Đáp án:
$\begin{array}{l}
a)\left( { - 2\dfrac{2}{5}x + 1} \right).\left( {x - 2020} \right) > 0\\
\Rightarrow \left( {\dfrac{{ - 12}}{5}x + 1} \right).\left( {x - 2020} \right) > 0\\
\Rightarrow \left( {\dfrac{{12}}{5}.x - 1} \right).\left( {x - 2020} \right) < 0\\
\Rightarrow \dfrac{5}{{12}}.\left( {\dfrac{{12}}{5}x - 1} \right).\left( {x - 2020} \right) < 0.\dfrac{5}{{12}}\\
\Rightarrow \left( {x - \dfrac{5}{{12}}} \right).\left( {x - 2020} \right) < 0\\
\Rightarrow \dfrac{5}{{12}} < x < 2020\\
\text{Vậy}\,\dfrac{5}{{12}} < x < 2020\\
b)\dfrac{{x - 2}}{{x + 5}} > 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 2 > 0\\
x + 5 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 2 < 0\\
x + 5 < 0
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 2\\
x > - 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 2\\
x < - 5
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x > 2\\
x < - 5
\end{array} \right.\\
\text{Vậy}\,x > 2\,hoac\,x < - 5\\
c)\left( {2x - 5} \right)\left( {3x - \dfrac{5}{3}} \right) < 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2x - 5 < 0\\
3x - \dfrac{5}{3} > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
2x - 5 > 0\\
3x - \dfrac{5}{3} < 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x < \dfrac{5}{2}\\
x > \dfrac{5}{9}
\end{array} \right.\\
\left\{ \begin{array}{l}
x > \dfrac{5}{2}\\
x < \dfrac{5}{9}
\end{array} \right.
\end{array} \right.\\
\Rightarrow \dfrac{5}{9} < x < \dfrac{5}{2}\\
\text{Vậy}\,\dfrac{5}{9} < x < \dfrac{5}{2}\\
d){x^2} + 2020x < 0\\
\Rightarrow x.\left( {x + 2020} \right) < 0\\
\Rightarrow - 2020 < x < 0\\
\text{Vậy}\, - 2020 < x < 0
\end{array}$
