Question

Решите пожалуйста, я тороплюсь ​

Answer 1

$\frac{x}{y} :\frac{6x^{2}}{y} =\frac{x}{y} *\frac{y}{6x^{2}} =\frac{1}{6x}$

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

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

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

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

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

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