50. Найдите значение выражения $$\frac{x^3y+xy^3}{2(y-x)} - \frac{5(x-y)}{x^2+y^2}$$ при $$x = -3$$ и $$y = \frac{1}{3}$$.

Решение: 1. Упростим выражение: $$\frac{x^3y+xy^3}{2(y-x)} - \frac{5(x-y)}{x^2+y^2} = \frac{xy(x^2+y^2)}{2(y-x)} - \frac{5(x-y)}{x^2+y^2}$$ 2. Подставим значения $$x = -3$$ и $$y = \frac{1}{3}$$: $$\frac{(-3)(\frac{1}{3})((-3)^2+(\frac{1}{3})^2)}{2(\frac{1}{3}-(-3))} - \frac{5(-3-\frac{1}{3})}{(-3)^2+(\frac{1}{3})^2} = \frac{-1(9+\frac{1}{9})}{2(\frac{1}{3}+3)} - \frac{5(-\frac{10}{3})}{9+\frac{1}{9}} = \frac{-(9+\frac{1}{9})}{2(\frac{10}{3})} - \frac{-\frac{50}{3}}{9+\frac{1}{9}}$$ 3. Продолжим упрощение: $$= \frac{-\frac{82}{9}}{\frac{20}{3}} + \frac{\frac{50}{3}}{\frac{82}{9}} = -\frac{82}{9} \cdot \frac{3}{20} + \frac{50}{3} \cdot \frac{9}{82} = -\frac{82}{60} + \frac{450}{246} = -\frac{41}{30} + \frac{225}{123} = \frac{-41\cdot41 + 225\cdot10}{30\cdot41} = \frac{-1681+2250}{1230} = \frac{569}{1230}$$ **Ответ: $$\frac{569}{1230}$$**