\(\Leftrightarrow\left(x^2-1\right)-\left(xy+y\right)=1\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)-y\left(x+1\right)=1\)
\(\Leftrightarrow\left(x+1\right)\left(x-y-1\right)=1\)
| x+1 | -1 | 1 |
| x-y-1 | -1 | 1 |
| x | -2 | 0 |
| y | -2 | -2 |
Vậy \(\left(x;y\right)=\left(-2;-2\right);\left(0;-2\right)\)