Answer
$\frac{\sqrt3-3}{6}$
Work Step by Step
We simplify as follows:
$\frac{-2}{\sqrt3+3}$
$=\frac{-2(\sqrt3-3)}{(\sqrt3+3)(\sqrt3-3)}$
$=\frac{-2(\sqrt3-3)}{3-9}$
$=\frac{-2(\sqrt3-3)}{-6}$
$=\frac{\sqrt3-3}{6}$
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