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(3x-1).y = -12<=> 3x-1 và y là Ư của -12 ={ \(\mp1;2;3;4;6;12\) }=> ta xét từng trường hợp : ....
a, 3x(y-1)-y=0
3x(y-1)-(y-1)-1=0
(y-1)(3x-1)=0+1
(y-1)(3x-1)=1 Vậy (y-1) và (3x-1) là ước của 1
Ư(1)+{1;-1}
th1 y-1=1 suy ra y=2 suy ra 3x-1=-1 suy ra x=0
th2 y-1=-1 suy ra y=0 suy ra 3x-1=1 suy ra x thuộc rỗng
b, 5x(y+1)+2y=16
5x(y+1)+2(y+1)-2=16
(y+1)(5x+2)=16+2
(y+1)(5x+2)=18
Vậy (y+1) và (5x+2) thuộc ước của 18
Ư(18)={1;18;2;9;3;6;-1;-18;-2;-9;-3;-6}
Cậu liệt kê nữa là xong
ngay xua co mot con chim. mui no o duoi dit. 1 hom no ngoi xuong dat va no chet.
(3x-1).y = -12<=> 3x-1 và y là Ư của -12 ={ 1;2;3;4;6;12∓1;2;3;4;6;12 }
=> ta xét từng trường hợp : ....
1. (3x-1)y=-12 suy ra \(3x-1\inƯ\left(-12\right)\)(em tự liệt kê nhé!)
Lại có 3x-1 chia 3 dư 2(thiếu 1) nên \(3x-1\in\left\{-1;2;-4;\right\}\)
Đến đây em lập bảng và tìm đáp số nhé!
2. \(5xy+5x+2y=-16\Rightarrow5x\left(y+1\right)+2y=-16\)
\(\Rightarrow5x\left(y+1\right)+2\left(y+1\right)=-16+2=-14\)
\(\Rightarrow\left(5x+2\right)\left(y+1\right)=14\)
\(\Rightarrow5x+2\inƯ\left(14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)mà 5x+2 lẻ nên \(5x+2\in\left\{\pm1;\pm7\right\}\)
Đến đây em hãy lập bảng và tìm ra đáp số nhé!
Chúc em học tốt
a) ( x - 1 ) . ( y + 2 ) = 7
Lập bảng ta có :
| x-1 | 1 | 7 | -1 | -7 |
| y+2 | 7 | 1 | -7 | -1 |
| x | 2 | 8 | 0 | -6 |
| y | 5 | -1 | -8 | -3 |
b) x . ( y - 3 ) = -12
Lập bảng ta có :
| y-3 | 12 | -12 | 2 | -2 | -3 | -4 |
| x | -1 | 1 | -6 | 6 | 4 | 3 |
| y | 15 | -9 | 5 | 1 | 0 | -1 |
c) xy - 3x - y = 0
x . ( y - 3 ) - y = 0
x . ( y - 3 ) - y + 3 = 3
x . ( y - 3 ) - ( y - 3 ) = 3
( x - 1 ) . ( y - 3 ) = 3
Lập bảng ta có :
| x-1 | 3 | 1 | -1 | -3 |
| y-3 | 1 | 3 | -3 | -1 |
| x | 4 | 2 | 0 | -2 |
| y | 4 | 6 | 0 | 2 |
d) xy + 2x + 2y = -16
x . ( y + 2 ) + 2y = -16
x . ( y + 2 ) + 2y + 4 = -12
x . ( y + 2 ) + 2 . ( y + 2 ) = -12
( x + 2 ) . ( y + 2 ) = -12
Lập bảng ta có :
| x+2 | 1 | -1 | -2 | -6 | -4 | -3 |
| y+2 | -12 | 12 | 6 | 2 | 3 | 4 |
| x | -1 | -3 | -4 | -8 | -6 | -5 |
| y | -14 | 10 | 4 | 0 | 1 | 2 |
Ta có : (x - 1).(y + 2) = 7
=> (x - 1) và y + 2 thuộc Ư(7) = {-7;-1;1;7}
Ta có bảng :
| x - 1 | -7 | -1 | 1 | 7 |
| y + 2 | -1 | -7 | 7 | 1 |
| x | -6 | 0 | 2 | 8 |
| y | -3 | -9 | 5 | -1 |
Vậy có 4 cặp x;y thoả mãn : (-6,-3) ; (0 , -9) ; (2 , 5) ; (8, -1)
a.
xy + 3x - 2y - 6 = 5
=>x(y + 3) - 2(y + 3) = 5
=>(x - 2)(y + 3) = 5.
Vì x, y thuộc Z nên x - 2, y + 3 thuộc Z
=> x - 2, y + 3 thuộc ước nguyên của 5
Lập bảng :
| x - 2 | -5 | -1 | 1 | 5 |
| y + 3 | -1 | -5 | 5 | 1 |
| x | -3 | 1 | 3 | 7 |
| y | -4 | -8 | 2 | -2 |
Vậy ......
b. Làm tương tự câu a.
c. Ta có x + y = 3 và x - y = 15
Bài này là tổng hiệu của cấp 1, áp dụng cách làm đó thì ta được số lớn là x = (3 + 15) : 2 = 9
Số bé là y = 9 - 15 = -6
d. Ta có : |x| + |y| = 1
=>|x| = 1 - |y|
Vì |x|, |y| >= 0 và |x| = 1 - |y| nên 0 =< |x|, |y| =< 1
Vì x, y thuộc Z nên x = 0 thì y = 1 hoặc -1 và ngược lại y = 0 thì x = 1 hoặc -1
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