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1. If $x = 3 + 2 \sqrt{2},$ then the value of $(\sqrt{x} - \frac{1}{\sqrt{x}})$ is:

A. 1

B. 2

C. $2 \sqrt{2}$

D. $3 \sqrt{3}$

Solution:

\begin{aligned} {(\sqrt{x} - \frac{1}{\sqrt{x}})}^{2} \\ = x + \frac{1}{x} - 2 \\ = (3 + 2 \sqrt{2}) + \frac{1}{(3 + 2 \sqrt{2})} - 2 \\ = (3 + 2 \sqrt{2}) + \frac{1}{(3 + 2 \sqrt{2})} \times \frac{(3 - 2 \sqrt{2})}{(3 - 2 \sqrt{2})} - 2 \\ = (3 + 2 \sqrt{2}) + (3 - 2 \sqrt{2}) - 2 \\ = 4 \\ ∴ (\sqrt{x} - \frac{1}{\sqrt{x}}) = 2 \end{aligned}

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