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1. $[\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} - \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}]$ simplifies to = ?

A. $2 \sqrt{6}$

B. $4 \sqrt{6}$

C. $2 \sqrt{3}$

D. $3 \sqrt{2}$

Solution:

[\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} - \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}]

= \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \times

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

\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} \times

\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}

\begin{aligned} = \frac{{(\sqrt{3} + \sqrt{2})}^{2}}{3 - 2} - \frac{{(\sqrt{3} - \sqrt{2})}^{2}}{3 - 2} \\ = (3 + 2 + 2 \sqrt{6}) - (3 + 2 - 2 \sqrt{6}) \\ = 4 \sqrt{6} \end{aligned}

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