Q.1

$$\frac{1}{\sqrt{15+\sqrt{60}-\sqrt{140}-\sqrt{84}}}=?$$

A. \(\frac{\sqrt{7}+\sqrt{5}-\sqrt{3}}{1+2\sqrt{15}}\)

B. \(\frac{\sqrt{7}+\sqrt{5}+\sqrt{3}}{1+2\sqrt{15}}\)

C. \(\frac{\sqrt{7}-\sqrt{5}+\sqrt{3}}{1+2\sqrt{15}}\)

D. \(\frac{\sqrt{7}+\sqrt{5}+\sqrt{3}}{1-2\sqrt{15}}\)

\(\sqrt{60}=2 \times \sqrt{3} \times \sqrt{5}\)

\(\sqrt{140}=2 \times \sqrt{5} \times \sqrt{7}\)

\(\sqrt{84}=2 \times \sqrt{3} \times \sqrt{7}\)

The RHS suggests the form: 2ab + 2bc + 2ac

In which identity do we have (2ab + 2bc + 2ac)?

B. \(\frac{\sqrt{7}+\sqrt{5}+\sqrt{3}}{1+2\sqrt{15}}\)