Evaluate: frac{sqrt{6 + 2sqrt{3} + 2sqrt{2} + | vedclass

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(A) First,simplify the numerator term $\sqrt{6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}}$.Note that $(1 + \sqrt{2} + \sqrt{3})^2 = 1^2 + (\sqrt{2})^2 + (\s

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(A) First,simplify the numerator term $\sqrt{6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}}$.Note that $(1 + \sqrt{2} + \sqrt{3})^2 = 1^2 + (\sqrt{2})^2 + (\sqrt{3})^2 + 2(1)(\sqrt{2}) + 2(1)(\sqrt{3}) + 2(\sqrt{2})(\sqrt{3}) = 1 + 2 + 3 + 2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6} = 6 + 2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6}$.Thus,$\sqrt{6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}} = 1 + \sqrt{2} + \sqrt{3}$.Next,simplify the denominator $\sqrt{5 + 2\sqrt{6}}$.Note that $(\sqrt{3} + \sqrt{2})^2 = 3 + 2 + 2\sqrt{6} = 5 + 2\sqrt{6}$.Thus,$\sqrt{5 + 2\sqrt{6}} = \sqrt{3} + \sqrt{2}$.Substituting these into the expression: $\frac{(1 + \sqrt{2} + \sqrt{3}) - 1}{\sqrt{3} + \sqrt{2}} = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = 1$.

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

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