Дано выражение:\[\frac{x^2y + xy^2}{2(y-x)} \cdot \frac{5(x-y)}{x^2+y^2}\]
Упростим выражение:\[\frac{xy(x+y)}{2(y-x)} \cdot \frac{5(x-y)}{x^2+y^2} = \frac{xy(x+y)}{2(y-x)} \cdot \frac{-5(y-x)}{x^2+y^2} = \frac{-5xy(x+y)}{2(x^2+y^2)}\]
Подставим x = -3 и y = 1/3:\[\frac{-5(-3)(\frac{1}{3})(-3+\frac{1}{3})}{2((-3)^2+(\frac{1}{3})^2)} = \frac{5(-\frac{8}{3})}{2(9+\frac{1}{9})} = \frac{-\frac{40}{3}}{2(\frac{82}{9})} = \frac{-\frac{40}{3}}{\frac{164}{9}} = -\frac{40}{3} \cdot \frac{9}{164} = -\frac{40 \cdot 3}{164} = -\frac{120}{164} = -\frac{30}{41}\]
Ответ: -30/41