Giải:
\(\left(2x-1\right)^{2018}+\left(2y+1\right)^{2020}=0\)
Vì:
\(\left\{{}\begin{matrix}\left(2x-1\right)^{2018}\ge0;\forall x\\\left(2y+1\right)^{2020}\ge0;\forall y\end{matrix}\right.\) (Lũy thừa số chẵn)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}2x-1=0\\2y+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=1\\2y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)
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