\(2xy-3y-x-1=0\Leftrightarrow4xy-6y-2x-2=0\Leftrightarrow2x\left(2y-1\right)-3\left(2y-1\right)=5\)
\(\Leftrightarrow\left(2x-3\right)\left(2y-1\right)=5\)
Ta có bảng :
2x-3 | -5 | -1 | 1 | 5 |
2y-1 | -1 | -5 | 5 | 1 |
x | -1 | 1 | 2 | 4 |
y | 0 | -2 | 3 | 1 |
Vậy (x;y) = (-1;0) ; (1;-2) ; (2;3) ; (4;1)