9.184. $\log _{0.25}\left|\frac{2 x+1}{x+3}+\frac{1}{2}\right|>\frac{1}{2}$.
## Solution.

The given inequality is equivalent to the inequality

$$
0 - \frac { 1 } { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x \neq - 1 , } \\
{ x \neq - 3 , } \\
{ \frac { 2 x + 1 } { x + 3 }  0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x \neq -1, \\
x \neq -3, \\
-3 - \frac{4}{3} \\
x < -3
\end{array}\right.}
\end{array} \Leftrightarrow\right.\right.\right. \\
& \Leftrightarrow x \in \left(-\frac{4}{3} ; -1\right) \cup \left(-1 ; -\frac{1}{2}\right)
\end{aligned}
$$

Answer: \( x \in \left(-\frac{4}{3} ; -1\right) \cup \left(-1 ; -\frac{1}{2}\right) \)