Answer
$=\displaystyle \frac{x-5}{2},\qquad$ $x\neq 1,-5$
Work Step by Step
$x^{2}-25=(x-5)(x+5)\quad $(diff. of squares)
$2x-2=2(x-1)$
$x^{2}+10x+25=(x+5)^{2}\quad $ (square of a sum)
$ x^{2}+4x-5\quad$ ... find factors of c whose sum is b...
... we find 5 and -1 ...
$=(x+5)(x-1)$
Expression $=\displaystyle \frac{(x-5)(x+5)}{2(x-1)}\div\frac{(x+5)^{2}}{(x+5)(x-1)}$
... Division with $\displaystyle \frac{a}{b}$ equals multiplication with $\displaystyle \frac{b}{a}.$
(neither a or b can be 0,
because of the denominator we exclude $-5$ and $+1$,
numerator: exclude $-5$ ... we already have)
$=\displaystyle \frac{(x-5)(x+5)}{2(x-1)}\cdot\frac{(x+5)(x-1)}{(x+5)^{2}}$
...exclude any additional values that yield 0 in the denominator:
$x\neq 1,-5$
... cancel common factors
$=\displaystyle \frac{x-5}{2},\qquad$ $x\neq 1,-5$
