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1. $\frac{3 \sqrt{2}}{\sqrt{6} + \sqrt{3}} -$ $\frac{2 \sqrt{6}}{\sqrt{3} + 1} +$ $\frac{2 \sqrt{3}}{\sqrt{6} + 2}$ is equal to = ?

A. 3

B. 2

C. 0

D. $\sqrt{3}$

Solution:

\frac{3 \sqrt{2}}{\sqrt{6} + \sqrt{3}} - \frac{2 \sqrt{6}}{\sqrt{3} + 1} + \frac{2 \sqrt{3}}{\sqrt{6} + 2}

= (\frac{3 \sqrt{2}}{\sqrt{6} + \sqrt{3}} \times \frac{\sqrt{6} - \sqrt{3}}{\sqrt{6} - \sqrt{3}}) -

(\frac{2 \sqrt{6}}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1}) +

(\frac{2 \sqrt{3}}{\sqrt{6} + 2} \times \frac{\sqrt{6} - 2}{\sqrt{6} - 2})

= \frac{3 \sqrt{2} (\sqrt{6} - \sqrt{3})}{3} -

\frac{2 \sqrt{6} (\sqrt{3} - 1)}{2} +

\frac{2 \sqrt{3} (\sqrt{6} - 2)}{2}

\begin{aligned} = \sqrt{12} - \sqrt{6} - \sqrt{18} + \sqrt{6} + \sqrt{18} - 2 \sqrt{3} \\ = \sqrt{12} - 2 \sqrt{3} \\ = 2 \sqrt{3} - 2 \sqrt{3} \\ = 0 \end{aligned}

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