Đáp án:
$\begin{array}{l}
\sqrt {1 + 2 + 3 + .. + \left( {n - 1} \right) + n + \left( {n - 1} \right) + ... + 3 + 2 + 1} \\
= 2020\\
\text{Xét}:A = 1 + 2 + 3 + .. + \left( {n - 1} \right)\\
= \dfrac{{\left( {n - 1 + 1} \right).\left( {n - 1} \right)}}{2}\\
= \dfrac{{n\left( {n - 1} \right)}}{2}\\
\Rightarrow 1 + 2 + 3 + .. + \left( {n - 1} \right) + n + \\
\left( {n - 1} \right) + 3 + 2 + 1\\
= 2.\dfrac{{n\left( {n - 1} \right)}}{2} + n\\
= n\left( {n - 1} \right) + n\\
= {n^2} - n + n\\
= {n^2}\\
\Rightarrow \sqrt {1 + 2 + 3 + .. + \left( {n - 1} \right) + n + \left( {n - 1} \right) + ... + 3 + 2 + 1} \\
= \sqrt {{n^2}} \\
= n = 2020\\
\text{Vậy}\,n = 2020\\
2)Do:\left| {x - \dfrac{3}{4}} \right| \ge 0\forall x\\
\Rightarrow \left| {x - \dfrac{3}{4}} \right| + 1 \ge 1\forall x\\
\Rightarrow A \ge 1\forall x\\
\Rightarrow GTNN:A = 1\\
\text{Dấu = xảy ra khi}:x = \dfrac{3}{4}\\
3)A = \left| {x - 20} \right| + \left| {x - 2040} \right|\\
= \left| {x - 20} \right| + \left| {2040 - x} \right| \ge \left| {x - 20 + 2040 - x} \right|\\
\Rightarrow A \ge \left| {2020} \right|\\
\Rightarrow A \ge 2020\\
\Rightarrow GTNN:A = 2020\\
\text{Dấu = xảy ra khi}:x - 20 = 2040 - x\\
\Rightarrow 2x = 2060\\
\Rightarrow x = 1030
\end{array}$
