2x - 3y = xy
=> xy - 2x + 3y = 0
=> (xy - 2x) + (3y - 6) = -6
=> x(y - 2) + 3(y - 2) = -6
=> (x + 3)(y - 2) = -6
Ta có bảng sau:
x + 3 | -1 | 1 | -2 | 2 | -3 | 3 | -6 | 6 |
x | -4 | -2 | -5 | -1 | -6 | 0 | -9 | 3 |
y - 2 | 6 | -6 | 3 | -3 | 2 | -2 | 1 | -1 |
y | 8 | -4 | 5 | -1 | 4 | 0 | 3 | 1 |
Vậy: (x;y) \(\in\){(-4;8);(-2;-4);(-5;5);(-1;-1);(-6;4);(0;0);(-9;3);(3;1)}