Ta có:
xy+x-y=4
<=> xy+x-y-1=3
<=> x(y+1)-(y+1)=3
<=>(y+1)(x-1)=3=1.3=3.1 (Do x, y nguyên dương)
=> \(\hept{\begin{cases}y+1=1\\x-1=3\end{cases}}\)=> \(\hept{\begin{cases}y=0\\x=4\end{cases}}\)
Và: \(\hept{\begin{cases}y+1=3\\x-1=1\end{cases}}\) => \(\hept{\begin{cases}y=2\\x=2\end{cases}}\)