9. Let $[x]$ denote the greatest integer not exceeding the real number $x$, and $\{x\}=x-[x]$. Then the solution to the equation
$$
[x]^{4}+\{x\}^{4}+x^{4}=2048
$$

is
9. $-3-\sqrt{5}$.
$$
\begin{array}{l}
\text { Let } a=[x], b=\{x\} \text {. Then } \\
x=[x]+\{x\}=a+b . \\
\text { Hence } x^{4}+[x]^{4}+\{x\}^{4} \\
=(a+b)^{4}+a^{4}+b^{4} \\
=2\left(a^{4}+2 a^{3} b+3 a^{2} b^{2}+2 a b^{3}+b^{4}\right) \\
=2\left(a^{2}+a b+b^{2}\right)^{2} \\
\Rightarrow a^{2}+a b+b^{2}=32 .
\end{array}
$$

If $a>0$, then
$$
\begin{array}{l}
a^{2} \leqslant a^{2}+a b+b^{2}<a^{2}+a+1<(a+1)^{2} \\
\Rightarrow a=5 .
\end{array}
$$

Thus, the equation (1) becomes
$$
b^{2}+5 b-7=0 \Rightarrow b=\frac{-5 \pm \sqrt{53}}{2} \text {, }
$$

but $0 \leqslant b<1$, no solution.
If $a<0$, then
$$
\begin{array}{l}
a^{2}+a<a^{2}+a b+b^{2}<a^{2}+1 \\
\Rightarrow a^{2}+2 a+1<a^{2}+a b+b^{2} \leqslant a^{2} \\
\Rightarrow a=-6 .
\end{array}
$$

Thus, the equation (1) becomes
$$
b^{2}-6 b+4=0 \Rightarrow b=3 \pm \sqrt{5} \text {. }
$$

Since $0 \leqslant b<1$, hence $b=3-\sqrt{5}$.
Therefore, $x=a+b=-3-\sqrt{5}$.