(9) The real solution to the equation $x^{[x]}=\frac{9}{2}$ is $\qquad$ . (where $[x]$ denotes the greatest integer not exceeding $x$)
(9) $\frac{3 \sqrt{2}}{2}$ Hint: Clearly $x>0$. If $x \geqslant 3$, then $[x] \geqslant 3$, thus $x^{[x]} \geqslant 3^{3}=27>\frac{9}{2}$.
If $0<x<2$, then $0 \leqslant[x]<2$, thus $x^{[x]}<2^{2}=4<\frac{9}{2}$. Therefore, $2 \leqslant x<3$, so $[x]=2$, hence $x^{2}=\frac{9}{2}$, thus $x=\frac{3 \sqrt{2}}{2}$.