Answer
$-\frac{2}{(x-1)(x+1)}$
Work Step by Step
Since the least common multiple (LCM) of the denominator is $(x-1)(x+1)$, then we have
$$
\frac{2x-3}{x-1}-\frac{2x+1}{x+1}=\frac{(2x-3)(x+1)}{(x-1)(x+1)}-\frac{(2x+1)(x-1)}{(x-1)(x+1)}
.$$
Now, since the denominators are the same, then we have
$$
\frac{(2x-3)(x+1)}{(x-1)(x+1)}-\frac{(2x+1)(x-1)}{(x-1)(x+1)}=\frac{2x^2-x-3-(2x^2-x-1)}{(x-1)(x+1)}\\
=\frac{-2}{(x-1)(x+1)}=-\frac{2}{(x-1)(x+1)}
.$$
