пж срочно! решите в течении 2 часов! номер 298 (3,4) даю 100 баллов

Решите задачу:

$(\frac{5}{x-y}- \frac{5x}{y^2-x^2})* \frac{x^2+2xy+y^2}{10x+5y}=(\frac{5}{x-y}+ \frac{5x}{(x-y)(x+y)})* \frac{(x+y)^2}{10x+5y}= \\ =\frac{5x+5y+5x}{(x-y)(x+y)}* \frac{(x+y)^2}{10x+5y}=\frac{10x+5y}{(x-y)(x+y)}* \frac{(x+y)^2}{10x+5y}= \frac{x+y}{x-y} \\ \frac{x+y}{x-y}= \frac{3.5-0.5}{3.5+0.5}= \frac{3}{4}=0.75$

$(\frac{xy}{y^2-x^2}- \frac{x}{2x-2y}): \frac{4x}{3y^2-3x^2}=(\frac{xy}{(y-x)(y+x)}+\frac{x}{2y-2x}): \frac{4x}{3(y^2-x^2)}= \\ =(\frac{xy}{(y-x)(y+x)}+\frac{x}{2(y-x)}): \frac{4x}{3(y^2-x^2)}=\frac{2xy+xy+x^2}{2(y-x)(y+x)}: \frac{4x}{3(y^2-x^2)}= \\ =\frac{3xy+x^2}{2(y-x)(y+x)}* \frac{3(y^2-x^2)}{4x}=\frac{x(3y+x)}{2(y-x)(y+x)}* \frac{3(y-x)(y+x)}{4x}= \frac{3(3y+x)}{8} \\ \frac{3(3y+x)}{8}= \frac{3(3*(-1.5)+(-2.5))}{8}= \frac{3(-4.5-2.5)}{8}= \frac{-21}{8}=-2.625$