

502

81421

dd

dd24

21

31

=

=

= ++

+

b) Nhn ct (-2) vi ct 1 ri cng vi ct 4

9432

4100

5010

0001

4

=

Khai trin nh thc theo dng 1 ta c

943

410

501

943

410

501

.)1.(1

9432

4100

5010

0001

114

=

=

= +

Nhn ct 1 vi 5 cng vo ct 3, khai trin nh thc theo dng 1 ta c

81624244

41.)1.(1

2443

410

00111

4 ====+

V d 3. Tnh nh thc ca ma trn tam gic trn v tam gic di

a)

nn

n1n1n1n

n21n222

n11n11211

n

a0...00

aa...00

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

aa...a0

aa...aa

= b)

nn1nn2n1n

1n1n21n11n

2221

11

n

aa...aa

0a...aa

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

00...aa

00...0a

=

Gii:


21/94


21

Ta ch cn xt a) Ln lt khai trin nh thc theo ct 1 :

nn2211

nn

n1n1n1n

n21n222

1111

nn

n1n1n1n

n21n222

n11n11211

n a...a.a...

a0...0

aa...0

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

aa...a

.)1.(a

a0...00aa...00

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

aa...a0

aa...aa

====

+

Tng t, ta c nn2211

nn1nn2n1n

1n1n21n11n

2221

11

n a...aa

aa...aa

0a...aa

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

00...aa

00...0a

==

b) Phng php bin i v dng tam gic:Dng cc tnh cht ca nh thc bin i nh thc a nh thc vnh thc ca

ma trn tam gic trn hoc di, sau p dng cng thc:

nn332211

nn

n222

n11211

a...a.a.a

a...00

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

a...a0

a...aa

= hoc nn2211

nn2n1n

2221

11

a...aa

a...aa

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

0...aa

0...0a

=

V d 1. Tnh cc nh thc

a)

04321

50321

54021

54301

54321

5

= b)

44321

43321

43221

43211

4321

4

baaaa1

abaaa1

aabaa1

aaaba1

aaaa1

+

+

+

+

=

Bn c tgii


22/94


22

V d 2. Tnh nh thc

a)

0xxxx1

x0xxx1xx0xx1

xxx0x1

xxxx01

111110

6 = b)

axxxxx

xaxxxxxxaxxx

xxxaxx

xxxxax

xxxxxa

6 =

Gii: a)

Nu x = 0, khai trin nh thc theo dng 1, suy ra 06 =

Nu x 0, nhn ct 1, dng 1 vi x, ri cng cc dng vo dng 1v t nhn t

chung (n -1) ra ngoi ta c:

0xxxxx

x0xxxx

xx0xxx

xxx0xx

xxxx0x

xxxxxx

.x

5

0xxxxx

x0xxxx

xx0xxx

xxx0xx

xxxx0x

xxxxx0

.x

1226

==

Nhn dng 1 vi (-1) ri cng vo cc dng khc ta c:

35

226x5)x(x.

x

5

x00000

0x0000

00x000

000x00

0000x0xxxxxx

.x

5==

=

b) Cng cc ct vo ct 1, ri t nhn t chung ra ngoi du nh thc ta c

[ ]

ax...xx1

xa...xx1

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

xx...ax1

xx...xa1

xx...xx1

.x5a

ax...xxx5a

xa...xxx5a

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

xx...axx5a

xx...xax5a

xx...xxx5a

6 +=

+

+

+

+

+

=


23/94


23

Nhn dng 1 vi (-1) v cng vo cc dng 2, dng 3, , dng n ta c

[ ] [ ] 6n )xa.(x5a

xa0...000

0xa...000..................

00...xa00

00...0xa0

xx...xx1

.x5a +=

+=


24/94


24

3. MA TRN NGHCH O CA MA TRN VUNG

Trong phn ny chng ta xem xt khi nim ma trn nghch o ca ma trn vung

cp n, iu kin tn ti v cch tm ma trn nghch o

1. nh thc ca tch hai ma trn vung

Cho hai ma trn vung cp n : A = [aij]n x n; B = [bij]n x n

nh l 1. nh thc ca tch hai ma trn vung bng tch cc nh thc ca ma trn

thnh phn: det(AB)= det(A)det(B)

V d 1. Cho A, B l ma trn vung cp 3 c det(A) = 2, det(B) = -2. Tnh det(AB),

det(A2B); det(2AB).

Bn c tgii

2. nh ngha ma trn nghch o

nh ngha 1. Cho A l ma trn vung cp n v E l ma trn n v cp n. Nu c ma

trn vung B cp n sao cho

A.B = B.A = E

th ta ni ma trn A l kh nghch v B c gi l ma trn nghch o ca ma trn A

(hay A c ma trn nghch o l B), v k hiu A-1 = B.

V d 2. a) Ma trn A =

40

01l kh nghch v c ma trn nghch o l

=

410

01

A1

.

V ta c

=

=

10

01

40

01.

4

10

01

4

10

01.

40

01.

b) Ma trn

=

00

00khng kh nghch v mi ma trn vung B cp 2 u c

E.BB. == .

Sduy nht ca ma trn nghch o

nh l 2. Ma trn nghch o A-1 ca ma trn vung A nu tn ti th duy nht

3. Stn ti ca ma trn nghch o

nh l 3. Ma trn vung A kh nghch khi v ch khi det(A) 0.


25/94


25

v A-1 =1

det(A). A =

1

det(A).

11 21 n1

12 22 n2

1n 2n nn

A A A

A A A

A A A

L

L

M M O M

L

V d 3. Tm A-1 ca

=

=

100

410121

B;62

31A

Bn c tgii

T khi nim v iu kin kh nghch ca ma trn, ta c mt s tnh cht sau:

nh l 4. Gi s A, B l cc ma trn vung cp n.

i) Nu A kh nghch th A-1, AT, kA (k 0), Am (m nguyn dng) cng kh nghch v

(A-1)-1 = A ; (AT)-1 = (A-1)T ; 11 Ak1)kA( = ; (Am)- 1 = (A-1)m

ii) Nu A, B kh nghch th AB cng kh nghch v (AB)-1 = B-1A-1

iii) Nu A kh nghch th cc phng trnh A.X = C, X.A = C c nghim duy nht

CAXCX.A 1==

1A.CXCXA ==

V d 4. Tm (A2)-1 vi

=

62

31A

Bn c tgii

4. Mt s phng php tm ma trn nghch o

a) Phng php nh thc

Da vo nh l 2.12, ta c cc bc tm ma trn nghch o ca ma trn A = [aij]nn nh

sau:

Bc 1: Tnh det(A)

Nu det(A) = 0 th A khng kh nghch.Nu det(A) 0 th A c ma trn nghch o.

Bc 2: Tm ma trn ph hp caA:


26/94


26

A =

11 21 n1

12 22 n2

1n 2n nn

A A A

A A A

A A A

L

L

M M O M

L

trong Aij l phn b i s ca a ij .

Bc 3: Tnh B = 1 Adet(A)

. Khi , ma trn B chnh l ma trn nghch o ca ma trn

A, tc l

A-1 = B

V d 1. Tm ma trn nghch o ca ma trn

a)1 2

A

3 4

=

b)

=

801

352

321

A

Gii.

a)

Bc 1: Ta c det(A) = 1.4 2.3 = -2 0 .

Nn ma trn A kh nghch v A.)Adet(

1A 1 =

Bc 2: Ta lp ma trn ph hp A ca ma trn A. Ta c

A11 = (-1)1+ 1.4 = 4; A12 = (- 1)1+ 2. 3 = - 3; A21 = (- 1)2 + 1.2 = - 2; A22 =(- 1)2 + 2.1 = 1

Nn

=

13

24A

Bc 3. Tnh ma trn nghch o

=

==

2

1

2

312

13

24.

2

1A.

)Adet(

1A 1

Vy

=

2

1

2

312

A 1

b)

Bc 1. Ta c det(A) = -1 0 nn A kh nghch v A.)Adet(

1A 1 =

Bc 2. Ta lp ma trn ph hp A ca ma trn A. Ta c


27/94


27

501

52.)1(A;13

81

32.)1(A;40

80

35.)1(A 2222

2112

1111 ======

+++

201

21.)1(A;5

81

31.)1(A;16

80

32.)1(A 3223

2222

1221 ======

+++

15221

.)1(A;33231

.)1(A;93532

.)1(A 333323

3213

31 ======+++

Nn

=

125

3513

91640

A

Bc 3. Tnh ma trn nghch o

==

125

3513

91640

A)Adet(

1A 1

Vy

=

125

3513

91640

A 1

V d 2. Gii phng trnh ma trn sau

a)

=

295

153X.

43

21b)

=

11

10

01

X.

801

352

321

Gii:

a) Ma trn

=

43

21A kh nghch nn phng trnh c nghim duy nht

=

=

=

2

532

011

295

153.

2

1

2

312

295

153.

43

21X

1

b) Ma trn

=

801

352

321

A kh nghch nn phng trnh c nghim duy nht

=

=

=

36

816

2549

11

10

01

.

125

3513

91640

11

10

01

.

