6.006. $x+\frac{1}{x}=2 \frac{m^{2}+n^{2}}{m^{2}-n^{2}}$.

6.006. $x+\frac{1}{x}=2 \frac{m^{2}+n^{2}}{m^{2}-n^{2}}$.
## Solution.

Domain of definition: $x \neq 0, m \neq \pm n$.

$$
x+\frac{1}{x}=2 \frac{m^{2}+n^{2}}{m^{2}-n^{2}} \Leftrightarrow x^{2}-2 \frac{m^{2}+n^{2}}{m^{2}-n^{2}} x+1=0,
$$

roots $x_{1}=\frac{m+n}{m-n}, x_{2}=\frac{m-n}{m+n}$.

Answer: $x_{1}=\frac{m+n}{m-n}, x_{2}=\frac{m-n}{m+n}$, where $m \neq|n|$.