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1-tipik hisob 1-kurs sirtqilar uchun-20-21 (кузги семестр)
- Bu sahifa navigatsiya:
- 4-misol.
a) Kramer usuli. (1) chi sistemaning determinantlarini topamiz: asosiy determinant: 3 -1 2 -2 1 1 = -9-1+8-2+6+6=8 0 ; 1 -2 -3
o„zgaruvchining determinanti: 36
11 -1 2 -6 1 1 = -33-5+24-10+22+18=16;
5 -2 -3 o„zgaruvchining determinanti:
3 11 2 -2 -6 1 = 54+11-20+12-15-66=-24;
1 5 -3
o„zgaruvchining determinanti:
3 -1 11 -2 1 -6 = 15+6+44-11-36-10=8.
1 -2 5 Kramer formulasiga asosan (1) ning echimi: 1 8 8 , 3 8 24 , 2 8 16 bo„ladi. b) Gauss usuli:
11 2 3 ,
6 2 ,
5 3 2 .
Sistemaning ikkinchi va uchinchi tenglamalaridagi larni yo„qotish uchun birinchi tenglamani 2- ga, ikkinchi tenglamani 3-ga ko„paytirib, ikkinchi tenglama bilan uchinchi tenglamani 3 ga ko„paytirib, birinchi tenglama bilan mos ravishda qo„shamiz:
11 2 3 ,
4 7 ,
(2)
4 11 5 . (2) sistemadagi uchinchi tenglamadan ni yo„qotish uchun ikkinchi tenglamani 5 ga ko„paytirib, uchinchi tenglama bilan qo„shamiz:
11 2 3
4 7
(3)
24 24 (3) sistemaning uchinchi tenglamasidan 1
ni topamiz, so„ngra uni ikkinchi tenglamaga qo„yib, =-3 va birinchi tenglamaga qo„yib, = 2 ni hosil qilamiz.
Javob:
= 2,
=-3,
1 , .
37
Ma‟lumki, AX= В matritsali tenglamaning echimi X=A -1 В bo„ladi. (1) chi sistema uchun A, X va В lar quyidagicha bo„ladi:
5 6 11 , , 3 2 1 1 1 2 2 1 3 B X A .
A -1 teskari matritsani aniqlaymiz. YUqorida A matritsaning determinanti det A= hisoblangan edi: ya‟ni 0 8 . Endi teskari matritsaning algebraik to„ldiruvchilari j i A - ni topamiz: , 1
3 3 2 1 1 11
, 5 1 6 3 1 1 2 12 A
, 3 1 4 2 1 1 2 13
, 7 4 3 3 2 2 1 21 A
, 11 2 9 3 1 2 3 22 A
, 5 1 6 2 1 1 3 23 A
, 3 2 1 1 1 2 1 31 A
, 7 4 3 1 2 2 3 32
.
2 3 1 2 1 3 33 A
Natijada, quyidagi teskari matritsa hosil bo„ladi:
. 1 5 3 7 11 5 3 7 1 8 1 1 A
U holda echim: 1 3 2 8 24 16 8 1 5 6 11 1 5 3 7 11 5 3 7 1 8 1 1 B A X
Javob: . 1 , 3 , 2
3-misol. 4 3 5 0 x y z tekislik bilan 1 2 3 3 1 4 x y z to‟g‟ri chiziqning kesishish nuqtasining koordinatalarini toping.
