9.196. $0.5^{2 \sqrt{x}}+2>3 \cdot 0.5^{\sqrt{x}}$.
## Solution.

Domain of definition: $x \geq 0$.

Let's rewrite the given inequality as $0.5^{2 \sqrt{x}} - 3 \cdot 0.5^{\sqrt{x}} + 2 > 0$ and solve it as a quadratic inequality in terms of $0.5^{\sqrt{x}}$. We get $\left[\begin{array}{l}0.5^{\sqrt{x}} > 2, \\ 0.5^{\sqrt{x}} < 1\end{array} \Leftrightarrow\right.$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0696.jpg?height=155&width=854&top_left_y=443&top_left_x=122)

Answer: $x \in (0 ; \infty)$.