Đáp án:
$\begin{array}{l}
A = \dfrac{{ - 2\left| {x - 2018} \right| - 2021}}{{2020 + \left| {x - 2018} \right|}}\\
= \dfrac{{ - 2\left| {x - 2018} \right| - 4040 + 2019}}{{2020 + \left| {x - 2018} \right|}}\\
= \dfrac{{ - 2.\left( {2020 + \left| {x - 2018} \right|} \right) + 2019}}{{2020 + \left| {x - 2018} \right|}}\\
= - 2 + \dfrac{{2019}}{{2020 + \left| {x - 2018} \right|}}\\
Do:\left| {x - 2018} \right| \ge 0\\
\Leftrightarrow \left| {x - 2018} \right| + 2020 \ge 2020\\
\Leftrightarrow \dfrac{1}{{2020 + \left| {x - 2018} \right|}} \le \dfrac{1}{{2020}}\\
\Leftrightarrow \dfrac{{2019}}{{2020 + \left| {x - 2018} \right|}} \le \dfrac{{2019}}{{2020}}\\
\Leftrightarrow - 2 + \dfrac{{2019}}{{2020 + \left| {x - 2018} \right|}} \le - 2 + \dfrac{{2019}}{{2020}}\\
\Leftrightarrow A \le \dfrac{{ - 2021}}{{2020}}\\
\Leftrightarrow GTLN:A = \dfrac{{ - 2021}}{{2020}}\\
Khi:x = 2018
\end{array}$
