12. Тип 10 № 11134 i Найдите значение выражения \frac{xy-xy^3}{2(y-x)} - \frac{3(x-y)}{x^2-y^2} при x = 4 и y = \frac{1}{4}.

Ответ: -\frac{255}{68}

1. Упрощаем первое слагаемое: \[\frac{xy-xy^3}{2(y-x)} = \frac{xy(1-y^2)}{2(y-x)} = \frac{xy(1-y)(1+y)}{2(y-x)} = -\frac{xy(1-y)(1+y)}{2(x-y)}\]
2. Упрощаем второе слагаемое: \[\frac{3(x-y)}{x^2-y^2} = \frac{3(x-y)}{(x-y)(x+y)} = \frac{3}{x+y}\]
3. Объединяем: \[-\frac{xy(1-y)(1+y)}{2(x-y)} - \frac{3}{x+y}\]
4. Подставляем значения x = 4 и y = 1/4: \[-\frac{4 \cdot \frac{1}{4} \cdot (1-\frac{1}{4})(1+\frac{1}{4})}{2(4-\frac{1}{4})} - \frac{3}{4+\frac{1}{4}}\] \[=-\frac{1 \cdot \frac{3}{4} \cdot \frac{5}{4}}{2 \cdot \frac{15}{4}} - \frac{3}{\frac{17}{4}}\] \[=-\frac{\frac{15}{16}}{\frac{30}{4}} - \frac{12}{17}\] \[=-\frac{15}{16} \cdot \frac{4}{30} - \frac{12}{17}\] \[=-\frac{1}{2 \cdot 4} - \frac{12}{17}\] \[=-\frac{1}{8} - \frac{12}{17}\] \[=-\frac{17 + 96}{8 \cdot 17}\] \[=-\frac{113}{136}\]

Ответ: -\frac{113}{136}