$\frac{2x}{x^2-y^2}-\frac{1}{x-y}-\frac{1}{x+y}=0$
$\frac{2x}{x^2-y^2}-\frac{1}{x-y}-\frac{1}{x+y}=\frac{2x}{(x-y)(x+y)}-\frac{1}{x-y}-\frac{1}{x+y}=\frac{2x-(x+y)-(x-y)}{(x-y)(x+y)}=\frac{2x-x-y-x+y}{(x-y)(x+y)}=\frac{0}{(x-y)(x+y)}=0$
Тождество доказано.

$\frac{2y}{x^2-y^2}-\frac{1}{x-y}+\frac{1}{x+y}=0$
$\frac{2y}{x^2-y^2}-\frac{1}{x-y}+\frac{1}{x+y}=\frac{2y}{(x-y)(x+y)}-\frac{1}{x-y}+\frac{1}{x+y}=\frac{2y-(x+y)+(x-y)}{(x-y)(x+y)}=\frac{2y-x-y+x-y}{(x-y)(x+y)}=\frac{0}{(x-y)(x+y)}=0$
Тождество доказано.

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

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

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

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