9.188. $\sqrt{x+3}+\sqrt{x-2}-\sqrt{2 x+4}>0$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}x+3 \geq 0, \\ x-2 \geq 0, \\ 2 x+4 \geq 0\end{array} \Leftrightarrow x \geq 2\right.$.

Let's rewrite the given inequality as $\sqrt{x+3}+\sqrt{x-2}>\sqrt{2 x+4}$ and square both sides. We have

$$
\begin{aligned}
& x+3+2 \sqrt{(x+3)(x-2)}+x-2>2 x+4 \Leftrightarrow 2 \sqrt{(x+3)(x-2)}>3 \Leftrightarrow \\
& \Leftrightarrow 4(x+3)(x-2)>9,4 x^{2}+4 x-33>0
\end{aligned}
$$

From this,

$$
\left[\begin{array}{l}
x>\frac{-1+\sqrt{34}}{2} \\
x\frac{-1+\sqrt{34}}{2} .\right.
$$

Answer: $x \in\left(\frac{-1+\sqrt{34}}{2} ; \infty\right)$.