Answer
$\displaystyle \frac{9-5\sqrt{3}}{3}$
Work Step by Step
We rationalize the denominator:
$\displaystyle \frac{\sqrt{3}-1}{2\sqrt{3}+3}=\frac{(\sqrt{3}-1)((2\sqrt{3}-3))}{(2\sqrt{3}+3)(2\sqrt{3}-3)}=\frac{2\sqrt{3}\sqrt{3}-3\sqrt{3}-2\sqrt{3}+3}{(2\sqrt{3})^2-3^2}=\frac{2*3-5\sqrt{3}+3}{(4*3-9)}=\frac{9-5\sqrt{3}}{3}$
