Найдите значение выражения \[\frac{x^3y-xy^3}{2(y-x)} : \frac{3(x-y)}{x^2-y^2}\] при \[x=4\] и \[y=\frac{1}{4}\]

Ответ: -\frac{65}{32}

Упрощаем выражение:

\[\frac{x^3y-xy^3}{2(y-x)} : \frac{3(x-y)}{x^2-y^2} = \frac{xy(x^2-y^2)}{2(y-x)} \cdot \frac{(x-y)(x+y)}{3(x-y)} = \frac{xy(x-y)(x+y)(x+y)}{2(y-x) \cdot 3(x-y)} = \frac{-xy(x+y)^2}{6(x-y)}\]

Подставляем значения \[x=4\] и \[y=\frac{1}{4}\]:

\[\frac{-4 \cdot \frac{1}{4} (4+\frac{1}{4})^2}{6(4-\frac{1}{4})} = \frac{-1 \cdot (\frac{17}{4})^2}{6 \cdot \frac{15}{4}} = \frac{-\frac{289}{16}}{\frac{90}{4}} = -\frac{289}{16} \cdot \frac{4}{90} = -\frac{289}{360} = -\frac{65}{32}\]

Ответ: -\frac{65}{32}