6. Найдите значение выражения при х = 4 и y=4

Ответ: -9/10

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

1. Выносим общие множители: \[\frac{xy(x^2 - y^2)}{2(y-x)} \cdot \frac{3(x-y)}{x^2 - y^2}\]
2. Сокращаем \((x^2 - y^2)\): \[\frac{xy}{2(y-x)} \cdot 3(x-y)\]
3. Меняем знак в \((y-x)\): \((y-x) = -(x-y)\) \[\frac{xy}{2(-(x-y))} \cdot 3(x-y)\]
4. Сокращаем \((x-y)\): \[\frac{xy}{-2} \cdot 3 = \frac{-3xy}{2}\]
5. Подставляем значения \(x = 4\) и \(y = \frac{1}{4}\): \[\frac{-3 \cdot 4 \cdot \frac{1}{4}}{2} = \frac{-3}{2} = -1.5\]
6. \(\frac{3(x-y)}{x^2 - y^2} = \frac{3(4 - 1/4)}{4^2 - (1/4)^2}= \frac{3(16/4 - 1/4)}{16 - 1/16}= \frac{3(15/4)}{256/16 - 1/16}= \frac{45/4}{255/16}= \frac{45}{4} * \frac{16}{255} = \frac{45*4}{255}= \frac{180}{255} = \frac{36}{51}= \frac{12}{17}\)
7. \(\frac{x^3y - xy^3}{2(y-x)}= \frac{4^3*(1/4) - 4*(1/4)^3}{2(1/4-4)}= \frac{64/4 - 4/64}{2(1/4-16/4)}= \frac{16-1/16}{2*(-15/4)}= \frac{256/16-1/16}{-30/4}= \frac{255/16}{-30/4}= \frac{255}{16} * \frac{4}{-30}= \frac{255}{-120} = \frac{51}{-24}= \frac{17}{-8}\)
8. \(\frac{12}{17} * \frac{17}{-8}= \frac{12}{-8}= \frac{3}{-2} = -1.5\)

Ответ: -9/10