Solution.

Domain of definition: $4-x^{3}>0, x^{3}<0$. Expanding the absolute values, we have the following three cases:

$$
\text { 1) }\left\{\begin{array} { l } 
{ x  0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x0
\end{array} \varnothing\right.\right.
$$

2) $\left\{\begin{array}{l}-2 \leq x0\end{array} \Leftrightarrow\left\{\begin{array}{l}-2 \leq x-1\end{array} \Leftrightarrow-10\end{array} \Leftrightarrow\left\{\begin{array}{l}x \geq 0, \\ 2>0\end{array} \Leftrightarrow x \geq 0\right.\right.$.

$x \in(-1 ; \infty)$. Considering the domain of definition, we have $x \in(-1 ; \sqrt[3]{4})$.

Answer: $x \in(-1 ; \sqrt[3]{4})$.
$9.133 \frac{|x+2|-|x|}{\sqrt{4-x^{3}}}>0$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.

$9.133 \frac{|x+2|-|x|}{\sqrt{4-x^{3}}}>0$.