АЗ. Найдите числовое значение выражения: x xy 4x² x-y-(x²-y²) : x²-2xy + y² при х = -2; у = -1.

$$ (\frac{x}{x-y} - \frac{xy}{x^2-y^2}) : \frac{4x^2}{x^2-2xy + y^2} = (\frac{x}{x-y} - \frac{xy}{(x-y)(x+y)}) : \frac{4x^2}{(x-y)^2} = (\frac{x(x+y) - xy}{(x-y)(x+y)}) : \frac{4x^2}{(x-y)^2} = \frac{x^2 + xy - xy}{(x-y)(x+y)} : \frac{4x^2}{(x-y)^2} = \frac{x^2}{(x-y)(x+y)} \cdot \frac{(x-y)^2}{4x^2} = \frac{x^2(x-y)^2}{4x^2(x-y)(x+y)} = \frac{x-y}{4(x+y)} $$

При $$x = -2$$ и $$y = -1$$:

$$ \frac{-2 - (-1)}{4(-2+(-1))} = \frac{-2+1}{4(-3)} = \frac{-1}{-12} = \frac{1}{12} $$

Ответ: $$ \frac{1}{12} $$