\(\left(x-3\right)^2+\left[y^2-25\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\\left[y^2-25\right]=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3^2=0\\y^2-5^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=3^2\\y^2=5^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\y=5\end{matrix}\right.\)
Vậy x=3;y=5
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