Question

a)\(\sqrt{12}+\dfrac{4}{\sqrt{3}-\sqrt{5}}\)

b) \(\dfrac{\sqrt{45}}{\sqrt{6}-2}-\sqrt{20}\)

Answer 1

\(a,\sqrt{12}+\dfrac{4}{\sqrt{3}-\sqrt{5}}=\dfrac{\sqrt{12}\left(\sqrt{3}-\sqrt{5}\right)+4}{\sqrt{3}-\sqrt{5}}\)

\(=\dfrac{6-2\sqrt{15}+4}{\sqrt{3}-\sqrt{5}}=\dfrac{10-2\sqrt{15}}{\sqrt{3}-\sqrt{5}}=-2\sqrt{5}\)

b, \(\dfrac{\sqrt{45}}{\sqrt{6}-2}-\sqrt{20}=\dfrac{\sqrt{45}-\sqrt{20}\left(\sqrt{6}-2\right)}{\sqrt{6}-2}=\dfrac{\sqrt{45}-2\sqrt{30}+4\sqrt{5}}{\sqrt{6}-2}=\dfrac{3\sqrt{30}+2\sqrt{5}}{2}\)