: 0 ( ) D Ax By Cz D a to‟g‟ri chiziqning kanonik tenglamasini parametrik ko‟rinishdagi 1 1 1 : , , ( )
L x x mt y y nt z z pt b tenglama ( ) a ga qo‟yish natijasida 1 1 1 ( ) ( ) 0 Am Bn Cp t Ax By Cz D ko‟rinishdagi tenglama hosil bo‟ladi. Bundan 1 1 1 Ax By Cz D t Am Bn Cp (1) 38
Agar 0 Am Bn Cp bo‟lsa, to‟g‟ri chiziq bilan tekislik kesishadi. U holda (1) ni ( )
natijasida kesishish nuqtasining koordinatalari topiladi. Agar 1
1 0 0 Am Bn Cp Ax By Cz D bo‟lsa
L D
bo‟ladi. Agar
1 1 1 0 0
Bn Cp Ax By Cz D bo‟lsa, u holda to‟g‟ri chiziq tekislikda yotadi. Berilgan misolda (17) formula yordamida
parametrni topish mumkin: 1 2
1 3 , 2 , 3 4 ( )
3 1 4 x y z t x t y t z t b
aniqlab olamiz: 4 ( 1)
( 3) 2 1 3 5 12 4 3 3( 1) 1 4 19 t
( ) b dan M nuqtaning koordinatalarini topamiz.
36 17 1 3 1 19 19 x t
26 2 19 19 y
48 105 3 19 19 z
17 26 105 ; ; 19 19 19 M
4-misol. (3; 4; 2) M nuqtadan o‟tib, 3 4 2 2 3 6 x y z to‟g‟ri chiziqqa parallel bo‟lgan to‟g‟ri chiziq tenglamasini tuzing. Yechish. (3; 4; 2) M nuqtadan o‟tuvchi to‟g‟ri chiziqning kanonik tenglamasi
3 4 2 x y z m n p Bu to‟g‟ri chiziq berilgan to‟g‟ri chiziqqa parallel bo‟lgani uchun ikki to‟g‟ri chiziqning parallellik shartiga asosan: 2, 3, 6 m n p bo‟lishi kerak. U holda to‟g‟ri chiziq tenglamasi 3 4
2 3 6 x y z
. 8
3 6 13 5 lim
2 2 2 x x x x x ko‟rsatilgan limitlarni toping
7 , 0 10 7 4 3 3 5 lim
) 4 3 )( 2 ( ) 3 5 )( 2 ( lim 8 2 3 6 13 5 lim
2 2 2 2 2 x x x x x x x x x x x x x b) . 64 5 21 lim 3 4 x x x ko‟rsatilgan limitlarni toping Yechish: ) 5 21 )( 64 ( ) 5 21 )( 5 21 ( lim
64 5 21 lim 3 4 3 4
x x x x x x x
) 5 21 )( 16 4 )( 4 ( 25 21 lim 2 4
x x x x x
. 480 1 ) 5 21 )( 16 4 ( 1 lim
) 5 21 )( 64 ( 4 lim
2 4 3 4
x x x x x x x
c) . 3 2 2 lim 5 2
x x x
39
x x x x x x x x 5 2 5 2 1 3 2 2 1 lim
3 2 2 lim
x x x x x x x 5 2 5 2 3 2 3 1 lim 3 2 3 2 2 1 lim
. lim
3 2 3 1 lim
2 / 15 ) 3 2 /( ) 5 2 ( 3 ) 3 2 /( ) 5 2 ( 3 3 / ) 3 2 (
e x x x x x x x x d) 1 arcsin
1 lim
3 1 x x x ko‟rsatilgan limitlarni toping Yechish: 3 1 1 arcsin
1 lim
1 arcsin
1 1 lim 1 arcsin
1 lim
2 1 2 1 3 1
x x x x x x x x x x x x 6-misol. a) . ) 1 ( 6 ) 1 3 2 ( 3 4 3 2
x x y Berilgan funksiyaning hosilasini toping Yechish. . ) 1 ( 18 1 3 2 3 4 4 3 ) 1 )( 3 ( 6 ) 3 4 ( ) 1 3 2 ( 4 3 ' 4 4 2 4 4 1 2
x x x x x x x y b) . 3 arccos ) 2 ( 2 5 x x tg y Berilgan funksiyaning hosilasini toping Yechish. . 6 9 1 1 ) 2 ( 3 arccos
) 2 ( cos 1 ) 2 ( 5 ' 4 5 2 2 4 x x x tg x x x tg y 4 5 2 2 4 9 1 6 ) 2 ( ) 2 ( cos 3 arccos ) 2 ( 5 x x x tg x x x tg Download 1.05 Mb. Do'stlaringiz bilan baham: |
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