18 tháng 9 2022 lúc 9:36
Vì \(\left\{{}\begin{matrix}\left(2x-y\right)^2\ge0\forall x,y\in R\\\left(x-1\right)^2\ge0\forall x\in R\end{matrix}\right.\) nên \(\left(2x-y\right)^2+\left(x+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)^2=0\\\left(x-1\right)^2-0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=y\\x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)
\(A=\left(x-2\right)^{2021}+\left(y-2\right)^{2020}\\ =\left(1-2\right)^{2021}+\left(2-2\right)^{2020}=-1^{2021}+0^{2020}=-1+0=-1\)
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