Question

нужно 7 и 8, желательно быстрее :D

Answer 1

Объяснение:

7.

$(\frac{5}{y+5}- \frac{5}{5-y}+\frac{y^2+25}{y^2-25})*\frac{y+5}{y^2+10y+25}=\\=(\frac{5}{y+5}+ \frac{5}{y-5}+\frac{y^2+25}{y^2-25})*\frac{y+5}{y^2+10y+25}=\\=\frac{5(y-5)+5(y+5)+y^2+25}{(y-5)(y+5)} *\frac{y+5}{y^2+10y+25}=\\=\frac{5y-25+5y+25+y^2+25}{(y-5)(y+5)} *\frac{y+5}{y^2+10y+25}=\\=\frac{5y+5y+y^2+25}{(y-5)(y+5)} *\frac{y+5}{y^2+10y+25}=\\=\frac{y^2+10y+25}{(y-5)(y+5)} *\frac{y+5}{y^2+10y+25}=\frac{1}{y-5}$

8.

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

$=\frac{y-x}{x-y} =-\frac{x-y}{x-y} =-1$