Question

Выражение 1/2xy^2 - (x/x-y-x/y-x)*(1/x^2-2/xy+1/y^2)

Answer 1

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

$= \frac{1}{2xy^2} - \frac{2(x-y)}{xy^2} = \frac{1 - 2*2(x-y)}{2xy^2} = \frac{1-4x+4y}{2xy^2} = \\ \\ =\frac{1}{xy^2} (0.5 - 2x + 2y) \\ \\$