Пожалуйста, помогите решить по действиям!!!

Решите задачу:

$( \frac{c+5}{5c-1}+ \frac{c+5}{c+1}): \frac{c^2+5c}{1-5c}+ \frac{c^2+5}{c+1} \\\\ \boxed{1} \ \ \frac{c+5}{5c-1} + \frac{c+5}{c+1}= \frac{(c+1)(c+5)+(5c-1)(c+5)}{(5c-1)(c+1)}= \frac{(c+5)(c+1+5c-1)}{y} = \frac{6c(c+5)}{(5c-1)(c+1)} \\\\ \boxed{2} \ \ \frac{6c(c+5)}{(5c-1)(c+1)} :\frac{c^2+5c}{1-5c}= \frac{6c(c+5)}{(5c-1)(c+1)} \cdot\frac{1-5c}{c(c+5)}= -\frac{6c(c+5)}{(1-5c)(c+1)} \cdot\frac{1-5c}{c(c+5)}=\\\\ =- \frac{6}{c+1}$

$\boxed{3} \ \ - \frac{6}{c+1}+\frac{c^2+5}{c+1}= \frac{-6+c^2+5}{c+1}= \frac{c^2-1}{c+1}= \frac{(c-1)(c+1)}{c+1}=c-1$