Toshkent davlat texnika universiteti "oliy matematika" kafedrasi
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1-tipik hisob 1-kurs sirtqilar uchun-20-21 (кузги семестр)
- Bu sahifa navigatsiya:
- Kramer, Gauss va
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56 – variant 1. 2 7 4 0 1 3 2 1 , 2, 3 3 9 8 3 4 0 1 2
j 2. 1 2 3 1 2 3 1 2 3 2 0 7 6 4 15 8 5 4 3
x x x x x x x x
1 3
( ) : 3 2 5 3 0, ( ) :
6 1 3 x y z T x y z l 4. 1 1 1 (0; 2;3) , 3 2
x y z M
5. a)
4 1 1 8 lim
2 3 2 1 x x x
b)
x x x x 5 5 3 lim 0
c) x x x x 3 2 1 lim
d) x x x x 2 0 5 arcsin lim
a) 2
2 5 8 ) 2 ( 7 x x x y b)
1 3 sin 5 x arctg x y
c)
x x y 5 3 cos ) 4 ( log
d) у x у x 2 2 3
7. 1 3 4 x x y 8.
. 3 ; 0 , 1 x e x y 33
57– variant 1. 2 4 5 1 0 1 3 2 , 3, 2 1 9 1 3
5 3 8 1 i j 2. 1 2 3 1 2 3 1 2 3 3 4 3 8 2 5 2 2 7 5 9 14 x x x x x x x x x
2 1
( ) : 3 4 0, ( ) : 4 3 2 x y z T x y z l
4. 1 1 1 (1;1;1) , 2 1 3 x y z M 5. a)
x x x x x 4 28 3 lim
2 2 4
b) x x x 3 5 7 2 lim 9
c)
x x x x 2 1 4 1 4 lim
d)
x x x x arcsin
2 cos
1 lim
2 0
6. a)
1 5 3 10 ) 4 ( 2 7
x x y
b) x arctg y x 3 2 5 2 sin
c) 4 2 ) 8 ( ) 2 lg( x x x y d)
3
3 2 3 2 a у x
2 ln 2 x x y 8.
0 ; 2 , x xe y 58 – variant 1. 1 3 2 1 0 1 3 2 2, , 5 3 4 1 2 3 2 0 1
j 2. 1 2 3 1 2 3 1 2 3 9 5 14 0 7 3 2 2 5 5 1 x x x x x x x x x
1 2
( ) : 2 5 16 0, ( ) :
2 5 2 x y z T x y z l 4. 4 2 1 (2; 2; 2) , 4 2
x y z M
5. a)
14 5 10 11 3 lim 2 2 2 x x x x x
b) 5 1 6 2 lim
4
x x
c) x x x x 2 3 4 3 lim
d) x x x x sin
4 cos
1 lim
0
a)
5 2 1 1 3 4 3 x x x y
b)
4 3 arccos 1 x x ctg y c)
1 2 lg 2 cos 2 2 x x x y d) у e x e x у cos
sin
7. 2 ) 1 ( 2 x x y 8. .
; 1 , 4 2 3 x x y 34
59– variant 1. 3 0 4 1 1 6 1 3 , 4, 3 0 4 2 5 2 4 7 10
j 2. 1 2 3 1 2 3 1 2 3 2 3 5 5 4 10 5 6 3 8 21 x x x x x x x x x
1 3
( ) : 3 7 2 7 0, ( ) :
1 0 2 x y z T x y z l 4. 1 1
(4;3; 2) , 3 3 4 x y z M
5. a)
4 27 7 64 lim
2 3 4 x x x x
b) 8 3 1 4 lim
3 2 x x x
c) 1 2 1 2 4 lim x x x x
d) 2 3 0 5 cos cos lim
x x x x
a) 6
2 ) 2 3 1 ( 4 3
x x y b)
x tg x arctg y 5 3
c) 2 2 4 cos
) 7 3 ( log
x x y d)
3 2 4 4 у x у x 7. 2 4 ) 1 ( x e x y 8. .
; 1 , 1 3
x x y 60 – variant 1. 5 2 1 4 1 1 1 1 4 2 1 4 3 2 1 3 4 1 |
i 2. 1 2
1 2 3 1 2 3 5 3 14 3 2 10 4 4 3 17 x x x x x x x x x 3. 7 3 1 ( ) : 2
7 3 0, ( ) : 3 1 2 x y z T x y z l
4. 6 4
(2;1; 4) , 2 4 5 x y z M 5. a)
12 20 3 6 11 2 lim 2 2 6 x x x x x
b) 64 4 20 lim 3 4
x x
c) x x x x 3 2 1 2 lim
d) 2 2 2 sin 1 lim
x x 6. a)
3 4 2 3 ) 3 1 ( ) 7 ( 4 x x x y b)
x arctg y x 4 2 3 3 5
c) ) 3 ( ln 3
ctg x y d) ctgxу у tg 2
1 4
2 x x y 8. .
; 1 , 1 2 3 x x x y 35
1- misol. |
| 4-tartibli determinantni j=3 ustun bo'yicha hisoblang.
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Shunday qilib, to‟rtinchi tartibli determinantning yoyilmasi to‟rtta qo‟shiluvchidan iborat. Birinchi va uchinchi hadlar tarkibida 0 ko‟paytuvchi bor, shu sababli bu qo‟shiluvchilarning qiymati nolga teng. Ikkinchi va to‟rtinchi hadlardagi uchinchi tartibli determinantlarning qiymatini hisoblaymiz:
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| Hisoblangan bu qiymatlarni yuqoridagi yoyilmaga qo‟yib, berilgan determinantning qiymatini topamiz:
11 2 3 ,
6 2 ,
(1)
5 3 2 .
Noma‟lum , va
o„zgarmas miqdorlarni aniqlash uchun Kramer, Gauss va teskari matritsalar usullaridan foydalanamiz.
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