Đáp án:
$\begin{array}{l}
A = - 2020{x^2} + 2021.x + 1\\
= - 2020.\left( {{x^2} - \dfrac{{2021}}{{2020}}.x - \dfrac{1}{{2020}}} \right)\\
= - 2020.\left( {{x^2} - 2.x.\dfrac{{2021}}{{4040}} + {{\left( {\dfrac{{2021}}{{4040}}} \right)}^2} - {{\left( {\dfrac{{2021}}{{4040}}} \right)}^2} - \dfrac{1}{{2020}}} \right)\\
= - 2020.{\left( {x - \dfrac{{2021}}{{4040}}} \right)^2} + \dfrac{{{{2021}^2}}}{{8080}} + 1\\
Do:{\left( {x - \dfrac{{2021}}{{4040}}} \right)^2} \ge 0\forall x\\
\Rightarrow - 2020.{\left( {x - \dfrac{{2021}}{{4040}}} \right)^2} \le 0\forall x\\
\Rightarrow - 2020.{\left( {x - \dfrac{{2021}}{{4040}}} \right)^2} + \dfrac{{{{2021}^2}}}{{8080}} + 1 \le \dfrac{{{{2021}^2}}}{{8080}} + 1\\
\Rightarrow A \le 506,5\\
\Rightarrow GTLN:A = 506,5 \Leftrightarrow x = \dfrac{{2021}}{{4040}}
\end{array}$
