\[\left( \frac{2ab}{a^{2} - b^{2}} + \frac{a - b}{2a + 2b} \right) \cdot \frac{2a}{a + b} + \frac{b}{b - a} = 1\]
\[1)\ \frac{2ab^{\backslash 2}}{(a - b)(a + b)} + \frac{a - b^{\backslash a - b}}{2(a + b)} =\]
\[= \frac{4ab + a^{2} - 2ab + b^{2}}{2(a - b)(a + b)} =\]
\[= \frac{a^{2} + 2ab + b^{2}}{2(a - b)(a + b)} =\]
\[= \frac{(a + b)^{2}}{2(a - b)(a + b)} = \frac{a + b}{2(a - b)}\]
\[2)\ \frac{a + b}{2(a - b)} \cdot \frac{2a}{a + b} = \frac{a}{a - b}\]
\[3)\frac{a}{a - b} + \frac{b}{b - a} =\]
\[= \frac{a}{a - b} - \frac{b}{a - b} = \frac{a - b}{a - b} = 1\]
\[Не\ зависит\ от\ переменных.\]
\[Что\ и\ требовалось\ доказать.\]