Đáp án:
\(0 > x > - 2020\)
Giải thích các bước giải:
\(\begin{array}{l}
b){x^2} + 2020x = x\left( {x + 2020} \right)\\
{x^2} + 2020x < 0\\
\to x\left( {x + 2020} \right) < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
x + 2020 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 0\\
x + 2020 > 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
x < - 2020
\end{array} \right.\left( l \right)\\
0 > x > - 2020
\end{array} \right.\\
\to 0 > x > - 2020
\end{array}\)
