(x-3)(2y+1)=7
=> x-3 ; 2y+1 thuộc Ư(7)={-1,-7,1,7}
Ta có bảng :
| x-3 | -1 | -7 | 1 | 7 |
| 2y+1 | -7 | -1 | 7 | 1 |
| x | 2 | -4 | 4 | 10 |
| y | -4 | -1 | 3 | 0 |
Vậy ta có các cặp x,y thõa mãn là : (2,-4);(-4,-1);(4,3);(10,0)
Tim x,y thuoc Z
(x-3).(2y+1)=7
(x-3)*(2y+1)=7
Tim x,y thuoc Z
Tim x,y thuoc Z biet
a}[x-3].[2y+1]=7
b}[2x+1].[3y-2]=-55
tim x,y thuoc Z biet
|y|.|2x+3|=8
|2x+4|+|y-3|=0
|x-1|+|2y+7|=3
|x+5|+|2y+6| nho hon hoac bang 0
tim x, y thuoc z biet xy-x+2y=3
1 )Tim x, y thuoc Z
x + y = x.y
2) Tim x thuoc Z
(x + 1)+(x+3)+(x+5)+...+(x+99)=0(x-3)+(x-2)+(x-1)+...+10+11=11-12(x-5)+7(3-x)=530(x+2)-6(x-5)-24x=100tim x,y thuoc z sao cho (x-7).(y+3)<0
Tim x;y thuoc Z biet (x+3)(y-5)=7
tim x,y thuoc z sao cho:
7x-xy+2y=18
