Ta có: \(\left(2x-3\right)\left(y+8\right)=10\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3\inƯ\left(10\right)\\y+8\inƯ\left(10\right)\end{matrix}\right.\)
Trường hợp 1:
\(\left\{{}\begin{matrix}2x-3=1\\y+8=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=4\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\left(nhận\right)\\y=2\left(nhận\right)\end{matrix}\right.\)
Trường hợp 2:
\(\left\{{}\begin{matrix}2x-3=-1\\y+8=-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=2\\y=-18\left(loại\right)\end{matrix}\right.\)
Trường hợp 3:
\(\left\{{}\begin{matrix}2x-3=10\\y+8=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=13\\y=-7\left(loại\right)\end{matrix}\right.\)
Trường hợp 4:
\(\left\{{}\begin{matrix}2x-3=-10\\y+8=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-7\\y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{7}{2}\left(loại\right)\\y=7\left(loại\right)\end{matrix}\right.\)
Trường hợp 5:
\(\left\{{}\begin{matrix}2x-3=2\\y+8=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=5\\y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\left(loại\right)\\y=-3\left(loại\right)\end{matrix}\right.\)
Trường hợp 6:
\(\left\{{}\begin{matrix}2x-3=-2\\y+8=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=1\\y=-13\left(loại\right)\end{matrix}\right.\)
Trường hợp 7:
\(\left\{{}\begin{matrix}2x-3=5\\y+8=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=8\\y=-6\left(loại\right)\end{matrix}\right.\)
Trường hợp 8:
\(\left\{{}\begin{matrix}2x-3=-5\\y+8=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-2\\y=-10\left(loại\right)\end{matrix}\right.\)
Vậy: (x,y)=(2;2)
