9.180. $\frac{6 x}{x-2}-\sqrt{\frac{12 x}{x-2}}-2 \sqrt[4]{\frac{12 x}{x-2}}>0$.
## Solution.

Domain of definition: $\frac{x}{x-2} \geq 0 ; x \neq 2$.

$$
\begin{aligned}
& \text { Let } \sqrt[4]{\frac{12 x}{x-2}}=y \geq 0 . \text { In terms of } y \text {, the inequality is } \\
& \frac{y^{4}}{2}-y^{2}-2 y>0 \Leftrightarrow y\left(y^{3}-2 y-4\right)>0 \Leftrightarrow y(y-2)\left(y^{2}+2 y+2\right)>0
\end{aligned}
$$

Since $y^{2}+2 y+2>0$ for $y \in R$, we obtain the inequality $y(y-2)>0 \Leftrightarrow$

$$
\begin{aligned}
\Leftrightarrow & y>2 \Leftrightarrow \sqrt[4]{\frac{12 x}{x-2}}>2, \frac{12 x}{x-2}>16, \frac{3 x}{x-2}-4>0, \frac{3 x-4+8}{x-2}>0 \\
& \frac{-x+8}{x-2}>0, \frac{x-8}{x-2}<0,(x-8)(x-2)<0,2<x<8
\end{aligned}
$$

Answer: $x \in(2 ; 8)$.