Решение:
- Первая скобка: \( \frac{(1-x)^{1/4}}{2(1+x)^{3/4}} + \frac{(1+x)^{1/4}}{2(1-x)^{3/4}} = \frac{(1-x)^{1/4}(1-x)^{3/4} + (1+x)^{1/4}(1+x)^{3/4}}{2(1+x)^{3/4}(1-x)^{3/4}} = \frac{(1-x) + (1+x)}{2((1+x)(1-x))^{3/4}} = \frac{2}{2(1-x^2)^{3/4}} = \frac{1}{(1-x^2)^{3/4}} \).
- Второе выражение: \( (1-x)^{-1/2} \left( \frac{1+x}{1-x} \right)^{-1/4} = \frac{1}{(1-x)^{1/2}} \cdot \frac{(1+x)^{-1/4}}{(1-x)^{-1/4}} = \frac{1}{(1-x)^{1/2}} \cdot \frac{1}{(1+x)^{1/4}} \cdot (1-x)^{1/4} = \frac{(1-x)^{1/4}}{(1-x)^{1/2}(1+x)^{1/4}} = \frac{(1-x)^{-1/4}}{(1+x)^{1/4}} \).
- Перемножим результаты: \( \frac{1}{(1-x^2)^{3/4}} \cdot \frac{(1-x)^{-1/4}}{(1+x)^{1/4}} = \frac{1}{((1-x)(1+x))^{3/4}} \cdot \frac{(1-x)^{-1/4}}{(1+x)^{1/4}} = \frac{1}{(1-x)^{3/4}(1+x)^{3/4}} \cdot \frac{(1-x)^{-1/4}}{(1+x)^{1/4}} = \frac{(1-x)^{-1/4}}{(1-x)^{3/4}(1+x)^{3/4}(1+x)^{1/4}} = \frac{(1-x)^{-1/4}}{(1-x)^{3/4}(1+x)^{1}} = \frac{1}{(1-x)^{3/4}(1-x)^{1/4}(1+x)} = \frac{1}{(1-x)(1+x)} = \frac{1}{1-x^2} \).
Ответ: \( \frac{1}{1-x^2} \)