Решите пример:

Решите задачу:

$( \frac{x+3y}{x^2y-3xy^2} + \frac{3}{x^2+3xy} )* \frac{9y^3-x^2y}{(9y+x)^2} =$ $( \frac{x+3y}{xy(x-3y)} + \frac{3}{x(x+3y)} )* \frac{y(9y^2-x^2)}{(9y+x)^2}$ $= \frac{(x+3y)^2+3y(x-3y)}{xy(x-3y)(x+3y)} * \frac{y(9y^2-x^2)}{(9y+x)^2}$ $= \frac{x^2+6xy+9y^2+3xy-9y^2}{xy(x-3y)(x+3y)}* \frac{y(9y^2-x^2)}{(9y+x)^2}$ $= \frac{x^2+9xy}{xy(x-3y)(x+3y)} *\frac{y(9y^2-x^2)}{(9y+x)^2}$ $= \frac{x(x+9y)}{xy(x-3y)(x+3y)} *\frac{y(9y^2-x^2)}{(9y+x)^2}$ $= \frac{x}{xy(x-3y)(x+3y)} *\frac{y(9y^2-x^2)}{(9y+x)}$ $= \frac{1}{y(x-3y)(x+3y)} *\frac{y(9y^2-x^2)}{(9y+x)}$ $= \frac{1}{(x-3y)(x+3y)} *\frac{(9y^2-x^2)}{(9y+x)}$
$= \frac{1}{(x-3y)(x+3y)} *\frac{(3y-x)(3y+x)}{(9y+x)}$ $= \frac{1}{(x-3y)} *\frac{(3y-x)}{(9y+x)}$ $= \frac{-(x-3y)}{(x-3y)(9y+x)} = - \frac{1}{9y+x}$