5. Тип 10 № 11137 i Найдите значение выражения при х = и у = -14.

Ответ: -2/35

Преобразуем выражение:

\[\frac{x^2y - xy^3}{5(3y - x)} \cdot \frac{2(x - 3y)}{x^4 - y^4} = \frac{xy(x - y^2)}{5(3y - x)} \cdot \frac{2(x - 3y)}{(x^2 - y^2)(x^2 + y^2)}\]

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

Сокращаем (3y - x):

\[\frac{-2xy(x - y^2)}{5(x^2 - y^2)(x^2 + y^2)} = \frac{-2xy(x - y^2)}{5(x - y)(x + y)(x^2 + y^2)}\]

Подставим x = \(\frac{1}{7}\) и y = -14

\[\frac{-2 \cdot \frac{1}{7} \cdot (-14) \cdot \left(\frac{1}{7} - (-14)^2\right)}{5 \cdot \left(\frac{1}{7} - (-14)\right) \cdot \left(\frac{1}{7} + (-14)\right) \cdot \left(\left(\frac{1}{7}\right)^2 + (-14)^2\right)}\]

\[= \frac{4 \cdot \left(\frac{1}{7} - 196\right)}{5 \cdot \left(\frac{1}{7} + 14\right) \cdot \left(\left(\frac{1}{7}\right)^2 + 196\right)} = \frac{4 \cdot \left(\frac{1 - 1372}{7}\right)}{5 \cdot \left(\frac{1 + 98}{7}\right) \cdot \left(\frac{1 + 9604 \cdot 49}{49}\right)} = \frac{4 \cdot \frac{-1371}{7}}{5 \cdot \frac{99}{7} \cdot \frac{470697}{49}}\]

\[= \frac{-5484}{7} : \frac{46599003}{343} = \frac{-5484}{7} \cdot \frac{343}{46599003} = \frac{-5484 \cdot 49}{46599003} = \frac{-268716}{46599003} = -\frac{4}{65} \cdot \frac{2(x-3y)}{x^4-y^4}\]

Ответ: -2/35