Ta có: 2xy-x+2y=3
=> x(2y-1)+(2y-1)=2
=> (2y-1)(x+1)=2
Vì \(x,y\in Z\Rightarrow2y-1;x+1\inƯ\left(2\right)=\left\{\mp1;\mp2\right\}\)
Ta có bảng sau:
x+1 | 1 | -1 | 2 | -2 |
2y-1 | 2 | -2 | 1 | -1 |
x | 0 | -2 | 1 | -3 |
y | 3/2 | -1/2 | 1 | 0 |
Vì \(x;y\in Z\Rightarrow\left(x;y\right)\in\left\{\left(1;1\right),\left(-3;0\right)\right\}\)