801

352

321

X

1


28/94


28

b) Phng php khGause-Jordan (Phng php bin i scp)

Thc t ta s p dng ng thi cc php bin i scp v dng a A v E v

a E v ma trn A-1. T, ta c quy tc tm ma trn nghch o bng php bin i s

cp (phng php Gauss Jordan):

Bc 1: Vit ma trn n v E cng cp vi ma trn A bn cnh pha phi ma trn A

c ma trn mi k hiu (A|E)

Bc 2: Dng cc php bin i scp trn dng i vi ma trn mi ny a dn

khi ma trn A v ma trn n v E, cn khi ma trn E thnh ma trn B, tc l (A|E)

(E|B). Khi , B chnh l ma trn nghch o ca A.

V d 1. Tm ma trn nghch o ca ma trn:

=

801

352

321

A

Gii

Bc 1: Lp ma trn (A|E) =

100

010

001

801

352

321

Bc 2: Bin i scp

++

+ 125

012

001

100

310

321

101

012

001

520

310

321

100

010

001

801

352

3213221

31

dd2dd2

dd

+

++

125

3513

91640

100

010

001

125

3513

3614

100

010

021123

23

13

dd2d

dd3dd3

.

Vy

=

125

3513

91640

A 1


29/94


29

4. HNG CA MA TRN

1. Khi nim

Cho ma trn n}{m,mink1;aAnxmij

= . Trc ht, ta nhc li khi nim nh thc con

cp k ca ma trn A.Ly ra k dng v k ct khc nhau . nh thc ca ma trn cp k c

cc phn t thuc giao im ca k dng v k ct c gi l nh thc con cp k ca

A , k hiu:

( )nj...jj1;ni...ii1D k21k21j...jji...iik21

k21


30/94


30

i) n}{m,min)A(r0

ii) r(A) = r(AT)

iii) Nu A l ma trn vung cp n th

* r(A) = n 0A hay A khng suy bin

* r(A) < n 0A = hay A suy bin

V d 2. Tm hng ca ma trn

=

12923

8632

4311

A

Gii:

Ta c nh thc con cp 2: 20122

83D2423 == 0 nn r(A) 2.

Xt cc nh thc con cp 3: c tt c 4C34 = nh thc con cp 3 ca A

0

923

632

311

D123123 =

= ; 0

1293

862

431

D134123 == ;

0

1223

832411

D124123 =

= ; 0

1292

863431

D234123 =

=

Hay mi nh thc con cp 3 ca A bng 0. Do r(A) = 2.

V d 3. Tm hng ca ma trn sau:

= +

+

+

0...00...00

.....................0...00...00

a...aa...00

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

a...aa...a0

a...aa...aa

A nr1rrrr

n21r2r222

n11r1r11211

vi a11a22 arr 0.

Gii:


31/94


31

Ta c nh thc con cp r : 0a...aa

a...00

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

a...a0

a...aa

D rr2211

rr

r222

r11211

r...12r...12 == v mi nh thc

cp cao hn r u cha t nht mt dng ton s khng nn nh thc bng 0. Do vy,r(A) = r.

T v d ny ta c kt qu sau:

nh l 2

(i) Cc php bin i scp khng lm thay i hng ca ma trn

ii) Hng ca ma trn dng bc thang bng s dng khc khng ca ma trn

nh l 3

(i) Nu A v B l hai ma trn cng cp m n bt k, ta lun c:

)B(r)A(r)BA(r ++

(ii) Vi A v B l hai ma trn bt k sao cho AB tn ti, ta lun c:

)A(r)AB(r v )B(r)AB(r hay }r(B){r(A),min)AB(r

(iii) Nu A l ma trn cp m x n, B l ma trn vung cp n x p th

r(A) + r(B) r(AB) + n

H qu : Nu A, B l cc ma trn vung cp n th ta c

)AB(rn)B(r)A(r ++

2. Cc phng php tm hng ca ma trn

a) Phng php nh thc

Trc ht, ta chng minh kt qu:

nh l 4. Cho ma trn A = [aij]m x n c mt nh thc con cp r khc 0 l Dr. Nu mi

nh thc con cp r + 1 cha Dru bng 0 th hng ca A bng r.

Tnh l ny, ta c phng php tm hng ca ma trn nhsau:

Bc 1: Tm mt nh thc con cp Dk khc 0 cp k ( { }n,mmink0


32/94


32

Bc 2: Ta tnh cc nh thc cp k + 1 cha Dk (nu c).

Trng hp 1: Nu cc nh thc cp k + 1 u bng 0 th ta kt lun r(A) = k.

Trng hp 2: Nu c mt nh thc cp k + 1 khc 0 th ta li tnh cc nh thc cp

k 2+ cha nh thc cp k 1+ khc 0 ny (nu c).Qu trnh c tip tc nh vy ta tm c hng ca A .

V d 1. Tm hng ca ma trn

=

12923

8632

4311

A

Gii:

Ta c nh thc con cp 2: 532

11

D12

12 =

= 0 nn r(A) 2.

Xt cc nh thc con cp 3 cha 1212D : c 2 nh thc con cp 3 ca A cha1212D

0

923

632

311

D123123 =

= ; ; 01223

832

411

D124123 =

= .

Nh vy, mi nh thc con cp 3 cha 1212D u bng 0 nn r(A) = 2.

a) Phng php bin i scpTnh l trn, ta c phng php bin i scp tm hng ca A:

Bc 1: S dng cc php bin i scp a ma trn A v dng bc thang B.

Bc 2:m s dng khc khng ca B, s l hng ca A.

V d 2. Tm hng ca ma trn sau:

=

2121

4112

2431

A

Gii:

Dng php bin i scp a ma trn A v dng bc thang


33/94


33

B

0000

0770

2431

0550

0770

2431

2121

4112

2431

A2

32

21

21

d7

1

dd5

dd2

dd=

=+

+

+

B l ma trn dng bc thang c 2 dng khc 0 nn r(A) = r(B) = 2

V d 3. Tm m ma trn sau c hng b nht

=

m711

1311

3211

A

Gii: Ta bin i a ma trn A v dng bc thang

Ly dng 1 cng vo dng 2, dng 1 nhn vi (- 1) cng vo dng 3, ta c:

+

+

3m500

2500

321121

31

dd

dd

Nhn dng 2 vi (- 1) cng vo dng 3 ta thu c ma trn dng bc thang:

+

+

5m000

2500

321121

31

dd

dd

T ma trn dng bc thang, ta c r(A) nh nht bng 2 khi m 5 = 0 5m =

V d 4. Tm hng ca ma trnn

30

21

Gii : Ta c

=

30

21A l ma trn vung cp 2 nn An cng l ma trn vung cp 2. Theo

nh l nhn nh thc ta c det(An) = [det(A)]n = 3n 0. Nn r(An) = 2


34/94


34

Chuyn 2. H PHNG TRNH TUYN TNH V NG DNG

1. KHI NIM V H PHNG TRNH TUYN TNH

1. Cc khi nim

nh ngha 1. H phng trnh tuyn tnh gm m phng trnh n n l h c dng

11 1 12 2 1n n 1

21 1 22 2 2n n 2

m1 1 m2 2 mn n m

a x + a x + ... + a x = b

a x + a x + ... + a x = b

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

a x + a x + ... + a x = b

(I)

trong aij (i 1,m; j 1,n)= = , bi (i 1,m)= l cc s thc cho trc; x1, x2, , xn l n n s

cn tm; cc bi (i 1,m)= c gi l cc h s t do.

Nu h (I) c s phng trnh bng sn (m = n) th h (I) c gi l h vung.

Nu b1 = b2 = = bm = 0 th h (I) c gi l h phng trnh tuyn tnh thun

nht.

Nghim ca h (I) l mt b n s (c1, c2, , cn) sao cho khi thay th

x1 = c1, x2 = c2, , xn = cn vo (I) th ta c m ng nht thc. C th vit nghim di

cc dng sau: (c1, c2, , cn) hoc

n

2

1

c...

c

c

.

Gii h (I) l ta i tm tt c cc nghim ca h (I).

Ta gi ma trn

A =

11 12 1n

21 22 2n

m1 m2 mn

a a ... a

a a ... a

...

a a ... a

l ma trn cc h s ca h (I).

Ma trn


35/94


35

%A =

mmn2m1m

2n22221

1n11211

b:a...aa

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

b:a...aa

b:a...aa

c gi l ma trn b sung ca h (I).nh ngha 2. Hai h phng trnh c gi l tng ng nu chng c cng tp

nghim. Cc php bin i ca h phng trnh m khng lm thay i tp nghim ca

h c gi l php bin i tng ng ca h phng trnh.

Trong qu trnh gii h phng trnh, chng ta thng dng cc php bin i sau :

- i ch hai phng trnh ca h cho nhau

- Nhn hai v ca mt phng trnh vi mt s khc 0

- Nhn hai v ca mt phng trnh vi mt s tu ri cng vo phng trnh khc v

theo v.

Ch 1. Cc php bin i tng ng ca h phng trnh trn chnh l cc php bin

i scp v dng i vi ma trn b sung ca h.

2. Dng ma trn, dng vc tca h phng trnh tuyn tnh

t

X =

1

2

n

x

x...

x

: ma trn ct n, B =

1

2

m

b

b...

b

: ma trn ct h s t do

Khi h phng trnh tuyn tnh (I) c biu din di dng ma trn

A.X = B (II)

t n,..,1j;

a...

a

a

C

nj

j2

j1

j =

= l cc ct ca ma trn A. Khi h (I) c vit di dng

x1C1 + x2C2 + +xnCn = B (III)

Nh vy, mt h phng trnh tuyn tnh c th vit tng ng di dng ma trn hoc

dng vc t.


