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1. $\frac{12}{3 + \sqrt{5} + 2 \sqrt{2}}$ is equal to = ?

A. $1 - \sqrt{5} + \sqrt{2} + \sqrt{16}$

B. $1 + \sqrt{5} + \sqrt{2} - \sqrt{10}$

C. $1 + \sqrt{5} + \sqrt{2} + \sqrt{10}$

D. $1 - \sqrt{5} - \sqrt{2} + \sqrt{10}$

Solution:

\begin{aligned} \frac{12}{3 + \sqrt{5} + 2 \sqrt{2}} \\ = \frac{12 (3 + \sqrt{5} - 2 \sqrt{2})}{[(3 + \sqrt{5}) + 2 \sqrt{2}] [(3 + \sqrt{5}) - 2 \sqrt{2}]} \\ = \frac{12 (3 + \sqrt{5} - 2 \sqrt{2})}{{(3 + \sqrt{5})}^{2} - {(2 \sqrt{2})}^{2}} \\ = \frac{12 (3 + \sqrt{5} - 2 \sqrt{2})}{9 + 5 + 6 \sqrt{5} - 8} \\ = \frac{12 (3 + \sqrt{5} - 2 \sqrt{2})}{6 \sqrt{5} + 6} \\ = \frac{2 (3 + \sqrt{5} - 2 \sqrt{2})}{\sqrt{5} + 1} \\ = \frac{2 (3 + \sqrt{5} - 2 \sqrt{2}) (\sqrt{5} - 1)}{(\sqrt{5} - 1) (\sqrt{5} + 1)} \\ = \frac{2 (3 \sqrt{5} + 5 - 2 \sqrt{10} - 3 - \sqrt{5} + 2 \sqrt{2})}{5 - 1} \\ = \frac{2 (2 \sqrt{5} + 2 \sqrt{2} - 2 \sqrt{10} + 2)}{4} \\ = \frac{2 \times 2 (\sqrt{5} + \sqrt{2} - \sqrt{10} + 1)}{4} \\ = \sqrt{5} + \sqrt{2} - \sqrt{10} + 1 \\ or, 1 + \sqrt{5} + \sqrt{2} - \sqrt{10} \end{aligned}

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