Ответ:

[tex](\frac{a+1}{a-1} - \frac{4a}{a^2-1}) : \frac{a-1}{a^2+a} = (\frac{a+1}{a-1} - \frac{4a}{(a-1)(a+1)}) * \frac{a^2+a}{a-1} = (\frac{(a+1)(a+1)-4a}{(a-1)(a+1)})* \frac{(a+1)a}{a-1} = \frac{a^2+2a+1-4a}{(a-1)(a+1)}* \frac{(a+1)a}{a-1} = \frac{a^2-2a+1}{(a-1)(a+1)}* \frac{(a+1)a}{a-1} = \frac{(a-1)^2}{(a-1)(a+1)} * \frac{a(a+1)}{(a-1)} = a[/tex]