Question

(1/x+1+1/x-1+1/x+2+1/x-2-2x/x^2-4)*(1/x+1/x^2)=2/x^2-x

Answer 1

$\left(\frac{1}{x+1} +\frac{1}{x-1} +\frac{1}{x+2}+\frac{1}{x-2}-\frac{2x}{x^2-4} \right) \left(\frac{1}{x} +\frac{1}{x^2}\right)= \frac{2}{x^2-x} \\ \\ \left(\frac{x-1+x+1}{(x+1)(x-1)} +\frac{x-2+x+2}{(x-2)(x+2)}-\frac{2x}{x^2-4} \right) *\frac{x+1}{x^2}= \frac{2}{x(x-1)}$

$\left(\frac{2x}{(x+1)(x-1)} +\frac{2x}{x^2-4}-\frac{2x}{x^2-4} \right) *\frac{x+1}{x^2}= \frac{2}{x(x-1)} \\ \\ \frac{2x}{(x+1)(x-1)} *\frac{x+1}{x^2}= \frac{2}{x(x-1)} \\ \\ \frac{2}{x(x-1)}=\frac{2}{x(x-1)}$