\(x^2-x-y^2+5y=9\\ \Leftrightarrow x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}-y^2+5y-\left(2,5\right)^2=9-6=3\\ \Leftrightarrow\left(x-\dfrac{1}{2}\right)^2-\left(y-2,5\right)^2=3\\ \Leftrightarrow\left(x-\dfrac{1}{2}+y-2,5\right)\left(x-\dfrac{1}{2}-y+2,5\right)=3\\ \Leftrightarrow\left(x+y-3\right)\left(x-y+2\right)=3\\ vì\:x,y\:nguyên\:nên\:\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y-3=-1\\x-y+2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}x+y-3=1\\x-y-3=3\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=2\\x-y=-5\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=4\\x-y=6\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=\dfrac{7}{2}\end{matrix}\right.\left(loại\right)\\\left\{{}\begin{matrix}x=5\\y=-1\end{matrix}\right.\left(nhận\right)\end{matrix}\right.\)
vậy cặp số x, y cần tìm là 5;-1
