5. If the equation with respect to $x$
$$
2 x|x|-a|x|=1
$$

has three different real solutions, then the range of the real number $a$ is $\qquad$.
5. $a<-2 \sqrt{2}$.

From the given information, $y=2 x-a$ intersects with $y=\frac{1}{|x|}(x \neq 0)$ at exactly three points. Considering the extreme case, $y=2 x-a$ is tangent to $y=-\frac{1}{x}(x<0)$, we know that $a<-2 \sqrt{2}$.