=> xy - x + y = 0
=> (xy - x) + (y - 1) = -1
=> x(y - 1) + (y - 1) = -1
=> (x + 1)(y - 1) = -1
=> x + 1 = 1 hoặc x + 1 = -1
+) x + 1 = 1 => x = 1 - 1 = 0
=> y - 1 = -1 => y = -1 + 1 = 0
+) x + 1 = -1 => x = -1 - 1 = -2
=> y - 1 = 1 => y = 1 + 1 = 2
Vậy: (x;y) \(\in\){(0;0);(-2;2)}