Đáp án: $f(2020)=1$
Giải thích các bước giải:
$f(2020)\Leftrightarrow x=2020$
$\Rightarrow x-2020=0$
$f(x)=x^{6}-2021x^{5}+2021x^{4}-2021x^{3}-2021x+2021$
$\Leftrightarrow f(x)=(x^{6}-2020x^{5})-(x^{5}-2020x^{4})+(x^{4}-2020x^{3})-(x^{3}-2020x^{2})+(x^{2}-2020x)-(x-2020)+1$
$\Leftrightarrow f(x)=x^{5}(x-2020)-x^{4}(x-2020)+x^{3}(x-2020)-x^{2}(x-2020)+x(x-2020)-(x-2020)+1$
$f(x)=1$