$(\frac{a^2}{a+1}-\frac{a^3}{a^2+2a+1}):(\frac{a}{a+1}-\frac{a^2}{a^2-1})=(\frac{a^2}{a+1}-\frac{a^3}{(a+1{)}^2}):(\frac{a}{a+1}-\frac{a^2}{(a-1)(a+1)})=\frac{a^2(a+1)-a^3}{(a+1{)}^2}:\frac{a(a-1)-a^2}{(a-1)(a+1)}=\frac{a^3+a^2-a^3}{(a+1{)}^2}:\frac{a^2-a-a^2}{(a-1)(a+1)}=\frac{a^2}{(a+1{)}^2}\ast \frac{(a-1)(a+1)}{-a}=\frac{-a(a-1)}{a+1}=\frac{a-a^2}{a+1}$
при a = −3:
$\frac{-3-(-3{)}^2}{-3+1}=\frac{-3-9}{-2}=\frac{-12}{-2}=6$

$(\frac{n-1}{n+1}-\frac{n+1}{n-1})\ast (\frac{1}{2}-\frac{n}{4}-\frac{1}{4n})=\frac{(n-1)(n-1)-(n+1)(n+1)}{(n+1)(n-1)}\ast \frac{2n-n^2-1}{4n}=\frac{n^2-2n+1-(n^2+2n+1)}{(n-1)(n+1)}\ast \frac{n^2-2n+1}{-4n}=\frac{n^2-2n+1-n^2-2n-1}{(n-1)(n+1)}\ast \frac{(n-1{)}^2}{-4n}=\frac{-4n(n-1)}{-4n(n+1)}=\frac{n-1}{n+1}$
при n = 3:
$\frac{3-1}{3+1}=\frac{2}{4}=0,5$