9.062. $\left(\frac{2}{5}\right)^{\log _{0.25}\left(x^{2}-5 x+8\right)} \leq 2.5$.
## Solution.

Domain of definition: $x \in R$.

## We have

$\left(\frac{2}{5}\right)^{\log _{0.25}\left(x^{2}-5 x+8\right)} \leq\left(\frac{2}{5}\right)^{-1}, \log _{0.25}\left(x^{2}-5 x+8\right) \geq-1 \Leftrightarrow x^{2}-5 x+8 \leq 4 \Leftrightarrow$

$\Leftrightarrow x^{2}-5 x+4 \leq 0,1 \leq x \leq 4$.

Answer: $\quad x \in[1 ; 4]$