9.208. $\sqrt{x^{2}-9 x+20} \leq \sqrt{x-1} \leq \sqrt{x^{2}-13}$.

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9.208. $\sqrt{x^{2}-9 x+20} \leq \sqrt{x-1} \leq \sqrt{x^{2}-13}$.
Solution.

Domain of definition: $\left\{\begin{array}{l}x^{2}-9 x+20 \geq 0, \\ x-1 \geq 0, \\ x^{2}-13 \geq 0\end{array} \Leftrightarrow\left[\begin{array}{l}\sqrt{13} \leq x \leq 4, \\ x \geq 5 .\end{array}\right.\right.$

The given inequality is equivalent to the system of inequalities

$\left\{\begin{array}{l}\sqrt{x^{2}-9 x+20} \leq \sqrt{x-1}, \\ \sqrt{x^{2}-13} \geq \sqrt{x-1}\end{array} \Leftrightarrow\left\{\begin{array}{l}x^{2}-9 x+20 \leq x-1, \\ x^{2}-13 \geq x-1\end{array} \Leftrightarrow\right.\right.$

$\Leftrightarrow\left\{\begin{array}{l}x^{2}-10 x+21 \leq 0, \\ x^{2}-x-12 \geq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}3 \leq x \leq 7, \\ {\left[\begin{array}{l}x \geq 4, \\ x \leq-3 .\end{array}\right.}\end{array}\right.\right.$

By the method of intervals, we have (taking into account the domain of definition) $\left[\begin{array}{l}x=4, \\ 5 \leq x \leq 7\end{array}\right.$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0701.jpg?height=109&width=443&top_left_y=1708&top_left_x=380)

Answer: $x \in[5 ; 7] \cup 4$.