\(\left|y+3\right|+5=\dfrac{10}{\left(2x-6\right)^2+2}\)
Để x,y nguyên thì \(\left\{{}\begin{matrix}10⋮\left(2x-6\right)^2+2\\\left|y+3\right|+5\in Z\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(2x-6\right)^2+2\in\left\{2;5;10\right\}\\\left|y+3\right|+5=\dfrac{10}{\left(2x-6\right)^2+2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(2x-6\right)^2\in\left\{0;3;8\right\}\\\left|y+3\right|+5=\dfrac{10}{\left(2x-6\right)^2+2}\end{matrix}\right.\)
mà x nguyên
nên \(\left\{{}\begin{matrix}\left(2x-6\right)^2=0\\\left|y+3\right|+5=\dfrac{10}{0+2}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-6=0\\\left|y+3\right|=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x=6\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\)