6.030. $\left(\frac{x^{2}+6}{x^{2}-4}\right)^{2}=\left(\frac{5 x}{4-x^{2}}\right)^{2}$.
## Solution.

Domain of definition: $x \neq \pm 2$.

Rewrite this equation as

$$
\begin{aligned}
& \left(\frac{x^{2}+6}{x^{2}-4}\right)^{2}=\left(\frac{5 x}{x^{2}-4}\right)^{2} \Rightarrow\left|\frac{x^{2}+6}{x^{2}-4}\right|=\sqrt{\left(\frac{5 x}{x^{2}-4}\right)^{2}} \Rightarrow\left|\frac{x^{2}+6}{x^{2}-4}\right|=\left|\frac{5 x}{x^{2}-4}\right| \Rightarrow \\
& \Rightarrow \frac{x^{2}+6}{x^{2}-4}=-\frac{5 x}{x^{2}-4} \text { or } \frac{x^{2}+6}{x^{2}-4}=\frac{5 x}{x^{2}-4}
\end{aligned}
$$

Considering the domain of definition, we get the equations $x^{2}+5 x+6=0$ or $x^{2}-5 x+6=0$,
from which $x_{1}=-3, x_{2}=-2, x_{3}=2, x_{4}=3 ; x_{2}=-2$ and $x_{3}=2$ do not satisfy the domain of definition.

Answer: $x_{1}=-3, x_{2}=3$.