Giải thích các bước giải:
$\begin{array}{l}
{x^4} - 5{x^2} + 2x + 3 \le 0\\
\Leftrightarrow {x^4} - {x^3} - {x^2} + {x^3} - {x^2} - x - 3{x^2} + 3x + 3 \le 0\\
\Leftrightarrow {x^2}\left( {{x^2} - x - 1} \right) + x\left( {{x^2} - x - 1} \right) - 3\left( {{x^2} - x - 1} \right) \le 0\\
\Leftrightarrow \left( {{x^2} - x - 1} \right)\left( {{x^2} + x - 3} \right) \le 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{x^2} - x - 1 \le 0\\
{x^2} + x - 3 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
{x^2} - x - 1 \ge 0\\
{x^2} + x - 3 \le 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\dfrac{{1 - \sqrt 5 }}{2} \le x \le \dfrac{{1 + \sqrt 5 }}{2}\\
\left[ \begin{array}{l}
x \ge \dfrac{{ - 1 + \sqrt {13} }}{2}\\
x \le \dfrac{{ - 1 - \sqrt {13} }}{2}
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge \dfrac{{1 + \sqrt 5 }}{2}\\
x \le \dfrac{{1 - \sqrt 5 }}{2}
\end{array} \right.\\
\dfrac{{ - 1 - \sqrt {13} }}{2} \le x \le \dfrac{{ - 1 + \sqrt {13} }}{2}
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{{ - 1 + \sqrt {13} }}{2} \le x \le \dfrac{{1 + \sqrt 5 }}{2}\\
\dfrac{{ - 1 - \sqrt {13} }}{2} \le x \le \dfrac{{1 - \sqrt 5 }}{2}
\end{array} \right.
\end{array}$
Vậy bất phương trình có tập nghiệm là: $S = \left[ {\dfrac{{ - 1 + \sqrt {13} }}{2};\dfrac{{1 + \sqrt 5 }}{2}} \right] \cup \left[ {\dfrac{{ - 1 - \sqrt {13} }}{2};\dfrac{{1 - \sqrt 5 }}{2}} \right]$
