9. The real solution to the equation $x^{[x]}=\frac{9}{2}$ is $\qquad$ $([x]$ denotes the greatest integer not exceeding the real number $x$).
9. $\frac{3 \sqrt{2}}{2}$.

Clearly, $x>0$.
If $x \geqslant 3$, then $[x] \geqslant 3$. Therefore,
$$
x^{[x]} \geqslant 3^{3}=27>\frac{9}{2} \text {; }
$$

If $0<x<2$, then $0 \leqslant[x]<2$. Therefore, $x^{[x]}<2^{2}=4<\frac{9}{2}$.
Thus, $2 \leqslant x<3 \Rightarrow[x]=2$
$$
\Rightarrow x^{2}=\frac{9}{2} \Rightarrow x=\frac{3 \sqrt{2}}{2} \text {. }
$$