Решение

B)

$$\begin{aligned} &\left(\frac{st}{s^2-t^2} + \frac{t}{2t-2s}\right) \cdot \frac{s+t}{2t} = \left(\frac{st}{(s-t)(s+t)} - \frac{t}{2(s-t)}\right) \cdot \frac{s+t}{2t} =\\ &= \left(\frac{2st - t(s+t)}{2(s-t)(s+t)}\right) \cdot \frac{s+t}{2t} = \frac{2st - ts - t^2}{2(s-t)(s+t)} \cdot \frac{s+t}{2t} = \frac{st - t^2}{2(s-t)(s+t)} \cdot \frac{s+t}{2t} = \\ &= \frac{t(s-t)}{2(s-t)(s+t)} \cdot \frac{s+t}{2t} = \frac{t(s-t)(s+t)}{2(s-t)(s+t)2t} = \frac{1}{4} \end{aligned}$$

Ответ: $$\frac{1}{4}$$

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Г)

$$\begin{aligned} &\frac{3a+b}{a^2b-ab^2} + \frac{b-a}{ab} : \frac{a^2-b^2}{3a-b} = \frac{3a+b}{ab(a-b)} + \frac{b-a}{ab} \cdot \frac{3a-b}{a^2-b^2} = \frac{3a+b}{ab(a-b)} - \frac{a-b}{ab} \cdot \frac{3a-b}{(a-b)(a+b)} = \\ &= \frac{3a+b}{ab(a-b)} - \frac{3a-b}{ab(a+b)} = \frac{(3a+b)(a+b)-(3a-b)(a-b)}{ab(a-b)(a+b)} = \frac{3a^2 + 3ab + ab + b^2 - (3a^2 -3ab - ab + b^2)}{ab(a-b)(a+b)} = \\ &= \frac{3a^2 + 4ab + b^2 - 3a^2 + 4ab - b^2}{ab(a-b)(a+b)} = \frac{8ab}{ab(a-b)(a+b)} = \frac{8}{(a-b)(a+b)} = \frac{8}{a^2 - b^2} \end{aligned}$$

Ответ: $$\frac{8}{a^2 - b^2}$$