36/94


36

V d 1. Cho h phng trnh tuyn tnh 3 phng trnh 4 n

=+

=+++

=+

3xx3xx

2xx2xx2

1xxx2x

4321

4321

4321

H c cc n l x1, x2, x3, x4 nn ma trn n s l

=

4

3

2

1

x

x

xx

X ; ma trn v phi l

=

3

2

1

B .

Ma trn h s ca h l

=

1311

1212

1121

A ; cc vct ct ca A l

=

1

2

1

C1 ;

=

11

2

C2 ;

=

32

1

C3 ;

=

11

1

C4 ; ma trn b sung ca h :

=

1:13112:1212

1:1121

A~

Khi h c dng ma trn v dng vc ttng ng l

A. X = B ; x1C1 + x2C2 + x3C3 = B

3. H c dng tam gic, h c dng bc thang

nh ngha 3. H n phng trnh, n n c dng

=

=++

=+++

nnnn

2nn2222

1nn1212111

bxa

...bxa...xa

bxa...xaxa

vi a11a22ann 0

c gi l h c dng tam gic

Ma trn h s A ca h chnh l ma trn dng tam gic trn. D thy rng h ny c

nghim duy nht. H ny gii bng cch gii t phng trnh th n tm n xn, ri gii

phng trnh th n -1 tm n xn-1, , qu trnh c tip tc cho n khi tm c n

x1. Cch gii ny gi l gii ngc t di ln trn tm nghim ca h phng trnh.

V d 2. Gii h sau :

=

=

=+

1x

2x2x

1xx2x

3

32

321

Gii: T phng trnh th 3 ta c x3 = -1 ; thay vo phng trnh th 2 ta c x2 = 2 + 2x3

= 0 ; thay x2, x3 vo phng trnh th nht ta c : x1 = 1 + 2x2 x3


37/94


37

Vy h c nghim duy nht :

=

2

0

1

X

nh ngha 4. H r phng trnh, n n s (r < n) c dng

=++

=++++

=+++++

rnrnrrr

2nn2rr2222

1nn1rr1212111

bxa....xa

....

bxa...xa...xabxa...xa...xaxa

vi a11a22... arr 0

c gi l h dng bc thang

Ma trn b sung ca h khi s c bc thang

=

0:0

...

b:a

...

...

...

0

...

a

...

...

...

0

...

0

0

...

0..................

b:a...a...a0

b:a...a...aa

A~rrmrr

2n2r222

1n1r11211

Vi h c dng bc thang, t phng trnh th r, ta tnh xr thng qua cc n xr+1, xr+2, ... ,

xn. Ri thay vo phng trnh th r -1 tnh xr 1 theo cc n xr+1, xr+2, ... , xn qu trnh

trn c tip tc cho n x2, x1.

V d 3. Gii h phng trnh

==+

=++

1x2x2xx2x

1x2xx2x

43

432

4321

Gii: T phng trnh th 3, ta c x3 = 2x4 1, thay vo phng trnh 2 ta c

x2 = - 2(2x4 -1) + x4 + 2 = -3x4 + 4

Cui cng, thay vo phng trnh th nht thu c

x1 = -2x2 + x3 2x4 + 1 = - 2(-3x4 + 4) + 2x4 1 2x4 + 1 = 6x4 8

Vy h c nghim X = (6k 8; - 3k + 4; 2k -1); k tu .


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38

2. PHNG PHP GII H PHNG TRNH TUYN TNH

1. iu kin tn ti nghim

nh l 1. (nh l Kronecker-Capeli). H phng trnh tuyn tnh (I) c nghim khi v

ch khi r(A) = r( %A ).

V d 1. Tm m h sau c nghim

=++

=++

=+

=++

mxx6x8x3

3x2x4x5x2

2xx2x3x

1xxx2x

4321

4321

4321

4321

Gii Ta c ma trn b sung ca h l

=

m:1683

3:2452

2:1231

:!1121

A~

Bin i scp a ma trn b sung v dng bc thang

=+

+

+

1m:0000

4:2100

3:2110

1:1121

m:1683

3:2452

2:1231

1:1121

A~ 321

21

432

ddddd

ddd

Ty ta c h c nghim 1m01m)A~(r)A(r ===

2. iu kin duy nht nghim

nh l 2. H phng trnh tuyn tnh (I) c nghim duy nht khi v ch khi

r(A) = r( %A ) = sn (= n)

Chng minh

H c nghim duy nht tn ti duy nht cc s x1o, x2

o, ... , xno sao cho

x1o C1 + x2

oC2 + +xnoCn = B

{ }( ) n)A~(r)A(rnC,...,C,Cr n21 === (pcm)

V d 2. Tm m h sau c nghim duy nht

=++

=++

=++

2321

321

321

mxxmx

mxmxx

1mxxx


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39

Gii: H c s phng trnh bng sn v bng 3 nn c nghim duy nht

0A3)A(r =

Ta c 2)1m)(2m(11m

1m1

m11

A +==

Nn h c nghim duy nht

2m

1m

3. Phng php gii h phng trnh tuyn tnh

a) Phng php gii h Cramer

nh ngha 5. H Cramer l h n phng trnh tuyn tnh n n (h vung) c ma trn hs A khng suy bin (det(A) 0).

nh l 3. H Cramer lun c nghim duy nht.

Cng thc nghim: xj =A

j (j = 1, n)

trong , j l ma trn nhn t A bng cch thay ct th j bi ct h s t do.

V d 3. Gii h sau

=+=+

=+

4xx3 5xx

7x2x3x4

31

21

321

Gii

Ta c nh thc ca ma trn h s A l 07

103

011

234

A =

= nn h l h Cramer. Do

c nghim duy nht:A

x;A

x;A

x 312

11

1

=

=

=

M 28

403

511

734

;35

143

051

274

;0

104

015

237

321 ===

==

=

Vy h c nghim (x1= 0/7 =0; x2 = 35/7 = 5; x3 = 28/7 = 4)


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40

V d 4. Tm iu kin h sau c nghim duy nht

=++

=++

=++

2321

321

321

mmxxx

mxmxx

1xxmx

Gii

Nhn xt: Cc h trn l cc h c s phng trnh bng sn. Nn h c nghim duy

nht 0A

Ta c 2)1m)(2m(m11

1m1

11m

A +==

Nn h c nghim duy nht

1m

2m

b). Phng php gii h tng qut

Gi s ta gii h tng qut m phng trnh tuyn tnh, n n dng:

11 1 12 2 1n n 1

21 1 22 2 2n n 2

m1 1 m2 2 mn n m

a x + a x + ... + a x = b

a x + a x + ... + a x = b

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

a x + a x + ... + a x = b

Cch gii:

Tnh r(A), r( %A )

So snh r(A) vi r( %A ):

+ Nu r(A) r( %A ) th h (I) v nghim.

+ Nu r(A) = r( %A ) = sn (= n) th h (I) c nghim duy nht cho bi cng thc

Cramer

xj = A

j

(j =1, n)

+ Nu r(A) = r( %A ) = r < n th h (4.1) c v s nghim (hay cn gi l h v

nh):

Ch ra mt nh thc con csca ma trn A, gi s ta ch ra mt nh thc

con csl Dr. Khi h phng trnh cho tng ng vi h gm r phng trnh


41/94


41

ca h cho m c h s ca cc n to nn Dr. r phng trnh ny c gi l cc

phng trnh cbn ca h (I). r n ca h (I) c h s to thnh r ct ca Drc gi l

cc n cbn, (n r) n cn li c gi l cc n t do. Ta gii h gm r phng trnh

cbn v r n cbn bng cch chuyn (n r) n t do sang v phi v coi nh l cc

tham s, ta c h Cramer. Gii h Cramer , ta c cng thc biu din r n cbn

qua (n r) n t do.

c) Phng php Gauss

Ni dung ca php kh Gauss nh sau:

Ta s dng cc php bin i scp v hng a ma trn b sung %A v dng

ma trn bc thang.

Khi h phng trnh cho tng vi h phng trnh mi c ma trn h s

b sung l ma trn bc thang va thu c. Sau ta gii h mi ny t phng trnh

cui cng, thay cc gi tr ca n va tm c vo phng trnh trn v c tip tc

gii cho n phng trnh u tin ta s thu c nghim ca h phng trnh cho.

V d 5. Gii h

=++

=+

=+

=++

3x5x4x4x

1x3x2x2x3

2xxx3x

1x4x3xx2

4321

4321

4321

4321

Gii

Trc ht, tm hng ca ma trn h s v ma trn b sung

Ta c

=+

++

5:6570

5:6570

5:6570

2:1131

3:5441

1:3223

1:4312

2:1131

3:5441

1:3223

2:1131

1:4312

A~ 21

41

31

21 dd2

dddd3

dd

+

+

+

++

0:0000

0:0000

5:6570

2:1131

5:6570

5:6570

5:6570

2:1131

32

42

21

41

31

dd

dd

dd2

dddd3

Khi r(A) = r( A~

) = 2 < n = 4. Do h c v s nghim

Khi h cho

=+

=+

5x6x5x7

2xxx3x

332

4321


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42

Chn 4 2 = 2 n t do l x3, x4 v x1, x2 khi l n cbn.

