Key concept: A rational expression (aka fraction) is undefined when the denominator equal zero

So let's check each denominator individually and see what values of x make that denominator equal to zero.

If \(x^2 - 1 = 0\), then \(x^2 = 1\), which means \(x=1\) and \(x=-1\)
In other words, the expression is not defined when \(x=1\) and when \(x=-1\)

If \(x^2 - 4x + 4 = 0\), then we can factor the left side to get: \((x-2)(x-2)= 0\)
This means \(x = 2\)
So, the expression is not defined when \(x=2\)

So the expression is undefined when \(x=1\), \(x=-1\) and \(x=2\)

Answer: E

Cheers,
Brent
_________________

GMATinsight

\(\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}\)

if i solve this further i get

\(\frac{x+1}{(x+1)(x-1)} - \frac{(x-2)}{(x-2)^2}\)

cancelling like terms

\(\frac{1}{x-1} - \frac{1}{x-2}\)

now the only values that give denominator 0 are 1 and 2
i know im missing -1 here, but could you help with where I went wrong? which term did I cancel that I wasn't supposed to cancel and why?

This helps a lot, thank you so much GMATinsight