2. Solve the inequality $\sqrt{\frac{x-4}{x+3}}-\sqrt{\frac{x+3}{x-4}}<\frac{7}{12}$.
# Solution:

$$
\begin{aligned}
& \sqrt{\frac{x-4}{x+3}}-\sqrt{\frac{x+3}{x-4}}=0 ; \frac{1}{y}-y=0 ; \\
& y_{1,2}=\frac{-7 \pm \sqrt{49+576}}{24}=\frac{-7 \pm 25}{24}, y_{1}=\frac{3}{4}, y_{2}=-\frac{4}{3} . \text { Therefore, } \\
& y>\frac{3}{4}, \frac{x+3}{x-4}>\frac{9}{16} \Leftrightarrow \frac{x+12}{x-4}>0 \Leftrightarrow\left[\begin{array}{c}
x<-12 \\
x>4 .
\end{array}\right.
\end{aligned}
$$

Answer: $x \in(-\infty ;-12) \cup(4 ;+\infty)$.