10) \frac{x^3y-xy^3}{2(y-x)} : \frac{3(x-y)}{x^2-y^2} при х = 4 и у = \frac{1}{4}

Ответ: -\(\frac{17}{8}\)

Разбираемся:

Упростим выражение:

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

Подставим значения х = 4 и у = \(\frac{1}{4}\):

\[-\frac{4 \cdot \frac{1}{4} \cdot (4+\frac{1}{4})^2}{6} = -\frac{1 \cdot (\frac{17}{4})^2}{6} = -\frac{\frac{289}{16}}{6} = -\frac{289}{16 \cdot 6} = -\frac{289}{96}\]

Ответ: -\(\frac{289}{96}\)