Question

With the rational denominator, the expression $\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$ is equivalent to:

• A. $\frac{3+\sqrt{6}+\sqrt{15}}{6}$
• B. $\frac{\sqrt{6}-2+\sqrt{10}}{6}$
• C. $\frac{2+\sqrt{6}-\sqrt{10}}{10}$
• D. $\frac{2+\sqrt{6}-\sqrt{10}}{6}$
• E. none of these

Answer 1

The correct option is A $\frac{3+\sqrt{6}+\sqrt{15}}{6}$

$\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$

$=\frac{\sqrt{2}}{(\sqrt{2}+\sqrt{3})-\sqrt{5}}\times \frac{(\sqrt{2}+\sqrt{3})+\sqrt{5}}{(\sqrt{2}+\sqrt{3})+\sqrt{5}}$

$=\frac{\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{5})}{(\sqrt{2}+\sqrt{3}{)}^2-5}$

$=\frac{2+\sqrt{6}+\sqrt{10}}{2\sqrt{6}}$

$=\frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{2\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$

$=\frac{\sqrt{6}+3+\sqrt{15}}{6}$