16. Найдите значение выражения (x³y + xy³) / (x-y)² + 5(x-y)/(x²+y²) при х = -3 и у= 1/3

Сначала упростим выражение:

$$\frac{x^3y + xy^3}{(x-y)^2} + \frac{5(x-y)}{x^2+y^2} = \frac{xy(x^2+y^2)}{(x-y)^2} + \frac{5(x-y)}{x^2+y^2}$$

Подставим значения x = -3 и y = 1/3:

$$\frac{-3 \cdot \frac{1}{3}((-3)^2+(\frac{1}{3})^2)}{(-3-\frac{1}{3})^2} + \frac{5(-3-\frac{1}{3})}{(-3)^2+(\frac{1}{3})^2} = \frac{-1(9+\frac{1}{9})}{(-\frac{10}{3})^2} + \frac{5(-\frac{10}{3})}{9+\frac{1}{9}} = \frac{-(9+\frac{1}{9})}{\frac{100}{9}} + \frac{-\frac{50}{3}}{9+\frac{1}{9}} = \frac{-\frac{82}{9}}{\frac{100}{9}} - \frac{\frac{50}{3}}{\frac{82}{9}} = -\frac{82}{100} - \frac{50}{3} \cdot \frac{9}{82} = -\frac{41}{50} - \frac{50 \cdot 3}{82} = -\frac{41}{50} - \frac{150}{82} = -\frac{41}{50} - \frac{75}{41} = \frac{-41 \cdot 41 - 75 \cdot 50}{50 \cdot 41} = \frac{-1681 - 3750}{2050} = \frac{-5431}{2050} = -2,64926829268$$

Ответ: -5431/2050 ≈ -2,65