x²³y-xy³ 3(x - y) 2(y-x) x2-y2 x=4y=1/4

Ответ: -3/2

\begin{aligned} &\frac{x^3y - xy^3}{2(y-x)} \cdot \frac{3(x-y)}{x^2 - y^2} = \frac{xy(x^2 - y^2)}{2(y-x)} \cdot \frac{3(x-y)}{x^2 - y^2} = \frac{xy \cdot 3(x-y)}{2(y-x)} = \frac{3xy(x-y)}{-2(x-y)} = \frac{3xy}{-2} = -\frac{3xy}{2}\\ &x = 4, y = \frac{1}{4}\\ &-\frac{3xy}{2} = -\frac{3 \cdot 4 \cdot \frac{1}{4}}{2} = -\frac{3}{2} \end{aligned}

Ответ: -3/2