17. Найдите значение выражения \frac{x^2y - xy^2}{5(3y - x)} \cdot \frac{2(x-3y)}{x^4 - y^4} при x = -\frac{1}{7} и y = -14.

Решение: 1. \frac{x^2y - xy^2}{5(3y - x)} \cdot \frac{2(x-3y)}{x^4 - y^4} = \frac{xy(x - y)}{5(3y - x)} \cdot \frac{2(x-3y)}{(x^2 - y^2)(x^2 + y^2)} = \frac{xy(x - y)}{-5(x - 3y)} \cdot \frac{2(x-3y)}{(x - y)(x + y)(x^2 + y^2)} = \frac{-2xy}{5(x + y)(x^2 + y^2)} 2. x = -\frac{1}{7}, y = -14 3. x + y = -\frac{1}{7} - 14 = -\frac{1}{7} - \frac{98}{7} = -\frac{99}{7} 4. x^2 + y^2 = (-\frac{1}{7})^2 + (-14)^2 = \frac{1}{49} + 196 = \frac{1}{49} + \frac{9604}{49} = \frac{9605}{49} 5. xy = -\frac{1}{7} * (-14) = 2 6. \frac{-2 * 2}{5(-\frac{99}{7})(\frac{9605}{49})} = \frac{-4}{5(-\frac{99}{7})(\frac{9605}{49})} = \frac{-4}{\frac{-4756425}{343}} = \frac{1372}{475475} \approx 0.00288 Ответ: \frac{1372}{475475}