Решите вообще не могу понять тему,если можно с объяснениями А и Б

Решите задачу:

$(\frac{x}{x+1}+\frac{x^2+1}{1-x^2}-\frac{x}{x-1}):\frac{x+x^2}{(1-x)^2} = (\frac{x}{x+1}-\frac{x^2+1}{x^2-1}-\frac{x}{x-1})\cdot\frac{(1-x)^2}{x+x^2} = \\ = (\frac{x}{x+1}^{(x-1}-\frac{x^2+1}{(x+1)(x-1)}-\frac{x}{x-1}^{(x+1})\cdot\frac{(x-1)^2}{x(1+x)} = \\ = \frac{x(x-1)-(x^2+1)-x(x+1)}{(x+1)(x-1)}\cdot\frac{(x-1)^2}{x(x+1)}= \frac{x(x-1-(x+1))-x^2-1}{x+1}\cdot\frac{x-1}{x(x+1)} =\\= \frac{x(x-1-x-1)-x^2-1}{x+1}\cdot\frac{x-1}{x(x+1)} = \frac{-x^2-2x-1}{x+1}\cdot\frac{x-1}{x(x+1)} =$
$\frac{-(x^2+2x+1)(x-1)}{x(x+1)^2} = \frac{-(x+1)^2(x-1)}{x(x+1)^2} = \frac{1-x}{x}$
$\frac{(a-5)^2}{a^2+5a}:(\frac{5}{a+5} - \frac{a^2+25}{a^2-25} - \frac{5}{5-a}) = \frac{(a-5)^2}{a(a+5)}:(\frac{5}{a+5} - \frac{a^2+25}{(a+5)(a-5)} + \frac{5}{a-5}) = \\ = \frac{(a-5)^2}{a(a+5)}:\frac{5(a-5)+(a^2+25)+5(a+5)}{(a+5)(a-5)} = \frac{(a-5)^2}{a(a+5)}\cdot\frac{(a+5)(a-5)}{5a-25-a^2-25+5a+25} = \\ = \frac{(a-5)^2}{a}\cdot\frac{a-5}{-a^2+10a-25} = \frac{(a-5)^3}{-a(a^2-10a+25)}=\frac{(a-5)^3}{-a(a-5)^2}=\frac{a-5}{-a}=\frac{5-a}{a}.$