Номер 6 и 7 срочнооооо

$\displaystyle 6)..( \frac{x-2}{x^{2}-2x+4}- \frac{xy-2y}{x^{3}+8}): \frac{x-y+2}{x^{3}y+8y}= \\ \\ =( \frac{x-2}{x^{2}-2x+4}- \frac{y(x-2)}{(x+2)(x^{2}-2x+4)})* \frac{y(x+2)(x^{2}-2x+4)}{x-y+2}= \\ \\ = \frac{(x+2)(x-2)-y(x-2)}{(x+2)(x^{2}-2x+4)}* \frac{y(x+2)(x^{2}-2x+4)}{x-y+2}= \\ \\ = \frac{y(x-2)(x-y+2)}{x-y+2}=y(x-2);$

Ответ: В).

$\displaystyle 7).. \frac{x/y-25y/x}{x/y+25y/x-10}= \frac{(x^{2}-25y^{2})/xy}{(x^{2}-10xy+25y^{2})/xy}= \\ \\ = \frac{(x-5y)(x+5y)}{(x-5y)^{2}}= \frac{x+5y}{x-5y};$

Ответ: В).