\(\dfrac{\sqrt{3}}{\sqrt{3}-2}=\dfrac{-\sqrt{3}}{2-\sqrt{3}}=\dfrac{-\sqrt{3}\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\dfrac{-2\sqrt{3}-3}{2^2-\left(\sqrt{3}\right)^2}\)
\(=\dfrac{-2\sqrt{3}-3}{4-3}=\dfrac{-2\sqrt{3}-3}{1}=-2\sqrt{3}-3\)