$11.30 x^{2}-4|x|+3>0$
11.30 Since $x^{2}=|x|^{2}$, we solve the inequality $|x|^{2}-4|x|+3>0$. The left side of this inequality is a quadratic trinomial in terms of $|x|$, which equals zero when $|x|=1$ and $|x|=3$. Using the interval method, we find $|x|<1$ or $|x|>3$. From this, it follows that $-1<x<1$ or $x<-3$ or $x>3$.

Answer: $(-\infty,-3) \cup(-1,1) \cup(3, \infty)$.