+ x(y-1)=3
=>x va y-1 \(\varepsilon\) {3}
Ma Ư(3)\(\varepsilon\){1,3,-1,-3}
=> Ta có bảng
x | 1 | 3 | -1 | -3 |
y-1 | 3 | 1 | -3 | -1 |
y | 4 | 2 | -2 | 0 |
Vậy (x,y)\(\varepsilon\){(3,4),(1,2),(-3,-2),(-1,0)}
b, (x+3)(y+1)=4.
=> x+3 va y+1 \(\varepsilon\) Ư(4)
Mà Ư(4)={1,2,4,-1,-2,-4}
Ta có bảng:
x-3 | 1 | 2 | 4 | -1 | -2 | -4 |
y+1 | 4 | 2 | 1 | -4 | -2 | -1 |
x | 4 | 5 | 7 | 2 | 1 | -1 |
y | 3 | 1 | 0 | -5 | -3 | -2 |
Vậy (x,y)\(\varepsilon\){(4,3),(5,1),(7,0),(2,-5),(1,-3),(-1,-2)}