Найдите значение выражения $$y = \frac{1}{4}$$

Найдем значение выражения $$\frac{x^3y-xy}{2(\frac{y}{x}-x)} \cdot \frac{3(x-y)}{x^2-y^2}$$ при $$x = 4$$ и $$y = \frac{1}{4}$$.

Упростим выражение:

$$\frac{x^3y-xy}{2(\frac{y}{x}-x)} \cdot \frac{3(x-y)}{x^2-y^2} = \frac{xy(x^2-1)}{2(\frac{y-x^2}{x})} \cdot \frac{3(x-y)}{(x-y)(x+y)} = \frac{xy(x^2-1)x}{2(y-x^2)} \cdot \frac{3}{x+y} = \frac{x^2y(x^2-1)}{2(y-x^2)} \cdot \frac{3}{x+y} = \frac{-x^2y(1-x^2)}{2(y-x^2)} \cdot \frac{3}{x+y} = \frac{-3x^2y(x^2-1)}{2(x+y)(y-x^2)}$$

$$\frac{x^3y-xy}{2(\frac{y}{x}-x)} \cdot \frac{3(x-y)}{x^2-y^2} = \frac{xy(x^2-1)}{2(\frac{y-x^2}{x})} \cdot \frac{3(x-y)}{(x-y)(x+y)} = \frac{xy(x^2-1)}{1} \cdot \frac{x}{2(y-x^2)} \cdot \frac{3}{x+y} = \frac{x^2y(x^2-1)}{2(y-x^2)} \cdot \frac{3}{x+y} = \frac{3x^2y(x^2-1)}{2(y-x^2)(x+y)}$$

Подставим $$x=4$$ и $$y=\frac{1}{4}$$:

$$\frac{3 \cdot 4^2 \cdot \frac{1}{4}(4^2-1)}{2(\frac{1}{4}-4^2)(4+\frac{1}{4})} = \frac{3 \cdot 16 \cdot \frac{1}{4} (16-1)}{2(\frac{1}{4}-16)(4+\frac{1}{4})} = \frac{3 \cdot 4 \cdot 15}{2(\frac{1-64}{4})(\frac{16+1}{4})} = \frac{12 \cdot 15}{2(\frac{-63}{4})(\frac{17}{4})} = \frac{180}{2 \cdot \frac{-63 \cdot 17}{16}} = \frac{180}{\frac{-1071}{8}} = \frac{180 \cdot 8}{-1071} = \frac{1440}{-1071} = -\frac{480}{357} = -\frac{160}{119}$$

Ответ: -1.3445378151