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ThS PHNG DUY QUANG (ch bin)
BI GING N THI CAO HC
Mn: TON KINH T
H NI, 2011
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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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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:
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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
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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 +=
+
+
+
+
+
=
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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 +=
+=
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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.
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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:
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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
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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
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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
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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
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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:
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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
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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
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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
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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
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%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.
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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
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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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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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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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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
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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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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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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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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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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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=++
++
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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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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= + = + +
= + = + +
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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[ ]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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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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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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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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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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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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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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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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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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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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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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* (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
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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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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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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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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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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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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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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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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