$1.15 \frac{\sqrt{a^{2}-4 a b+4 b^{2}}}{\sqrt{a^{2}+4 a b+4 b^{2}}}-\frac{8 a b}{a^{2}-4 b^{2}}+\frac{2 b}{a-2 b}, 0<a<2 b$.
### 1.15 We have

$\sqrt{a^{2}-4 a b+4 b^{2}}=\sqrt{(a-2 b)^{2}}=|a-2 b|=2 b-a$,

$\sqrt{a^{2}+4 a b+4 b^{2}}=\sqrt{(a+2 b)^{2}}=|a+2 b|=2 b+a ;$

Here, formulas (1.19), (1.10), and (1.23) were used. Therefore,

$\frac{\sqrt{a^{2}-4 a b+4 b^{2}}}{\sqrt{a^{2}+4 a b+4 b^{2}}}=\frac{2 b-a}{2 b+a}$.

## Now we find

$$
\begin{aligned}
& \frac{2 b-a}{2 b+a}-\frac{8 a b}{a^{2}-4 b^{2}}+\frac{2 b}{a-2 b}= \\
& =\frac{(2 b-a)(a-2 b)-8 a b+2 b(2 b+a)}{a^{2}-4 b^{2}}=\frac{a}{2 b-a}
\end{aligned}
$$

Answer: $\frac{a}{2 b-a}$.