Question

Помогите дам 99балов​

Answer 1

Ответ:

А)  1) 1 - $\frac{1}{1-y}$ = $\frac{1-y-1}{1-y}$  = $\frac{-y}{1-y}$

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

Б)   1) $\frac{x}{xy-y^{2} }$ - $\frac{y}{x^{2} -xy}$ = $\frac{x}{y(x-y)}$ - $\frac{y}{x(x-y)}$ = $\frac{x^{2} -y^{2} }{xy(x-y)}$ = $\frac{(x-y)(x+y)}{xy(x-y)}$ = $\frac{x+y}{xy}$

      2) $\frac{x+y}{xy}$ ÷ $\frac{x^{2} -y^{2} }{8xy}$ = $\frac{x+y}{xy}$ × $\frac{8xy}{x^{2} -y^{2} }$ = $(x+y)$ × $\frac{8}{(x-y)(x+y)}$ = $\frac{8}{x-y}$

Объяснение: