\(\dfrac{4}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{6}{\sqrt{3}-3}\\ =\dfrac{4\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\dfrac{\sqrt{3}+\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+\dfrac{6\left(\sqrt{3}+3\right)}{\left(\sqrt{3}+3\right)\left(\sqrt{3}-3\right)}\\ =\dfrac{4\sqrt{3}-4}{3-1}+\dfrac{\sqrt{3}+\sqrt{2}}{3-2}+\dfrac{6\sqrt{3}+18}{3-9}\\ =\dfrac{4\sqrt{3}-4}{2}+\sqrt{3}+\sqrt{2}-\dfrac{6\sqrt{3}+18}{6}\\ =2\sqrt{3}-2+\sqrt{3}+\sqrt{2}-\sqrt{3}-3\\ =2\sqrt{3}+\sqrt{2}-5\)