Question

The value of $\frac{3+\sqrt{6}}{8\sqrt{3}-2\sqrt{12}-\sqrt{32}+\sqrt{50}-\sqrt{27}}$ is :

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $\sqrt{6}$
• D. $\sqrt{18}$

Answer 1

The correct option is B $\sqrt{3}$

$\frac{3+\sqrt{6}}{8\sqrt{3}-2\sqrt{12}-\sqrt{32}+\sqrt{50}-\sqrt{27}}$

$\Rightarrow \frac{3+\sqrt{2}.\sqrt{3}}{8\sqrt{3}-4\sqrt{3}-4\sqrt{2}+5\sqrt{2}-3\sqrt{3}}$

$\Rightarrow \frac{\sqrt{3}.\sqrt{3}+\sqrt{2}.\sqrt{3}}{(8\sqrt{3}-4\sqrt{3}-3\sqrt{3})+(5\sqrt{2}-4\sqrt{2})}$

$\Rightarrow \frac{\sqrt{3}(\sqrt{3}+\sqrt{2})}{(\sqrt{3}+\sqrt{2})}$

$\Rightarrow \sqrt{3}$

$\therefore \frac{31\sqrt{6}}{2\sqrt{3}-2\sqrt{12}-\sqrt{32}+\sqrt{50}-\sqrt{27}}=\sqrt{3}$