H trthnh

Rx,x;

7 x6x55x

7

x11x81x

5x6x5x7

2xxx3x43

432

431

332

4321

+=

+=

=

+=

Vy h c v s nghim )R;(;;7

655;

7

1181X

++=

V d 6. Gii v bin lun h phng trnh sau

=++

=++

=++

1xxx

1mxmxx

1mxxmx

321

321

321

GiiNhn xt:y l h c 3 phng trnh, 3 n s c ma trn b sung l

=

1:111

1:mm1

1:11m

A~

Ta c 2)1m(

111

mm1

m1m

A ==

* Nu 1m th A 0 nn h l h Cramer nn c nghim duy nht.

Mt khc ta li c

23

22

21 )1m(

111

1m1

11m

;)1m(

111

m11

m1m

;)1m(

111

mm1

m11

======

Do , nghim ca h l

=

==

==

= 1

Ax;1

Ax;1

Ax 31

22

11

* Nu m = 1 khi ma trn b sung c dng

=

0:000

0:000

1:111

1:111

1:111

1:111

A~

Suy ra r(A) = 3n1)A~

(r =


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43

H x1 + x2 + x3 = 1 Rx,x;xx1x 21213 =

Vy h c nghim X = ( ) R,;;;1

4. H phng trnh tuyn tnh thun nht

nh ngha 6. H phng trnh tuyn tnh thun nht l h phng trnh tuyn tnh ccc h s t do bng khng, tc l h c dng

11 1 12 2 1n n

21 1 22 2 2n n

m1 1 m2 2 mn n

a x + a x + ... + a x = 0

a x + a x + ... + a x = 0

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

a x + a x + ... + a x = 0

(IV)

Nhn xt.

+ H (IV) lun c nghim l x1 = x2 = = xn = 0. Nghim ny c gi l

nghim tm thng ca h (IV).

+ H (IV) c nghim duy nht ( l nghim tm thng) r(A) = n.

+ H (IV) c nghim khng tm thng r(A) < n.

+ H thun nht vung (h c s phng trnh bng sn) c nghim khng tm

thng khi v ch khi det(A) = 0.

V d 1. Gii v bin lun h phng trnh sau

3x + y + 10z = 0

2x + ay + 5z = 0

x + 4y + 7z = 0

Gii

y l h thun nht vung. Ta c

det(A) =

3 1 10

2 a 5

1 4 7

= 11(a + 1)

+ Vi a -1, det(A) 0 nn h cho ch c nghim tm thung x = y = z = 0.

+ Vi a = -1, h cho trthnh

3x + y + 10z = 0

2x - y + 5z = 0

x + 4y + 7z = 0


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44

V3 1

2 -1= - 5 0 nn h cho tng ng vi h

3x + y + 10z = 0

2x - y + 5z = 0

3x + y = -10z

2x - y = -5z

5x = -15z x = -3z y = 2x + 5z = - z

Vy, vi a = -1, h cho c v s nghim l

x = -3t

y = -t

z = t

; t R

V d 2. Tm m h sau c nghim khng tm thng v tm cc nghim

=++=+

=+

0x)m1(x3x 0xmxx

0xx)m2(

321

321

21

Gii

Cch 1. Dng phng php Gauss

Ta c, ma trn h s l

+

+

=

2m3m5m30

m23m0

m131

01m2

1m1

m131

m131

1m1

01m2

A2

* Nu m + 3 = 0 3m = , ta c

3)A(r

100

240

m131

240

100

m131

A =

nn h ch c nghim tm thng

* Nu 3m ta c

+

m3

4m3m00

m3m210 m131

2m3m5m30m3m210 m131A

232

h c nghim tm thng th m3 3m2 4 = 0 2)A(r1m

2m=

=

= .


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45

Khi , h c nghim tm thng

+) Nu m = 2 th

000

010

131

A

Khi h1

1 2 3 *2

23

x x 3x x 0x 0;

x 0x

=+ =

= = =

+) Nu m = - 1 th

0004

310

131

A

Khi h*

1

1 2 3

2

2 3

3

x

4

x 3x x 0 3x ; 34x x 0

4 x

=

+ = = + = =

Cch 2.

V h c s phng trnh bng sn nn ta tnh nh thc ma trn h s A.

)1m()2m(

m131

1m1

01m2

A 2 +=

=

* Nu

1m

2m0A th h c nghim tm thng duy nht

* Nu

=

==

1m

2m0A . Khi h c nghim khng tm thng v s dng phng

php Gauss ta tm c nghim ca h.

V d 3. Tm nghim tng qut ca h sau

=

=+

=++

=+++

0xx3x2x

0xx4xx3

0x2xx3x2

0xx3x2x

4321

4321

4321

4321

Gii

Ta c ma trn h s cc n


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46

=++

++

0000

0000

4570

1321

0000

4570

4570

1321

1321

1413

2132

1321

A2121

41

31

dddd2

dddd3

Suy ra r(A) = 2 < n = 4 nn h c v s nghim. Chn x3, x4 lm n t do v x1, x2 l ncbn. Khi , h trthnh

=

+=++=

=++

=+++

432

434321

432

4321

x7

4x

7

5x

x7

3x

7

11xx3x2x

0x4x5x7

0xx3x2x(x3, x4 tu )

Vy nghim tng qut ca h R,;;;7

4

7

5;

7

3

7

11

+ .


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47

TON CAO CP 2

Chuyn 3. GII HN, LIN TC VI PHN, TCH PHN HM 1 BIN

1. GII HN HM S

I. Cc khi nim cbn v hm mt bin s

1) nh ngha:nh x f : X R

x a y = f x( ) vi , X R X c gi l hm s.

* Ta gi X l tp xc nh ca hm f x( ) , k hiu D(f); f(X) l tp gi tr ca hm f, k

hiu R(f); xD(f) gi l bin c lp.

V d 1: Hm y = 24 x c = =f fD [- , ], R [ , ]2 2 0 4 .

* Hm chn : Gi s X R , X nhn gc O lm tm i xng. Hm sf x( ) c gi lchn nu =f x f x( ) ( ) x X , l hm l nu = f x f x( ) ( ) x X.

* Ch : th hm s chn nhn trc tung lm trc i xng. th hm s l nhn gcta O lm tm i xng.

* Hm l : Hm sf x( ) c gi l hm s tun hon nu T > 0 sao cho

ff ( x T ) f( x ), x D+ =

S T nh nht sao cho c ng thc trn c gi l chu k ca hm sf x( ) .

V d 2: Cc hm s y = sinx, y = cosx l tun hon vi chu k 2 ; cc hm s y = tgx, y= cotgx l tun hon vi chu k .

Hm n iu: Hm y f x= ( ) c gi l tng (tng ngt) trn khong I Dfnu x1,

x2 I, x1 < x2 th f(x1) f(x2) (f(x1) < f(x2)); gim (gim ngt) trn I nu x1, x2 I th

(x1) f(x2) (f(x1) > f(x2)).

Hm s tng hoc gim trn I c gi l hm n iu trn I.

Hm b chn: Cho hm s f(x) xc nh trn X. Hm s f(x) c gi l b chn trnnu M : f( x ) M x X; b chn di nu M : f( x ) M x X; b chn

nu M : f( x ) M , x X.

2) Hm s hp:nh ngha.

Cho X, Y, Z R, cho hm s f : X Y, g: Y Z. Khi , hm s h: X Z c nh

ngha bi h(x):= g(f(x)), x X c gi l hm s hp ca hm s f v g. K hiu l:hay og[ f( x )] ( g f )( x ), x X .

V d 3. Xt cc hm s = + = +f( x ) x , g( x ) x .22 1 4 Khi :


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48

= + = + +

= + = + +

g[ f( x )] f ( x ) ( x )

f [g( x )] g( x ) ( x )

2 2

2

4 2 1 4

2 1 2 4 1

3) Hm s ngc:

* nh ngha. Cho hai tp X, Y R; cho hm s

f: X Y

x a y = f(x)

Nu tn ti hm s g: Y X tho mn:

+ (g o f)(x) = X1

+ (fo g)(y) = Y1

th g(x) c gi l hm s ngc ca hm s f(x).

K hiu: g = f-1

* Ch :

1. f: X Y l song nh g = f-1: Y X. Tc f c hm s ngc khi v ch khi fl song nh.

2. Nu hm s y = f(x) n iu nghim ngt th n c hm ngc

3. D 1f = Rf, R 1f = Df.

4. th ca hm ngc y = f-1(x) i xng vi th hm s y = f(x) qua ngphn gic ca gc th nht.

V d 4. Tm hm ngc ca hm y = 4 x

Gii. Ta c Df = [0, + ), Rf = [0, + ). Hm y =4 x l hm tng nghim ngt trn Df

nn n c hm ngc. Rt x theo y, ta c: x y , y= 4 0 , i vai tr ca vx y ta c:hm y 4 x= c hm ngc l y = x4, x 0.

4) Cc hm s thng gp:. Cc hm s scp cbn

* Hm s lu tha y = x, l mt s thc cho trc

f fD R; R R= =

* Hm s m: y = ax (a>0, a 1)*

f fD R; R R+= =

* Hm s logarit: y = logax ( a > 0 v a 1 )*f fD R ; R R+= =

* Cc hm s lng gic:+) Hm f(x) = sinx

[ ]f fD R; R 1,1= =

+) Hm y = cosx


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49

[ ]f fD R; R 1,1= =

+) Hm y = tgx

f fD x R / x k ,k Z , R R2

= + =

+) Hm y = cotgx{ }f fD x R / x k ,k Z ,R R= =

* Cc hm s lng gic ngc:

+) Hm y = arcsinx l hm ngc ca hm s y = sinx trn ,2

2

c

- Min xc nh Dy = [-1, 1]

- Min gi tr Ry = ,2

2

.

