9. Найдите значение выражения $$\frac{x^2y-xy^3}{2(y-x)} \cdot \frac{3(x-y)}{x^2-y^2}$$ при x = 4 и y = 1/4.

$$\frac{x^2y-xy^3}{2(y-x)} \cdot \frac{3(x-y)}{x^2-y^2} = \frac{xy(x-y^2)}{2(y-x)} \cdot \frac{3(x-y)}{(x-y)(x+y)} = \frac{xy(x+y)(x-y)}{2(y-x)} \cdot \frac{3(x-y)}{(x-y)(x+y)} = \frac{xy}{2(y-x)} \cdot 3 = -\frac{3xy}{2(x-y)}$$ Подставим x = 4 и y = 1/4: $$\frac{-3(4)(\frac{1}{4})}{2(4-\frac{1}{4})} = \frac{-3}{2(\frac{16}{4}-\frac{1}{4})} = \frac{-3}{2(\frac{15}{4})} = \frac{-3}{\frac{30}{4}} = \frac{-3 \cdot 4}{30} = \frac{-12}{30} = -\frac{2}{5} = -0.4$$ Ответ: -0.4