Solution
$$x=2+\sqrt{3}$$
$$\frac{1}{x}=2-\sqrt{3}$$
$$(\sqrt{x} + \frac{1}{\sqrt{x}})^{2}$$ = $$x+\frac{1}{x}$$ + 2
$$(\sqrt{x} + \frac{1}{\sqrt{x}})^{2}$$ = 4 + 2 = 6
$$\sqrt{x} + \frac{1}{\sqrt{x}}$$ = $$\sqrt{6}$$
so the answer is option B.
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