7. The solution set of the inequality $|x|^{3}-2 x^{2}-4|x|+3<0$ is
$$
\left(-3, \frac{1-\sqrt{5}}{2}\right) \cup\left(\frac{-1+\sqrt{5}}{2}, 3\right)
$$
7.【Analysis and Solution】From the original inequality, we can derive $(|x|-3)\left(x^{2}+|x|-1\right)<0$, from which we get the solution set of the inequality as $\left(-3,-\frac{\sqrt{5}-1}{2}\right) \cup\left(\frac{\sqrt{5}-1}{2}, 3\right)$.
When $x \geqslant 0$, the original inequality is $x^{3}-2 x^{2}-4 x+3<0$,
which is $\left(x-\frac{-1-\sqrt{5}}{2}\right)\left(x-\frac{-1+\sqrt{5}}{2}\right)(x-3)<0$.
Noting that $x \geqslant 0$, we get $\frac{\sqrt{5}-1}{2}<x<3$.
Since $f(x)=|x| x^{3}-2 x^{2}-4|x|+3$ is an even function,
when $x<0$, the solution is $-3<x<\frac{1-\sqrt{5}}{2}$.
Therefore, the solution set of the original inequality is $\left(-3, \frac{1-\sqrt{5}}{2}\right) \cup\left(\frac{-1+\sqrt{5}}{2}, 3\right)$.