Answer
$ -\dfrac{x^2-23x-38}{(x-2)(x+2)^2}$
Work Step by Step
$\dfrac{5}{x-2}+\dfrac{3}{x^2+4x+4}-\dfrac{6}{x+2} = \dfrac{5(x+2)-6(x-2)}{(x-2)(x+2)} + \dfrac{3}{(x+2)^2} = \dfrac{(x+2)^2(5x+10-6x+12)+3(x-2)(x+2)}{(x-2)(x+2)(x+2)^2} = \dfrac{(x+2)(22-x)+3(x-2)}{(x-2)(x+2)^2} = \dfrac{22x-x^2+44-2x+3x-6}{{(x-2)(x+2)^2}} = \dfrac{-x^2+23x+38}{(x-2)(x+2)^2} = -\dfrac{x^2-23x-38}{(x-2)(x+2)^2}$
