sqrt{6 + 2sqrt{3} + 2sqrt{2} + 2sqrt{6}} - fr | vedclass

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(D) First,simplify the term $\sqrt{6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}}$.This is of the form $\sqrt{a^2 + b^2 + c^2 + 2ab + 2bc + 2ca} = a + b + c$.

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$	ext{None of these}$

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(D) First,simplify the term $\sqrt{6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}}$.This is of the form $\sqrt{a^2 + b^2 + c^2 + 2ab + 2bc + 2ca} = a + b + c$.Here,$a^2 = 1, b^2 = 2, c^2 = 3$,so $a=1, b=\sqrt{2}, c=\sqrt{3}$.Thus,$\sqrt{6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}} = 1 + \sqrt{2} + \sqrt{3}$.Next,simplify $\frac{1}{\sqrt{5 + 2\sqrt{6}}}$.Note that $5 + 2\sqrt{6} = 3 + 2 + 2\sqrt{3 	imes 2} = (\sqrt{3} + \sqrt{2})^2$.So,$\frac{1}{\sqrt{5 + 2\sqrt{6}}} = \frac{1}{\sqrt{3} + \sqrt{2}}$.Rationalizing the denominator: $\frac{1}{\sqrt{3} + \sqrt{2}} 	imes \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2}$.Finally,calculate the expression: $(1 + \sqrt{2} + \sqrt{3}) - (\sqrt{3} - \sqrt{2}) = 1 + \sqrt{2} + \sqrt{3} - \sqrt{3} + \sqrt{2} = 1 + 2\sqrt{2}$.

$\sqrt{6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}} - \frac{1}{\sqrt{5 + 2\sqrt{6}}} = $

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