+) Hm y = arccosx l hm ngc ca hm s y = cosx trn [ ]0, c

- Min xc nh Dy = [-1, 1]

- Min gi tr Ry = [ ]0, .

+) Hm y = arctgx l hm ngc ca hm s y = tgx trong ,2

2

c

- Min xc nh Dy = R

- Min gi tr Ry = ,2

2

.+) Hm y = arccotgx l hm ngc ca hm s y = cotgx trong ( )0, c

- Min xc nh Dy = R - Min gi tr Ry = ( )0, .

* Hm s scp:

Ta gi cc hm s scp l nhng hm sc to thnh bi mt s hu hn ccphp ton s hc v php ton hp trn cc hm s scp cbn, v cc hng s.

5) Mt s hm s kinh t thng gp trong kinh t

* Hm cung v hm cu:

Cc nh kinh t s dng khi nim hm cung v hm cu biu din s ph thuc

ca lng cung v lng cu ca mt loi hng ha vo gi ca hng ha . Hm cung

v hm cu c dng:

Hm cung: QS = S(p)

Hm cu: QD = D(p)


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50

Trong p l gi hng ha; QS l lng cung: tc l lng hng ha ngi bn bng

lng bn mi mc gi; QD l lng cu: tc l lng hng ha ngi mua bng lng

mua mi mc gi.

Khi xem xt m hnh hm cung, hm cu ni trn ta gi thit rng cc yu t khc

khng i.

Hm sn xut ngn hn

Cc nh kinh t hc s dng khi nim hm sn xut m t s ph thuc ca

sn lng hng ha (tng s lng sn phm hin vt) ca mt nh sn xut vo cc yu

tu vo, gi l yu t sn xut.

Khi phn tch sn xut, ta thng quan tm n hai yu t sn xut quan trng l

vn v lao ng c k hiu tng ng l K v L.

V d: Hm sn xut dng Cobb Douglas vi hai yu t vn (K) v lao ng(L): = LaKQ

Hm doanh thu, hm chi ph v hm li nhun

Hm doanh thu l hm s biu din s ph thuc ca tng doanh thu (TR) vo sn

lng (Q): TR = TR(Q).

Hm chi ph l hm s biu din s ph thuc ca tng chi ph sn xut (TC) vo

sn lng (Q): TC = TC(Q).

Hm li nhun l hm s biu din s ph thuc ca tng li nhun ( ) vo snlng (Q): )Q(= . Hm li nhun c th xc nh bi )Q(TC)Q(TR = .

II. Gii hn ca dy s

1. nh ngha. Ta ni rng dy s {xn} c gii hn l a (hu hn ) nu vi mi s > 0

nh tu , tn ti mt s t nhin n 0 sao cho nx a < , vi mi n n 0

K hiu: nnlim x a

= hoc khinx a n +

Nu dy {xn} c gii hn l a (hu hn) th ta ni dy ny hi t va . Ngc li, nudy {xn} khng c gii hn, ta ni dy ny phn k.

V d 1: Xt dy s nx c, n= . Ta c n, x c c c n > = = < 0 0 Vy theo nh ngha

nlim c c

+= .

V d 2: ( )knlim k n+= >

10 0 v khin

nlim q q

+=


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51

Ch :+) sao chon n

nx A n N x A n n

= + > > 0 00lim ,

+) sao chon nn

x A n N x A n n

= + > < 0 00lim , , .

V d 3: Cho >k 0 ta c

khi

khik

n

A

A n A

+ >= 0, n0N* sao cho n

n0, k N* th n k nx x .+ <


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52

2. GII HN HM S1. nh ngha:nh ngha 1.Cho hm s f(x) xc nh trong khong (a, b). Ta ni rng f(x) c gi hn l A (hu hn)

khi x x0, nu vi mi dy {xn} trong (a,b)\{x0} m n 0x x nkhi th dy gi tr

tng ng {f(xn)} hi tn A.K hiu:

0xlim

xf(x) = A hay f(x) A khi x x0.

nh ngha 2.Cho hm s f(x) xc nh trong khong (a, b). Ta ni rng f(x) c gi hn l A (hu hn)

khi x x0, nu vi bt k > 0, tn ti s > 0 sao cho vi mi x (a, b) tho mn 0 0 v0x x

lim g(x)

=0x

limx

h(x) = A

th0x x

lim

f(x) = A.

Tnh cht 5. Nu ( )0 0;f(x) g(x), x x x , v i > 0 no + v

0 0x x x xlim f (x) A, lim g(x) B = = th A B Tnh cht 6. Nu

0 0x x x xlim f (x) A 0 lim g(x) Bv

= > = th [ ]0

g(x) B

x xlim f (x) A

= .

3. Gii hn mt pha

Khi nim x x0 cn c xt trong hai trng hp:

Th1. x x0, x > x0: tc l x dn n x0 t bn phi ( x x0+).

Th2. x x0, x < x0: tc l x dn n x0 t bn tri (x x0- ).


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53

Gii hn ca hm s f(x) khi x x0+ v khi x x0

-c gi tng ng l gii

hn bn phi v gii hn bn tri ca hm s ti im x0:

Gii hn bn phi:0x x

lim+f(x) =

0

0

x xx x

lim>

f(x);

Gii hn bn tri:0x x

lim

f(x) =0

0x xx x

lim 0), sinx, tgx l cc VCB khi x 0

Hm s tgx l VCL khi x2

b) So snh cc VCB

Cho f(x), g(x) l cc VCB khi x a. Gi sx a

f(x)lim k

g(x)=

i) Nu k = 0 th f(x) c gi l VCB bc cao hn so vi g(x) khi x a. K hiu

f(x) = 0(g(x)),

ii) Nu k 0 v hu hn th f(x) v g(x) c gi l nhng VCB cng bc khi x a. K

hiu f = 0*(g) khi x a.

c bit, khi k = 1, th f(x) v g(x) c gi l nhng VCB tng ng khi x a v vit: f(x) g(x), x a.

Nhn xt: Nu p q> > 0 th p qx x= 0( ) .

Nu [ ]f x h x= 0( ) ( ) , [ ]g x h x= 0( ) ( ) v a l hng s th

i) [ ]a f x h x= 0. ( ) ( )


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54

ii) [ ]f x g x h x = 0( ) ( ) ( )

iii) [ ]f x g x h x= 0( ). ( ) ( )

V d 3 : Vi k>0, khi x 0 ta c: k k k k p k pa x a x a x a x x+ + + ++ + + + =1 2 31 2 3 0... ( )

V d 4: Cc hm sinx v x l hai VCB tng ng khi x 0 v

0lim

x

sinxx

= 1.

V d 5: Cc hm tg2x v sinx l hai VCB cng bc khi x 0, v

0lim

x 0

2lim . .2 2

2 s

tg2xsinx inx

= =x

tg x x

x

c) nh l:

nh l 1. Hm s f(x) c gii hn l L khi x a khi v ch khi f x L x= + ( ) ( ) , vi

mi x( ) l hm v cng b khi x a.

nh l 2. Gis khi x x0, ta c cc cp VCB tng ng:(x) 1(x), (x) 1(x)

Khi , nux x

( x )lim

( x )

01

1

tn ti (hu hn hoc v hn) thx x x x

( x ) ( x )lim lim

( x ) ( x )

= 0 0

1

1

.V d 6. Tnh0

limx

2

3x + 2x

sinx + tg x

Gii. V x + 2x2 x, x 0 v sin2x + tg5 x sin2x, x 0. p dng nh l trn, ta c

0lim

x

25

x + 2xsin2x + tg x

=0

limx

2xsin2x

= 2.


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55

3. HM S LIN TC1. nh nghanh ngha 1. Cho hm s f(x) xc nh trn tp hp Df, Ta ni hm s f(x)

-Lin tc ti im x0Df nu:0

limx x

f(x) = f(x0).

- Lin tc phi ti x0 nu f(x0+)=0

lim+x x f(x) = .f(x0)

- Lin tc tri ti x0 nu f(x0-) =

0

limx x

f(x) = f(x0)

Hm s khng lin tc ti x0 th ta ni hm s gin on ti x .0

VD 1: Xt tnh lin tc ca hm sx khi x

f (x)x a khi x

+


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56

H qu: Nu Nu hm s f (x ) lin tc trn on [a;b] v f(a).f(b) < 0 th tn ti( )c a;b f( c) o.= Tc l phng trnh f( x ) o= c t nht mt nghim thuc ( )a;b .

O

f(a)


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57

4 O HM, VI PHN V NG DNGI. o hm v vi phn cp 1:1. Khi nima. o hm ti mt imnh ngha 1:

Cho hm y = f(x) xc nh trn khong (a, b) v x0 l im cnh thuc khong (a,

b). Nu tn ti gii hn hu hno

o

xx xx

)x(f)x(fLim

0

th gii hn gi l o hm ca hm

s y=f(x) ti im x0.

