\(\Leftrightarrow\left(2x-1\right)^{2022}-\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(2x-1\right)^2\left[\left(2x-1\right)^{2020}-1\right]=0\)
TH1 : x = 1/2
TH2 : \(\left[{}\begin{matrix}2x-1=1\\2x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)