\(x^2+4x-y^2=1\)
\(\Leftrightarrow x^2+4x+4-y^2=5\)
\(\Leftrightarrow\left(x+2\right)^2-y^2=5\)
\(\Leftrightarrow\left(x+2-y\right)\left(x+2+y\right)=5\)
*Trường hợp 1:
\(\left\{{}\begin{matrix}x+2-y=5\\x+2+y=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
*Trường hợp 2: \(\left\{{}\begin{matrix}x+2-y=1\\x+2+y=5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
*Trường hợp 3: \(\left\{{}\begin{matrix}x+2-y=-1\\x+2+y=-5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-5\\y=-2\end{matrix}\right.\)
*Trường hợp 4: \(\left\{{}\begin{matrix}x+2-y=-5\\x+2+y=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-5\\y=2\end{matrix}\right.\)
Vậy:...........
