9.170. $\frac{10(5-x)}{3(x-4)}-\frac{11}{3} \cdot \frac{6-x}{x-4} \geq \frac{5(6-x)}{x-2}$.
## Solution.

Domain of definition: $x \neq 2, x \neq 4$.

We have $\frac{x-16}{3(x-4)}-\frac{5(6-x)}{x-2} \geq 0 \Leftrightarrow \frac{(x-16)(x-2)-15(6-x)(x-4)}{3(x-4)(x-2)} \geq 0$,

$\frac{x^{2}-34 x+76}{(x-4)(x-2)} \geq 0, \frac{\left(x-\frac{7}{2}\right)(x-7)}{(x-4)(x-2)} \geq 0 \Leftrightarrow$

$$
\Leftrightarrow\left\{\begin{array}{l}
\left(x-\frac{7}{2}\right)(x-7)(x-4)(x-2) \geq 0 \\
x \neq 4 \\
x \neq 2
\end{array}\right.
$$

Using the interval method, we get $x \in(-\infty ; 2) \cup\left[\frac{7}{2} ; 4\right) \cup[7 ; \infty)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0684.jpg?height=134&width=447&top_left_y=1096&top_left_x=465)

Answer: $x \in(-\infty ; 2) \cup\left[\frac{7}{2} ; 4\right) \cup[7 ; \infty)$.