\(xy-2x+y=1\)
\(\Leftrightarrow xy-2x+y-2=1-2\)
\(\Leftrightarrow x\left(y-2\right)+y-2=-1\)
\(\Leftrightarrow\left(y-2\right)\left(x+1\right)=-1\)
Ta có bảng:
y-2 | -1 | 1 |
x+1 | 1 | -1 |
y | 1 | 3 |
x | 0 | -2 |
Vậy \(\left(x;y\right)=\left(0;1\right);\left(-2;3\right)\)