Упростить выражение.

$\frac{x-4}{x^3-x} : (\frac{x-1}{2x^2+3x+1}-\frac{1}{x^2-1})=\frac{2x+1}{x^{2}}$

1) $\frac{x-1}{2x^2+3x+1}-\frac{1}{x^2-1}=\frac{x-1}{(2x+1)(x+1)}-\frac{1}{(x+1)(x-1)}= \frac{(x-1)(x-1)-(2x+1)}{(2x+1)(x+1)(x-1)}=\frac{(x^{2}-x-x+1)-(2x+1)}{(2x+1)(x+1)(x-1)}=$ = $\frac{x^{2}-2x+1-2x-1}{(2x+1)(x+1)(x-1)}=\frac{x^{2}-(2x+2x)+(1-1)}{(2x+1)(x+1)(x-1)}=\frac{x^{2}-4x}{(2x+1)(x+1)(x-1)}$

2) $\frac{x-4}{x^3-x} : \frac{x^{2}-4x}{(2x+1)(x+1)(x-1)}=\frac{x-4}{x(x^{2}-1)} : \frac{x^{2}-4x}{(2x+1)(x+1)(x-1)}=\frac{x-4}{x(x-1)(x+1)} : \frac{x(x-4)}{(2x+1)(x+1)(x-1)}=$ = $\frac{x-4}{x(x-1)(x+1)}\cdot\frac{(2x+1)(x+1)(x-1)}{x(x-4)}=\frac{2x+1}{x^{2}}$