K hiu:f ( x )0 hoc y ( x )0

nh ngha 2 : Hm s f(x) c gi l hm kh vi ti im x0 nu tn ti s thc k sao

cho

f( x ) f( x ) k. x ( x )0 0 = + = + = + = + .

Khi biu thc k. x c gi l vi phn ca hm s ti x 0 ,K hiu l df( x )0 , tc l: df( x ) k. x0 = = = =

Ch :

Nu tn ti gii hnx x

f ( x ) f( x )lim

x x00

0 + + + +

th n c gi l o hm phi ti x0 k hiu

l:f ( x )0++++ .

Nu tn ti gii hnx x

f( x ) f( x )lim

x x00

0

th n c gi l o hm tri ti x0 k hiu

l:f ( x )0 .

c. o hm trn mt minnh ngha: Nu hm sy f( x )==== c o hm ti mi im thuc min X, th quy tc

cho tng ng mi gi tr x X , mt gi tr xc nh f ( x ) , cho ta mt hm s

y f ( x ) ==== xc nh trn X. Ta gi hm s ny l o hm ca hm sy f( x )==== trn

min X.

V d 1:o hm ca hm sy x 2==== l hm sy x.2====

o hm ca hm sy sin x==== l hm sy cos x.====

2. Tnh cht

nh l 1. Nu hm sy f( x )==== c o hm ti x 0 th n lin tc ti im .nh l2.o hm f ( x )0 l h s gc ca tip tuyn ca ng cong (C): y = f(x) ti

im M( x ;y )0 0 . V nh vy n cng l so dc ca ng cong (C) ti im

M( x ;y )0 0 .


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58

3. o hm v vi phn ca cc hm scp cbn: Gio trnh4. Cc quy tc tnh o hm v vi phna) nh l 1. Nu u = u(x), v = v(x) c o hm ti im x0 th ti im :

(u v) = u v ( )d u v du dv =

(ku) = k u d(ku) = kdu(uv) = uv + u v d(uv) = vdu + udv

'uv

=2

u v - uv

v

(v 0) d(

uv

) = 2vdu - udv

v(v 0)

V d 2: Tnh o hm cc hm s sau:

y x x x4 22 3 4 5= + += + += + += + +

y x ln x3==== ln x

yx

====

x sin x cos xy

x cos x sin x

++++====

b) o hm ca hm s hpnh l: Nu hm su g( x )==== c o hm ti x0 , hm sy f( u )==== c o hm ti

u g( x )0 0==== th hm sy f( g( x))==== c o hm ti x0 v v c tnh tho cng thc:

y ( x ) f ( u ).u ( x )0 0 ====

hay c th vit ngn gn l: x u xy f . u ====

V d 3: Tnh o hm ca y = cos4xHm s y = cos4x l hm hp ca hai hm cbn y = u4 v u = cosx.

Theo nh l trn ta c y = 4u cosx)( ) ( = 4u3. (-sinx) = -4cos3x.sinx

V d 4: ( )sin x sin x sin xe e .(sin x ) e .cos x = =

5. Tnh bt bin ca biu thc vi phnVi phn cp 1 bt bin qua php i bin

II. o hm v vi phn cp cao1) nh ngha

+) Nu hm s y = f(x) c o hm ti mi im thuc khong X th o hm y =

f(x) l mt hm ca i s x, xc nh trn khong X, do ta c th ly o hm ca

hm s y = f(x). o hm ca f(x) nu tn ti c gi l o hm cp hai ca f(x), k

hiu l y, hoc2

2

d y

dx, hoc f(x), hoc

2

2

( )d f x

dx.


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59

Tng qut, o hm ca o hm cp (n 1) ca y = f(x) c gi l o hm cp

n ca hm s. K hiu: y(n), hocn

n

d y

dx, hoc

( )nn

d f x

dx, hoc f(n)(x).

Nh vy: y(n) = f(n)(x) = [f(n 1) (x)]

V d 5. Tnh o hm cp n ca hm s y = sinx

Gii. Ta c y = cosx = cos(x + 1.2

)

y = -sin = sin(x + 2.2

)

Gi s, y(n 1) = sin(x + (n -1)2

). Ta tnh y(n)

y(n) = cos(x + (n -1)2

) = sin(x + (n -1)

2

+

2

) = sin(x + n

2

)

+) Vi phn cp hai ca hm s y = f(x) l vi phn ca vi phn ( cp 1) ca hm , tc ld2y = d(dy)

Tng qut, vi phn cp n ca hm s y = f(x) l vi phn ca vi phn cp n 1 cahm , tc l

dny = d(dn 1y)

Ch .

* Trong cng thc vi phn dy = ydx, o hm y ph thuc vo x, cn

dx = x l s gia bt k ca bin c lp x, khng ph thuc x. Do , khi ly vi phn

theo x th dx c xem nh l hng s. Ta c

d2y = d(dy) = d(ydx) = dxdy = dx(ydx) = ydx2, dx2 = (dx)2d3y = d(d2y) = d(ydx2) = dx2d(y) = dx2(ydx) = ydx3, dx3 = (dx)3

Bng phng php quy np, ta c th chng minh cng thc tnh vi phn cp n

ca hm s y = f(x): dny = y(n)(dx)n = y(n)dxn* Biu thc vi phn cp cao khng c tnh cht bt bin nh biu thc vi phn cp

1.V d 6. Cho hm s y = x2

Nu x l bin c lp th d2y = 2dx2Nu x = t2, th y = t4 . Khi d2y = 12t2dt2

Nu thay dx = 2tdt vo biu thc d2y = 2dx2, ta c d2y = 2(2tdt)2 = 8t2(dt)2 = 8t2dt2 12t2dt2

2) Cc quy tc tnh o hm v vi phn cp caoa. Quy tc tnh o hm

* (u v)(n) = u(n) v(n)* (ku)(n) = ku(n)


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60

* (uv)(n) = (n

k k) (n - k)n

k = 0C u v (Cng thc Leibnitz)

b. Quy tc tnh vi phn* dn(u v) = dnu dnv

* dn

(ku) = kdn

u

* dn(uv) =n

k n - k kn

k = 0C d ud v (Cng thc Leibnitz)

* o hm cp cao ca mt s hm thng gp:

1)

2) .

3) .

4) ).5) .

= +

=

=

= +

= +

( n ) -n

ax ( n ) n ax

( n ) n-n

( n ) n

( n ) n

( x ) ( )...( n )x .

( e ) a e

( n - ) !(ln x ) (- )

x

(sin bx ) b sin( bx n

(cos bx ) b cos( bx n )

1

1 1

11

2

2

III. ng dng ca php tnh vi phnQuy tc Lpitan:nh l: Gi s cc hm s f(x) v g(x) tho mn cc iu kin sau:

i) Gii hn0

( )lim

( )x x

f x

g xc dng v nh

0

0hoc

(

0x xlim

f(x) =

0x xlim

g(x) =0 hoc

0x xlim

f(x)=

0x xlim

g(x)= );

ii) Tn ti gii hn

0x x

f ( x )limg ( x )

(hu hn hay v hn).

Khi :

=

0 0x x x x

f( x ) f ( x )lim lim

g( x ) g ( x )

V d 1. Tnh2

2

2 -lim

- 2=

x

x

xA

x

Gii. Gii hn ny c dng

0

0. p dng Quy tc Lpitan ta c:

( )

( )

2

2 2

2 2 2 24 2 4

12

= = =

x x

x x

- x ln - xA lim lim ln

x -

V d 2: Tnh+

=x

x

eB lim

x


61/94


61

Gii hn ny c dng

. p dng quy tc Lpitan ta c:

( )+ +

= = = +

xx

x x

eB lim lim e

x

V d 3: Tnh 30=

xtgx xC lim

x

Gii hn ny c dng

0

0. p dng Quy tc Lpitan ta c:

22

2 2 20 0

11

1 1

33 3

= = =x x

sin xcos xC lim limx cos x x

Ch .a) Ta c th p dng quy tc Lpitan nhiu ln, tc l:

Nu mt ln cn no ca im a, cc hm s f(x) v g(x) c o hm n cp n

v cc gii hnx a

f(x)lim

g(x),

x a

f '(x)lim

g'(x),

x a

f ''(x)lim

g''(x),...,

( )

( )

n 1

n 1x a

f (x)lim

g (x)

c dng v nh

0

0( hoc

). Khi , nu tn ti gii hn

( )lim

( )

(n)

(n)x a

f x= A

g xth ta cng c

( )lim

( )x a

f x= A

g x

V d 4. Tnh0

limx -x

x

e - e 2xx - sinx

Gii. Xt f(x) = ex e-x 2x, g(x) = x sinx. Ta cf(x) = ex + e-x 2, g(x) = 1- cosx

f(x) = ex e-x, g(x) = sinx

f(x) = ex + e-x, g(x) = cosx

Ta thy0

limx

f(x)g(x)

,0

limx

f (x)g(x)

=

0limx 1

x -xe + e - 2- cosx

,0

limx

f (x)g (x)

=

0limx s

x -xe - einx

c dng

0

0,

0limx

f (x)g (x)

=

0lim

x -x

x

e + ecosx

= 2. Vy0

limx

f(x)g(x)

=0

limx

f (x)g (x)

= 2.

b) Quy tc Lpitan vn ng nu cc gii hn c xt trong cc nh l trn l gii hn

mt pha v trng hp khi x , hoc x .

