Ta có:
$x^3 - 2022x^2 + 2021x = 0$
$\to x(x^2 - 2022x + 2021)=0$
$\to \left[\begin{array}{l}x = 0\\x = 1\\x = 2021\end{array}\right.$
Ta được:
$B =\{x\in \Bbb Z\Big|\, x^3 - 2020x^2 + 2021x = 0\}$
$\to B =\{0;1;2021\}$
$A = (-\infty;2020)$
$+)\quad A\cap B = \{0;1\}$
$+)\quad B\backslash A = \{2021\}$