Giải thích các bước giải:
Em xem lại yêu cầu đề bài nhé
\(\begin{array}{l}
x\left( {{x^2} - 4x + 5} \right)\left( {2x - {x^2} + 3} \right) < 0\\
Vi\left\{ \begin{array}{l}
{x^2} - 4x + 5 = {\left( {x - 2} \right)^2} + 1 > 0\\
2x - {x^2} + 3 = - \left( {{x^2} - 2x + 1} \right) + 4 < 4
\end{array} \right.\\
\Rightarrow x\left( { - {x^2} + 2x + 3} \right) < 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x < 0\\
- {x^2} + 2x + 3 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x > 0\\
- {x^2} + 2x + 3 < 0
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x < 0\\
\left( {x + 1} \right)\left( {x - 3} \right) < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x > 0\\
\left( {x + 1} \right)\left( {x - 3} \right) > 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x < - 1\\
x > 3
\end{array} \right.
\end{array}\)
