Answer
$-2+\sqrt3$
Work Step by Step
Multiply the numerator and denominator by the conjugate and then simplify the radicals:
$\frac{\sqrt 3-3}{\sqrt 3+3}\times\frac{(\sqrt 3-3)}{(\sqrt 3-3)}=\frac{\sqrt 9-3\sqrt 3-3\sqrt 3+9}{\sqrt 9-3\sqrt 3+3\sqrt 3-9}=\frac{12-6\sqrt 3}{-6}=-2+\sqrt3$
