b) [imath]B= \dfrac{ \sqrt{18}}{ \sqrt{6}}+ \dfrac{4}{ \sqrt{5}-1} - \dfrac{3+ \sqrt{3}}{1+ \sqrt{3}}[/imath]
[imath]=\sqrt{\dfrac{18}{6}}+ \dfrac{4 ( \sqrt{5}+1)}{ (\sqrt{5}-1)( \sqrt{5}+1)} - \dfrac{ \sqrt{3}(\sqrt{3}+1)}{1+ \sqrt{3}} \\
= \sqrt{3} + \dfrac{4 ( \sqrt{5}+1)}{ 5-1} - \sqrt{3} \\
= \dfrac{4 ( \sqrt{5}+1)}{ 4} \\
= \sqrt{5}+1[/imath]