Задача 73111 Нужно решение ...
Условие
Решение
[m]((\frac{x}{y-x})^{-2}-\frac{(x+y)^2-4xy}{x^2-xy})^2 \cdot \frac{x^4}{x^2y^2-y^4} = ((\frac{y-x}{x})^{2}-\frac{x^2+2xy+y^2-4xy}{x(x-y)})^2 \cdot \frac{x^4}{y^2(x^2-y^2)} = [/m]
[m]=(\frac{x^2-2xy+y^2}{x^2}-\frac{x^2-2xy+y^2}{x(x-y)})^2 \cdot \frac{x^4}{y^2(x^2-y^2)} = (\frac{(x-y)^2(x-y)}{x^2(x-y)}-\frac{x(x-y)^2}{x^2(x-y)})^2 \cdot \frac{x^4}{y^2(x^2-y^2)} =[/m]
[m]= (\frac{(x-y)^3 - x(x^2-2xy+y^2)}{x^2(x-y)})^2 \cdot \frac{x^4}{y^2(x^2-y^2)} = (\frac{x^3-3x^2y+3xy^2-y^3 - x^3+2x^2y-xy^2}{x^2(x-y)})^2 \cdot \frac{x^4}{y^2(x^2-y^2)} =[/m]
[m]= (\frac{-x^2y+2xy^2-y^3}{x^2(x-y)})^2 \cdot \frac{x^4}{y^2(x^2-y^2)} =\frac{(-y(x^2-2xy+y^2))^2}{x^4(x-y)^2} \cdot \frac{x^4}{y^2(x-y)(x+y)} =[/m]
[m]= \frac{y^2((x-y)^2)^2}{x^4(x-y)^2} \cdot \frac{x^4}{y^2(x-y)(x+y)} = \frac{(x-y)^4}{(x-y)^2} \cdot \frac{1}{(x-y)(x+y)} =\frac{(x-y)^2}{(x-y)(x+y)} = \frac{x-y}{x+y}[/m]