\(\Leftrightarrow x^2-y^2=9\\ \Leftrightarrow\left(x-y\right)\left(x+y\right)=9=9.1=\left(-9\right)\left(-1\right)=3.3=\left(-3\right)\left(-3\right)\)
\(\left\{{}\begin{matrix}x-y=9\\x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=-4\end{matrix}\right.\); \(\left\{{}\begin{matrix}x-y=1\\x+y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=4\end{matrix}\right.\); \(\left\{{}\begin{matrix}x-y=-9\\x+y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=4\end{matrix}\right.\); \(\left\{{}\begin{matrix}x-y=-1\\x+y=-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=-4\end{matrix}\right.\); \(\left\{{}\begin{matrix}x-y=-3\\x+y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=0\end{matrix}\right.\); \(\left\{{}\begin{matrix}x-y=3\\x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=0\end{matrix}\right.\)
Vậy các cặp \(\left(x;y\right)\) là \(\left(5;-4\right);\left(5;4\right);\left(-5;4\right);\left(-5;-4\right);\left(-3;0\right);\left(3;0\right)\)
