2.137. $\left(\frac{x+2 b}{x-2 b}+\frac{x+2 a}{x-2 a}\right): \frac{x}{2} ; \quad x=\frac{4 a b}{a+b}$.
## Solution.

Domain of definition: $a \neq-b \neq 0$.

$$
\begin{aligned}
& \left(\frac{\frac{4 a b}{a+b}+2 b}{\frac{4 a b}{a+b}-2 b}+\frac{\frac{4 a b}{a+b}+2 a}{\frac{4 a b}{a+b}-2 a}\right): \frac{4 a b}{2(a+b)}= \\
& =\left(\frac{4 a b+2 a b+b^{2}}{a+b}: \frac{4 a b-2 a b-2 b^{2}}{a+b}+\frac{4 a b+2 a^{2}+2 a b}{a+b}: \frac{4 a b-2 a^{2}-2 a b}{a+b}\right) \times \\
& \times \frac{a+b}{2 a b}=\left(\frac{2 b(3 a+b)}{a+b} \cdot \frac{a+b}{2 b(a-b)}+\frac{2 b(3 b+a)}{a+b} \cdot \frac{a+b}{2 b(b-a)}\right) \cdot \frac{a+b}{2 a b}= \\
& =\left(\frac{3 a+b}{a-b}+\frac{3 b+a}{b-a}\right) \cdot \frac{a+b}{2 a b}=\left(\frac{3 a+b}{a-b}-\frac{3 b+a}{a-b}\right) \cdot \frac{a+b}{2 a b}= \\
& =\frac{3 a+b-3 b-a}{a-b} \cdot \frac{a+b}{2 a b}=\frac{2 a-2 b}{a-b} \cdot \frac{a+b}{2 a b}=\frac{2(a-b)(a+b)}{2(a-b) a b}=\frac{a+b}{a b} \text {. }
\end{aligned}
$$

Answer: $\frac{a+b}{a b}$.