9.110. Find the domain of the function $f$, if

$f(x)=\sqrt{9-\left(\frac{4 x-22}{x-5}\right)^{2}}$
## Solution.

We obtain the inequality $9-\left(\frac{4 x-22}{x-5}\right)^{2} \geq 0 \Leftrightarrow\left(\frac{4 x-22}{x-5}\right)^{2} \leq 9 \Leftrightarrow$

$$
\begin{aligned}
& \Leftrightarrow\left|\frac{4 x-22}{x-5}\right| \leq 3 \Leftrightarrow-3 \leq \frac{4 x-22}{x-5} \leq 3 \Leftrightarrow\left\{\begin{array}{l}
\frac{4 x-22}{x-5} \leq 3, \\
\frac{4 x-22}{x-5} \geq-3
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ \frac { 4 x - 2 2 } { x - 5 } - 3 \leq 0 , } \\
{ \frac { 4 x - 2 2 } { x - 5 } + 3 \geq 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ \frac { x - 7 } { x - 5 } \leq 0 , } \\
{ \frac { 7 x - 3 7 } { x - 5 } \geq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x-7)(x-5) \leq 0, \\
(7 x-37)(x-5) \geq 0, \\
x \neq 5
\end{array}\right.\right.\right.
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0650.jpg?height=204&width=575&top_left_y=1509&top_left_x=411)

Using the interval method, we get $x \in\left[\frac{37}{7} ; 7\right]$.

Answer: $x \in\left[\frac{37}{7} ; 7\right]$.