Question

can someone please figure this out?
(x^2 + 2x + 1)/(x - 2)/(x^2 - 1)/(x^2 - 4)

Answer 1

\[\huge {{(x^2 + 2x + 1)\over(x - 2)}\over{(x^2 - 1)\over(x^2 - 4)}}\]?

Answer 2

yep

Answer 3

the rule to use here is \[{\frac ab\over\frac cd}=\frac ab\cdot\frac dc\]apply this, then factor everything

things should cancel...

Answer 4

in your case the rule gives\[\huge {{(x^2 + 2x + 1)\over(x - 2)}\over{(x^2 - 1)\over(x^2 - 4)}}={(x^2 + 2x + 1)\over(x - 2)}\cdot{x^2 - 4\over x^2-1}\]now factor as much as possible