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

$$\frac{x}{x-y}-(\frac{xy}{x^2-y^2}):\frac{4x^2}{x^2-2xy+y^2}=\frac{x}{x-y}-(\frac{xy}{(x-y)(x+y)}):\frac{4x^2}{(x-y)^2}=\frac{x}{x-y}-(\frac{xy}{(x-y)(x+y)} \cdot \frac{(x-y)^2}{4x^2})=\frac{x}{x-y}-(\frac{xy(x-y)^2}{(x-y)(x+y)4x^2})=\frac{x}{x-y}-\frac{y(x-y)}{4x(x+y)}=\frac{x}{x-y}-\frac{xy-y^2}{4x^2+4xy}=\frac{4x^2+4xy-xy+y^2}{(x-y)(4x(x+y))}=\frac{4x^2+3xy+y^2}{4x(x-y)(x+y)}$$

при x=-2; y=-1

$$\frac{4 \cdot (-2)^2 + 3 \cdot (-2) \cdot (-1) + (-1)^2}{4 \cdot (-2) ((-2)-(-1))((-2)+(-1))}=\frac{4 \cdot 4 + 3 \cdot 2 + 1}{(-8)(-2+1)(-2-1)}=\frac{16+6+1}{(-8)(-1)(-3)}=\frac{23}{-24}$$

Ответ: $$\frac{-23}{24}$$