Question

$\frac{3+\sqrt{6}}{\sqrt{75}-\sqrt{48}-\sqrt{32}+\sqrt{50}}=$.......

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $\sqrt{3}+\sqrt{2}$
• D. $\sqrt{3}-\sqrt{2}$

Answer 1

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

$\frac{3+\sqrt{6}}{\sqrt{75}-\sqrt{48}-\sqrt{32}+\sqrt{50}}$

$=\frac{3+\sqrt{6}}{5\sqrt{3}-4\sqrt{3}-4\sqrt{2}+5\sqrt{2}}$

$=\frac{3+\sqrt{6}}{\sqrt{3}+\sqrt{2}}$

$=\frac{\left(\sqrt{3}\right)\left(\sqrt{3}\right)+\left(\sqrt{3}\right)\left(\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}$

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

$=\sqrt{3}$.

Therefore, option $B$ is correct.