Với \(n\in\mathbb{N^*}\), ta có: \(\left\{{}\begin{matrix}\left(x+1\right)^{2n}\ge0\forall x\\\left(y-1\right)^{2n}\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x+1\right)^{2n}+\left(y-1\right)^{2n}\ge0\forall x,y\)
Mà: \(\left(x+1\right)^{2n}+\left(y-1\right)^{2n}=0\)
nên: \(\left\{{}\begin{matrix}x+1=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
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