Ta có: (x-1)(y+2)=4
\(\Leftrightarrow\left(x-1\right);\left(y+2\right)\inƯ\left(4\right)\)
*Trường hợp 1:
\(\left\{{}\begin{matrix}x-1=1\\y+2=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)(thỏa mãn)
*Trường hợp 2:
\(\left\{{}\begin{matrix}x-1=4\\y+2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=-1\end{matrix}\right.\)(thỏa mãn)
*Trường hợp 3:
\(\left\{{}\begin{matrix}x-1=-1\\y+2=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-6\end{matrix}\right.\)(thỏa mãn)
*Trường hợp 4:
\(\left\{{}\begin{matrix}x-1=-4\\y+2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-3\end{matrix}\right.\)(thỏa mãn)
*Trường hợp 5:
\(\left\{{}\begin{matrix}x-1=2\\y+2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=0\end{matrix}\right.\)(thỏa mãn)
*Trường hợp 6:
\(\left\{{}\begin{matrix}x-1=-2\\y+2=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-4\end{matrix}\right.\)
Vậy: x∈{0;5;-3;3;-1}; y∈{2;-1;-6;-3;0;-4}
