\(\left(x+3\right)^2+\left(x-1\right)^2=0\)
Vì \(\left\{{}\begin{matrix}\left(x+3\right)^2\ge0\\\left(x-1\right)^2\ge0\end{matrix}\right.\)=> \(\left(x+3\right)^2+\left(x-1\right)^2\ge0\)
Khi \(\left\{{}\begin{matrix}x+3=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0-3\\x=0+1\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)
