\(xy-2x+3y=1\\ \Rightarrow x\left(y-2\right)+3y-6=-5\\ \Rightarrow x\left(y-2\right)+3\left(y-2\right)=-5\\ \Rightarrow\left(x+3\right)\left(y-2\right)=-5\)
Ta lập bảng sau:
x+3 | 1 | -1 | 5 | -5 |
x | -2 | -4 | 2 | -8 |
y-2 | -5 | 5 | -1 | 1 |
y | -3 | 7 | 1 | 3 |
Vậy \(\left(x;y\right)\in\left\{\left(-2;-3\right);\left(-4;7\right);\left(2;1\right);\left(-8;3\right)\right\}\)