$(\frac{1}{x-1}-\frac{1}{x+1})\ast (x^2-2x+1)=\frac{2x-2}{x+1}$
$(\frac{1}{x-1}-\frac{1}{x+1})\ast (x^2-2x+1)=\frac{x+1-(x-1)}{(x-1)(x+1)}\ast (x-1{)}^2=\frac{x+1-x+1}{(x-1)(x+1)}\ast (x-1{)}^2=\frac{2}{x+1}\ast (x-1)=\frac{2(x-1)}{x+1}=\frac{2x-2}{x+1}$
Тождество доказано.

$(\frac{1}{x-2}-\frac{1}{x+2})\ast (x^2-4x+4)=\frac{4x-8}{x+2}$
$(\frac{1}{x-2}-\frac{1}{x+2})\ast (x^2-4x+4)=\frac{x+2-(x-2)}{(x-2)(x+2)}\ast (x-2{)}^2=\frac{x+2-x+2}{(x-2)(x+2)}\ast (x-2{)}^2=\frac{4}{x+2}\ast (x-2)=\frac{4(x-2)}{x+2}=\frac{4x-8}{x+2}$
Тождество доказано.