Question

Answer 1

1: \(=\dfrac{\sqrt{3}+\sqrt{2}-1}{3-3+2\sqrt{2}}=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{2}-1\right)}{4}\)

2: \(=\dfrac{\sqrt{5}+\sqrt{3}+2}{8+2\sqrt{15}-4}=\dfrac{\sqrt{5}+\sqrt{3}+2}{4+2\sqrt{15}}\)

\(=\dfrac{\left(\sqrt{5}+\sqrt{3}+2\right)\left(4-2\sqrt{15}\right)}{-44}\)

3: \(=\dfrac{1+\sqrt{3}-\sqrt{5}}{4+2\sqrt{3}-5}=\dfrac{\left(1+\sqrt{3}-\sqrt{5}\right)\left(2\sqrt{3}+1\right)}{11}\)