\(a,=8\sqrt{3}+2\sqrt{3}-8\sqrt{3}=2\sqrt{3}\\ b,=\left(\sqrt{3}-\sqrt{2}\right)\cdot\dfrac{\sqrt{2}}{\sqrt{5}-1}=\dfrac{\sqrt{6}-2}{\sqrt{5}-1}\\ =\dfrac{\left(\sqrt{6}-2\right)\left(\sqrt{5}+1\right)}{4}=\dfrac{\sqrt{30}+\sqrt{6}-2\sqrt{5}+1}{4}\)
\(\dfrac{2}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}-\dfrac{5}{\sqrt{5}}=\dfrac{2\left(\sqrt{5}-1\right)}{4}+\sqrt{\dfrac{4}{6-2\sqrt{5}}}-\sqrt{5}\\ =\dfrac{-\sqrt{5}-1}{2}+\dfrac{2}{\sqrt{5}-1}\\ =\dfrac{-\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)+4}{2\left(\sqrt{5}-1\right)}=\dfrac{-4+4}{2\left(\sqrt{5}-1\right)}=0\)
\(\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}+\sqrt{\dfrac{2}{4+\sqrt{15}}}=\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\sqrt{\dfrac{4}{8+2\sqrt{15}}}\\ =\sqrt{3}+\dfrac{2}{\sqrt{5}+\sqrt{3}}=\sqrt{3}+\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{2}\\ =\sqrt{3}+\sqrt{5}-\sqrt{3}=\sqrt{5}\)
