\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
hay \(x\in\left\{0;1;-1\right\}\)
\(x^{2022}-x^{2020}=0\\ x^{2020}\left(x^2-1\right)=0\\ x^{2020}\left(x-1\right)\left(x+1\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x^{2020}=0\Rightarrow x=0\\x-1=0\Rightarrow x=1\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
x2022 - x2020 = 0
x2020. x2 - x2020 = 0
x2020. (x2-1) = 0
Xét 2 trường hợp
TH1: x2020 = 0
=> x = 0
TH2: x2 - 1 = 0
=> x = 1
vậy x = {0;1}