\((x-y)^{2016^{ }}+|x^2-2y|\) = 0
Ta thấy : \((x-y)^{2016^{ }}\ge0\) (1)
\(|x^2-2y|\ge0\) (2)
Từ (1) và (2) ta suy ra \(\left\{{}\begin{matrix}\left(x-y\right)^{2016}=0\\|x^{2^{ }}-2y|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x^2-2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\x^2=2y\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=y\\x.x=2.x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=2\end{matrix}\right.\)
\(\Rightarrow x=y=2\)
Vậy \(x=y=2\)