V d 5. Tnh0

limx

xlnx+


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62

Gii. Ta c xlnx =ln x1x

Ta thy khi x 0 th lnx - . p dng quy tc Lpitan, ta c:

0limx xlnx+ = 0limx +lnx1x

= 0limx + (-x) = 0.

c) Trng hp

0x x

f ( x )lim

g ( x )khng tn ti th cha th kt lun g v s tn ta ca gii hn

0

( )lim

( )x x

f x

g x.

V d 6: Xt gii hn+

+

x

sin x xlim

x

Ta thy ( )( )

( )+ = +

1sin x x cosxx khng c gii hn khi

+x .

Trong khi + +

+ = + =

1 1x x

sin x x sin xlim lim

x x

d) Ngoi vic dng quy tc Lpitan kh cc dng v nh0

0v

, quy tc Lpitan

cn c dng kh cc dng v nh khc nh - , 0, , 00, 1, 0 bng cch

chuyn v hai dng v nh trn.

V d 7. Tnh1

1 1lim -

ln - 1

x x x

(c dng v nh - )

Gii. Ta c

1

1 1

1

x

lim -ln x x -

=1

1

1x

x - - ln xlim

( x - )ln x=

x 1

11

1 +

-xlimx -

ln xx

=1

1

1 +x

x -lim

x ln x x -=

1

1

1 1 + +xlim

ln x=

1

2.

V d 8. Tnh

1

x

2+

ln xlim - arctgx (c dng v nh 00)

Gii.t y =

1ln-

2

x arctgx , ta c

lny =ln( )

2lnln

=

arctgx

yx

(c dng v nh

)


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63

Theo quy tc Lpitan, ta c

xlim ln+

y = limx+

21 1

.1-

21

+ xarctgx

x

= limx+

21

2

+

+

x

x

arctgx= lim

x+

2

2

2

1 -

(1 )1

1

+

+

x

x

x

=2

2

1 -lim 1

1+=

+x

x

x.

T suy ra -1lim+

=x

y e = 1e

V d 9. Tnh2

0

cot g x

xlim(cos x ) (dng ( )1 .

Gii. Trong trng hp ny:

y = (cosx)cotg2

x

lny = cotg2x.ln(cosx)Ta c

0lim (ln )x

y

=0

limx

cotg2x.ln(cosx) =0

limx

ln(c

x2osx)

tg(dng

0

0) =

0limx

2

tgx1

2tgx.cos x

=0

limx

2 1

2 2

=

cos x

V d 10. Tnh1x

x +l im x

(dng 0 )

Gii. Ta c:

y =1xx lny =

1

xlnx lim

x+(lny) = lim

x+

lnxx

(dng

) =

1lim

x x+= 0.

1x

x +l im x

= limx+

y = e0 = 1.

V d 11. Tnhx 0lim

+xlnx ( > 0) (dng 0. ).

Gii. Ta c

x 0lim

+xlnx =

x 0

ln xlim

1x

+= 10

2

1

+ xxlimx

x

=

x 0

xlim

+ = 0.


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64

5. TCH PHN HM MT BIN S

I. Tch phn bt nh:1. Khi nim nguyn hm* nh ngha:

Hm s F(x) c gi l nguyn hm ca hm s f(x) trn mt khong (a, b) nu:F(x) = f(x), hay dF(x) = f(x)dx, ( , )x a b

V d 1. Hm s sinx l nguyn hm ca hm s cosx trn R, v (sinx) = cosx , vi mis thc x.

nh l:Nu F(x) l mt nguyn hm ca f(x) trn khong (a, b) th:

i) F(x) + C, vi C l mt hng s bt k, cng l nguyn hm ca hm s f(x).

ii) Ngc li, mi nguyn hm ca hm s f(x) u biu din c di dng F(x) +C, vi C l mt hng s.

2. Tch phn bt nha) nh ngha:Tch phn bt nh ca hm s f(x) l biu thc nguyn hm tng qut F(x) + C, trong F(x) l mt nguyn hm ca hm s f(x) v C l hng s bt k.K hiu:

( ) ( )= +f x dx F x C

V d 2: += Cxcosxdxsin b)Tnh cht:

* ( )

'

f (x)dx f (x)= hay ( )d f (x)dx f (x)dx= ;* '( ) ( )F x dx F x C = + hay ( ) ( )dF x F x C = + ;

[ ]f (x) g(x) dx f (x)dx g(x)dx =

* kf (x)dx k f (x)dx, k= l hng s.c) Cng thc tch phn cbn:

1) kdx = kx + C

2) x dx =

1

, khi -11ln | | , khi = -1.

+

+ + +

x

C x C

3) xa dx =xa

lna+ C (a > 0, a 1)

c bit xe dx = ex + C


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65

4) sin xdx = -cosx + C

5) cos xdx = sinx + C

6)2

dx

cos

x= tgx + C

7)2

dxsin x

= -cotgx + C

8)2

dx

1 + x = arctgx + C = -arccotgx + C

9)2

dx

1 x = arcsinx + C = - arccosx + C

10)( )d cos x

tgxdx ln | cos x | Ccosx

= = +

11)

dsinx

cotgxdx sinx= = ln|sinx| + C3. Cc phng php tnh tch phna) Phng php bin i: bin i, nhn lin hpb) Phng php i bin: Hai dng i bin sb) Phng php tch phn tng phn:Cng thc tch phn tng phn: Cho u(x) v v(x) l cc hm s c o hm lin tc.

Khi , ta c

= udv uv vdu Cc dng tch phn tng phnd) p dng: S dng cc phng php trn, ta c th d dng tnh c cc tch phncbn sau:

V d 3.2 2

1 x= arctg + C

a a+ x

dx

a, (i bin x = atgt).

V d 4.2 2

1 + x= ln

2a a - x- x

dx a

a+ C.

II. Tch phn cc hm thng dng1. Tch phn ca cc phn thc hu tnh ngha. Mt phn thc hu t l mt hm s c dng:

R(x) = 0 1 m

0 1 n

b + b + ... + b( )=

( ) + a + ... + a

m

n

x xP x

Q x a x x

vi ai, bi R v an, bn 0.

- Nu m < n th R(x) c gi l phn thc thc s.- Nu m n th R(x) c gi l phn thc khng thc s.


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66

Nu R(x) khng l phn thc thc s th bng cch chia t cho mu ca R(x) thbao gita cng c th biu din R(x) di dng:

R(x) = f(x) + R1(x)trong f(x) l mt a thc, cn R1(x) l mt phn thc thc s.

a) Tch phn cc phn thc n gin:

Dng 1: I =ax

dx 1= ln|ax + b| + C+ b a

, a 0

Dng 2: I =dx

.(ax k k - 1

1 1= + C

a(1 - k)+ b) (ax + b)( a 0 , k = 2, 3, )

Dng 3: I =2

dx

x 2- a=

1 xln

2a

- a+ C (a 0)

x + a>

Dng 4: I =2

dx

x 2+ a=

1

a

xarctg + C (a 0)

a>

Dng 5: I =2

1dx

ax + bx + c=

2

2

dtk ,

tdu

,u

22

22

nu = b - 4ac 0-

nu = b - 4ac 0+

0, t 2ax + bx + c = t - x a hoc t + x a

+) Nu c > 0, t

2

ax + bx + c = tx + c hoc tx - c +) Nu tam thc ax 2 + bx + c c hai nghim phn bit x1, x2. Khi

ax 2 + bx + c = a(x x1)(x x2)

th ta t 2ax + bx + c = t(x x1).

V d 4. Tnh tch phn I = 2x dx+ 2x + 2


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70

Gii.t 2x + 2x + 2 = t x x2 + 2x + 2 = (t x)2

x =2t - 2

2(t + 1) dx =

2(tdt

2(t 2+ 2t + 2)

+ 1)

Khi :

I = 2x dx+ 2x + 2 = [t dt2(t

2 2

2t - 2 (t + 2t + 2)

- ]2(t + 1) + 1)

=41 (t

dt4 3

+4)

(t + 1)

S dng ng nht thc:t4 + 4 = [(t + 1) - 1]4 + 4 = (t + 1)4 4(t + 1)3 + 6(t + 1)2 4(t + 1) + 5

Ta tnh c I

f) Dng R(x, )dx2 2a x+ , vi a > 0

t x =

atgt,

- t 0-x - a + -x - b, nu x + a < 0 v x + b < 0.

3. Tch phn ca mt s biu thc lng gic

a) Xt tch phn I = R(s inx, cosx)dx +) Phng php chung:

t t = tgx

2( - < x < ) x = 2arctgt, dx =

2dt

1 2+ t

Sau , ta c th p dng cc cng thc

sinx = 2t1 2+ t

, cosx = 11

2

2- t+ t, tgx = 2t

1 2- t

V d 1.

a) Tnh I =dx

1 + sinx + cosx

b) Tnh I =dx

3cosx - 5

+) Mt s trng hp c bit:- Nu R(-sinx, cosx) = -R(sinx, cosx) th t t = cosx

-Nu R(sinx, -cosx) = -R(sinx, cosx) th t t = sinx-Nu R(-sinx, -cosx) = R(sinx, cosx) th t t = tgx

V d 2: Tnh

a)I = 4sin x.c xdx3os

b) I =2

dx

sin x 2s x2inx.cosx-cos+(t t = tgx)

b) Cc tch phn dng: sinax.cosbxdx, cosax.cosbxdx, sinax.sinbxdx D dng tnh c cc tch phn bng cch bin i tch thnh tng:

sinax.cosbx = 1 (sin(a2

+ b)x + sin(a - b)x)

cosax.cosbx =1

(c (a2

os + b)x + cos(a - b)x)

sinax.sinbx =1

(c (a2

os - b)x - cos(a + b)x)


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72

V d 3. Tnh sin2xcos5xdx .

c) Dng msin xc xdxnos ; m, n N, m, n chn v m2 + n2 0.

