ĐK: \(y\ge-1\)
hpt \(\Leftrightarrow\left\{{}\begin{matrix}x^2+2x+6=y^2+2y+1\\\dfrac{1}{4}\left[3\left(x+y\right)^2+\left(x-y\right)^2\right]=7\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left(x-y\right)\left(x+y+2\right)=-5\\3\left(x+y\right)^2+\left(x-y\right)^2=28\end{matrix}\right.\)(1)
đặt \(\left\{{}\begin{matrix}t=x+y\\u=x-y\end{matrix}\right.\) hpt (1) trở thành:
\(\left\{{}\begin{matrix}u\left(t+2\right)=-5\\3t^2+u^2=28\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t=-1\\u=-5\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}t=3\\u=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\x-y=-5\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x+y=3\\x-y=-1\end{matrix}\right.\)
giải các hệ trên ta đc:
\(\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
Vậy.....
