7. The solution set of the inequality $|x|^{3}-2 x^{2}-4|x|+3<0$ is $\qquad$
$$
\begin{array}{l}
|x|^{3}-2 x^{2}-4|x|+3=(|x|-3)\left(x^{2}+|x|-1\right) \\
=(|x|-3)\left(|x|-\frac{-1+\sqrt{5}}{2}\right)\left(|x|+\frac{1+\sqrt{5}}{2}\right)<0 \Rightarrow \frac{-1+\sqrt{5}}{2}<|x|<3 \\
\Rightarrow x \in\left(-3, \frac{1-\sqrt{5}}{2}\right) \cup\left(\frac{-1+\sqrt{5}}{2}, 3\right) .
\end{array}
$$

The solution is as follows:

$$
\begin{array}{l}
|x|^{3}-2 x^{2}-4|x|+3=(|x|-3)\left(x^{2}+|x|-1\right) \\
=(|x|-3)\left(|x|-\frac{-1+\sqrt{5}}{2}\right)\left(|x|+\frac{1+\sqrt{5}}{2}\right)<0 \Rightarrow \frac{-1+\sqrt{5}}{2}<|x|<3 \\
\Rightarrow x \in\left(-3, \frac{1-\sqrt{5}}{2}\right) \cup\left(\frac{-1+\sqrt{5}}{2}, 3\right) .
\end{array}
$$