Ta s dng cc cng thc h bc:

sin2x =1 - cos2x

2

cos2x =1 + cos2x

2.

V d 4. Tnh I = 2sin x.c xdx4os

III. Tch phn xc nh

1. nh ngha = b

a

I f( x )dx

Khi , f(x) c gi l kh tch trn [a, b] v vit f K[a, b].

* Ch : Tch phn xc nh ca mt hm s f(x) kh tch trn [a, b] l mt s xc nh(trong khi tch phn bt nh ca f(x) l hm s ca bin s x). Do , tch phn xc nh

khng ph thuc vo bin s di du tch phn (ta c th dng ch s bt k thay cho

x): ( )b

a

f x dx =b

a

( ) = f(t)dt = ... b

a

f u du

*nh l tn ti tch phn xc nh.

Nu f(x) lin tc trn on [ ]a;b th f kh tch trn [ ]a;b

2. Cc tnh cht cbn ca tch phn xc nh:TC1: Nu hm s f(x) kh tch trn on [a, b] th

b

a

f(x)dx = -a

b

f(x)dx

TC2: Nu f(x) kh tch trn on cha c ba im a, b, c thb c b

a a c

f (x)dx f (x)dx f (x)dx= +

TC3: Nu cc hm s f(x), g(x) kh tch trn on [a, b] thb b

a a

[ ( ) g(x)]dx = f(x)dx g(x)+ + b

a

f x dx ,

Nu hm s f(x) kh tch trn on [a, b] thb b

a a

kf (x)dx k f (x)dx=


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TC4: Cho f, g kh tch trn [a, b], nu f(x) g(x), x [a, b] thb

a

f(x)dxb

a

g(x)dx.

TC5: (nh l v gi tr trung bnh) Nu f(x) lin tc trn [ ]a;b , th c (a, b) :

b

a

f(x)dx = f(c)(b - a)

3. Cch tnh tch phn xc nha) Mi lin h gia tch phn xc nh v nguyn hm

Cho f(x) lin tc trn [a, b] x [a, b], f(t) cng kh tch trn [a, x]

(x)x

a

= f(t)dt, x [a, b]

Vi mi x X, tch phn (x) l mt s xc nh, do (x) l mt hm s ca

cn trn x. Ta gi hm s ny l hm cn trn.

nh l 1. (nh l vo hm ca hm cn trn)Nu f lin tc trn khong X v a Xth ti mi im x X ta c:

' (x) =

'x

a

f(t)dt

= f(x).

b) Cng thc Newton-Leibnitznh l 2. Nu hm f(x) lin tc trn [a, b] v F(x) l mt nguyn hm ca f(x) trongon , th

b

a

f(x)dx =ba

F(x) = F(b) F(a)

Mt s tnh chtChng minh rng

TC1: Nu f kh tch trn [-a, a] v l hm s chn th:a a

a 0

f (x)dx 2 f (x)dx =

Nu f kh tch trn [-a, a] v l hm s l th:a

af (x)dx 0

=

TC2: Nu f l hm lin tc trn R v tun hon vi chu k l T tha

a

+ T T

0

f(x)dx = f(x)dx

TC3: Nu f kh tch trn [-, ] v l hm chn th


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x

f(x)dx

a 0= f(x)dx

+ 1

, a > 0.

TC4: Nu f [0, 1]

2

0

f (s2

0

inx)dx = f(cosx)dx .

TC5: Nu f kh tch trn [a, b] v f(a + b x) = f(x) thTC6: Nu f kh tch trn [a, b] v f(a + b x) = -f(x)

b

a

f(x)dx = 0.

TC7: Nu f kh tch trn [0, 2a] vi a > 0

2a

0

f(x)dx =a

0

[f(x) + f(2a - x)]dx .

4. Cc phng php cbn tnh tch phn xc nh

a) Phng php bin ib) Phng php i bin s

Tnhb

a

f(x)dx , vi f(x) lin tc trn [a, b]

Dng 1.i bin x = (t), vi gi thit:

* (t) c o hm lin tc trn [, ]

* [, ] [a, b]

* ( ) = a, ( ) = b

Khi b

a

f(x)dx (t)dt'= f[ (t)]

V d 1. Tnha

2 2

0

x a dx2- x

Gii. t x = asint dx = acostdt

Vi x = 0 t = 0

Vi x = a t =

2

a2 2

0

x a dx2- x =

2

2 2

0

(a sin t)(ac tc tdt2

4 2 2

0

ost)(acostdt) = a sin os

=4a

4

2

2

0

sin 2tdt =4a

8

2

0

(1 dt- cos4t) =4a

8

2

0

sin4t)1

(t -4

=4a

16


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Dng 2.i bin t = (x), vi gi thit

* (x) n iu nghim ngt v c o hm lin tc trn [a, b]

* f(x)dx tr thnh g(t)dt, trong g(t) l mt hm s lin tc trong [ (a), (b) ]

hoc [ ](b), (a) th

b

a f(x)dx

(b)

(a)= g(t)dt

V d 2. Tnh I =1

21

dx

x - 2xcos + 1 (0 < )

Gii.1

21

dx

x - 2xcos + 1 =

1

21

dx

(x ) 2

- cos + sin

t t = x - cos dt = dx,

Vi x = -1 t = -1 - cos, x = 1 t = 1- cos

Vy I =1

1 - cos

2 2- cos

dtt + sin

=

1

1 - cos

- cos

1 tarctg

sin sin =

1

sin (arctg

1

- cossin

+ arctg1

+ cossin

)

V1

sin

- cos=

2 2sin2

sin = tg

2,

1

+ cossin

=

2c

2sin

2os= cotg

2

I = 1sin

( 2

+ -2 2

) = 2sin

c) Phng php tch phn tng phn:Cho u(x), v(x) l cc hm s c o hm lin tc trn [a, b]. Khi , ta c

b

a

udv =ba

uv -b

a

vdu

V d 3. Tnh I =3

0

dxx

arcsin1 + x

Gii. t u = arcsin x1 + x

, dv = dx du = 12 x(1 + x)

dx, v = x

p dng cng thc tch phn tng phn, ta c

I =3

0

xx

arcsin1 + x

-3

0

xdx

2(1 + x) x= - 23

20

( x ) d x

1 )+ ( x

= -3

0

d x +3

20

d x

1 )+ ( x


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= -3

0x +

3

0arctg x = - 3 +

3=

4

3- 3


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Chuyn 4. PHP TNH VI PHN HM NHIU BIN S V NG DNG

1. GII HN HM NHIU BIN SI. Cc khi nim1 nh ngha:

nh ngha 1: Xt khng gian Euclide n chiuRn = {M = (x1, x2, ..., xn)|xi R, i = 1, n}

Ly M0 = (x 01 , x02 , ..., x

0n ), M = (x1, x2, ..., xn), N(y1, y2, ..., yn) R

n, D Rn

Khong cch gia hai im M v N: d(M, N) =n

2i

i 1

(x )i- y=

Cho dy im {Mk (xk1 , x

k2 , ..., x

kn )} D. Ta ni dy im {Mk} dn ti M0 khi k

, nu kklim d(M , ) 00M

=

K hiu: kklimM = M0, hay Mk M0, k Ta c

Mk M0 d(Mk, M0) =n

k 2i

i 1

(x )0i- x=

0 k 0i ix x , i = 1,n .

Hnh cu tm M0, bn knh r (r > 0) trong Rn, k hiu l S(M0, r):

S(M0, r) = {M Rn: d(M0, M) < r}

S(M0, r) cn c gi l r_ln cn ca im M0.

Mi tp trong Rn cha mt r_ln cn no ca im M0c gi l mt ln cnca im M0.

M0c gi l im bin ca D nu mi ln cn ca im M0 va cha nhngim thuc D v va cha nhng im khng thuc D.Ch .im bin ca tp D c th thuc D, cng c th khng thuc D.

Tp hp tt c nhng im bin ca D c gi l bin ca n.

Tp D Rnc gi l tp ng nu n cha mi im bin ca n.

im M D c gi l im trong ca D nu r > 0, S(M, r)

D.

Tp hp D c gi l mnu mi im ca n u l im trong.

nh ngha 2. Cho D Rn, nh x f : D R

xc nh bi M = (x1, x2, ..., xn) D a f(M) = f(x1, x2, ..., xn)c gi l hm s ca n bin s xc nh trn D.

+) D c gi l min xc nh ca hm s f

+) x1, x2, ..., xnc gi l cc bin sc lp.

+) {f (D) , , ) : , , )1 2 n 1 2 nR| (x x ..., x D f(x x ..., x = }= c gi l min gi tr

ca f.


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V d 1. Hm s )2 2 2f(x, y, z) = 1 - (x + y + z c tp

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Bai Giang Toan Kinh Te Quang 2012