Đáp án:
$\begin{array}{l}
\left| {x\left( {{x^2} - \dfrac{5}{4}} \right)} \right| = x\\
\Rightarrow \left[ \begin{array}{l}
x\left( {{x^2} - \dfrac{5}{4}} \right) = x\\
x\left( {{x^2} - \dfrac{5}{4}} \right) = - x
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x\left( {{x^2} - \dfrac{5}{4}} \right) - x = 0\\
x\left( {{x^2} - \dfrac{5}{4}} \right) + x = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x\left( {{x^2} - \dfrac{5}{4} - 1} \right) = 0\\
x\left( {{x^2} - \dfrac{5}{4} + 1} \right) = 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x\left( {{x^2} - \dfrac{9}{4}} \right) = 0\\
x\left( {{x^2} - \dfrac{1}{4}} \right) = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} = \dfrac{9}{4}\\
{x^2} = \dfrac{1}{4}
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 0\\
x = \pm \dfrac{3}{2}\\
x = \pm \dfrac{1}{2}
\end{array} \right.\\
\text{Vậy}\,x = 0;x = \pm \dfrac{3}{2};x = \pm \dfrac{1}{2}
\end{array}$
