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