Answer
$\displaystyle \frac{x-5}{2}$
Work Step by Step
Dividing with $\displaystyle \frac{P}{Q}$ equals multiplying with the reciprocal, $\displaystyle \frac{Q}{P}.$
$\displaystyle \frac{x^{2}-25}{2x-2}\div\frac{x^{2}+10x+25}{x^{2}+4x-5}=\frac{x^{2}-25}{2x-2}\cdot\frac{x^{2}+4x-5}{x^{2}+10x+25}\qquad$
... factor what you can
$x^{2}-25=(x+5)(x-5)$
$2x-2=2(x-1)$
$x^{2}+10x+25=(x+5)^{2}$
$x^{2}+4x-5=(x+5)(x-1)$
$=\displaystyle \frac{(x+5)(x-5)}{2(x-2)}\cdot\frac{(x+5)(x-1)}{(x+5)^{2}} \qquad$... divide out the common factors
$=\displaystyle \frac{(1)(x-5)}{2(1)}\cdot\frac{(1)(1)}{(1)^{1}} $
= $\displaystyle \frac{x-5}{2}$
