Question

\(b.\dfrac{3\sqrt{3}-1}{\sqrt{3}+\sqrt{2}}\)

Answer 1

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

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

\(=9-3\sqrt{6}-\sqrt{3}+\sqrt{2}\)

Answer 2

 

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