Answer
$\dfrac{x^{2}-1}{x^{2}-2x+1}=\dfrac{x+1}{x-1}$
Work Step by Step
$\dfrac{x^{2}-1}{x^{2}-2x+1}$
Factor the numerator:
$\dfrac{x^{2}-1}{x^{2}-2x+1}=\dfrac{(x-1)(x+1)}{x^{2}-2x+1}=...$
Factor the perfect square trinomial in the denominator. Then, simplify:
$...=\dfrac{(x-1)(x+1)}{(x-1)^{2}}=\dfrac{x+1}{x-1}$
