let's take a peek at the denominators
(x² - 4), (x + 2), (x - 2)
now x² - 4 is really x² - 2², which is a difference of squares, and therefore (x-2)(x+2).
from that, we can simply use the LCD of all denominators, which will then be just (x-2)(x+2), and multiply both sides by that LCD to do away with the denominators.
[tex]\bf \cfrac{2}{x^2-4}-\cfrac{1}{x+2}=\cfrac{3}{x-2} \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{(x-2)(x+2)}}{(x-2)(x+2)\left( \cfrac{2}{(x-2)(x+2)}-\cfrac{1}{x+2} \right)=(x-2)(x+2)\left( \cfrac{3}{x-2} \right)} \\\\\\ 2-(x-2)=(x+2)3\implies 2-x+2=3x+6\implies 4-x=3x+6 \\\\\\ -2=4x\implies \cfrac{-2}{4}=x\implies \boxed{\cfrac{-1}{2}=x}[/tex]
