Answer
$\dfrac{9 -5 \sqrt 3}{3}$
Work Step by Step
Rationalize the denominator by multiplying $2\sqrt3-3$ to both the numerator and the denominator:
$\dfrac{\sqrt 3-1}{2\sqrt 3+3}=\dfrac{\sqrt 3-1}{2\sqrt 3+3} \cdot \dfrac{2\sqrt 3-3}{2\sqrt 3-3}$
Use FOIL method in the numerator and special formula $(a+b)(a-b)=a^2-b^2$ in the denominator.
$=\dfrac{\sqrt 3\cdot 2\sqrt 3 -3\cdot \sqrt 3-1\cdot 2\sqrt 3+(-1)(-3)}{(2\sqrt 3)^2-3^2}$
$=\dfrac{2(\sqrt 3)^2 -3\sqrt 3-2\sqrt 3+3}{4(\sqrt 3)^2-9}$
Use the rule $(\sqrt{a})^2=a, a>0$ then simplify.
$=\dfrac{2(3) -3 \sqrt 3-2\sqrt 3+3}{4(3)-9}$
$=\dfrac{6 -5 \sqrt 3+3}{12-9}$
$=\dfrac{9 -5 \sqrt 3}{3}$
Hence, the correct answer is $\frac{9 -5 \sqrt 3}{3}$.
