\(\left(x-1\right)^2+\left(y+3\right)^2=0\left(1\right)\)
Ta thấy \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0,\forall x\\\left(y+3\right)^2\ge0,\forall y\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left(x+1\right)^2=0^2\\\left(y+3\right)^2=0^2\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x+1=0\\y+3=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\)
