1) \(x^3-2x^2+x-xy^2=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x+y-1\right)\left(x-y-1\right)\)2) -Áp dụng định lí Bezout ta có:
\(A\left(1\right)=1^{2022}+a.1^{2021}-3\left(a+2\right).1^2+2020=-5\)
\(\Rightarrow1+a-3\left(a+2\right)+2020+5=0\)
\(\Rightarrow1+a-3a-6+2020+5=0\)
\(\Rightarrow-2a+2020=0\)
\(\Rightarrow a=1010\).
-Vậy hệ số a là 1010.