Question

Giải hệ phương trình sau

Answer 1

\(\left\{{}\begin{matrix}x^2+y^2+6x+2y=0\\x+y+8=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+\left(x+8\right)^2+6x-2\left(x+8\right)=0\\y=-x-8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+10x+24=0\\y=-x-8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=-4\\x=-6\end{matrix}\right.\\y=-x-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-4\\y=-4\end{matrix}\right.\\\left\{{}\begin{matrix}x=-6\\y=-2\end{matrix}\right.\end{matrix}\right.\)

Kết luận: Hệ phương trình đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(-4;-4\right);\left(-6;-2\right)\right\}\)