9.210. $\frac{5 x-7}{x-5}<4-\frac{x}{5-x}+\frac{3 x}{x^{2}-25}<4$.

9.210. $\frac{5 x-7}{x-5}<4-\frac{x}{5-x}+\frac{3 x}{x^{2}-25}<4$.
Solution.

Domain of definition: $x \neq \pm 5$.

Rewrite the inequality as a system of inequalities

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ \frac { 5 x - 7 } { x - 5 } < 4 - \frac { x } { 5 - x } + \frac { 3 x } { x ^ { 2 } - 2 5 } } \\
{ 4 - \frac { x } { 5 - x } + \frac { 3 x } { x ^ { 2 } - 2 5 } < 4 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{5 x-7}{x-5}-4-\frac{x}{x-5}-\frac{3 x}{x^{2}-25}<0, \\
\frac{3 x}{(x-5)(x+5)}+\frac{x}{x-5}<0
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ \frac { 2 x + 1 3 } { ( x - 5 ) ( x + 5 ) } < 0 , } \\
{ \frac { x ^ { 2 } + 8 x } { ( x - 5 ) ( x + 5 ) } < 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(2 x+13)(x-5)(x+5)<0, \\
x(x+8)(x-5)(x+5)<0 .
\end{array}\right.\right.
\end{aligned}
$$

We will solve the obtained system using the interval method

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0703.jpg?height=208&width=487&top_left_y=150&top_left_x=370)

The common solutions to the two inequalities in the system are $x \in\left(-8 ;-\frac{13}{2}\right) \cup(0 ; 5)$.

Answer: $x \in\left(-8 ;-\frac{13}{2}\right) \cup(0 ; 5)$.