Question

Solve $\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}+\frac{3\sqrt{2}}{\sqrt{6}+\sqrt{3}}+\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}$

• A. $0$
• B. $1$
• C. $-1$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $0$

$\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}+\frac{3\sqrt{2}}{\sqrt{6}+\sqrt{3}}+\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}$
Rationalising the denominators we get
$=\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}+\frac{3\sqrt{2}}{\sqrt{6}+\sqrt{3}}\times \frac{\sqrt{6}-\sqrt{3}}{\sqrt{6}-\sqrt{3}}+\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}\times \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}$
$=-\sqrt{6}(\sqrt{2}-\sqrt{3})+\frac{3\sqrt{2}(\sqrt{6}-\sqrt{3})}{3}-\frac{4\sqrt{2}(\sqrt{6}-\sqrt{2})}{4}$
$=-\sqrt{12}+\sqrt{18}+\sqrt{12}-\sqrt{6}-\sqrt{18}+\sqrt{6}=0$