If x = 2 + sqrt{3} and xy = 1,then frac{x}{sq | vedclass

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(A) Given $x = 2 + \sqrt{3}$ and $xy = 1$,we have $y = \frac{1}{x} = \frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3}$.Let $S = \frac{x}{\sqrt{2} + \sqrt{x}} + \

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$\sqrt{2}$

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(A) Given $x = 2 + \sqrt{3}$ and $xy = 1$,we have $y = \frac{1}{x} = \frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3}$.Let $S = \frac{x}{\sqrt{2} + \sqrt{x}} + \frac{y}{\sqrt{2} - \sqrt{y}}$.Rationalizing the denominators:$S = \frac{x(\sqrt{2} - \sqrt{x})}{2 - x} + \frac{y(\sqrt{2} + \sqrt{y})}{2 - y}$.Since $x - 2 = \sqrt{3}$ and $2 - y = \sqrt{3}$,we have:$S = \frac{x(\sqrt{2} - \sqrt{x})}{-\sqrt{3}} + \frac{y(\sqrt{2} + \sqrt{y})}{\sqrt{3}} = \frac{1}{\sqrt{3}} [y\sqrt{2} + y\sqrt{y} - x\sqrt{2} + x\sqrt{x}]$.$S = \frac{1}{\sqrt{3}} [\sqrt{2}(y - x) + (x\sqrt{x} + y\sqrt{y})]$.Since $y - x = (2 - \sqrt{3}) - (2 + \sqrt{3}) = -2\sqrt{3}$,and $x\sqrt{x} + y\sqrt{y} = (2 + \sqrt{3})^{3/2} + (2 - \sqrt{3})^{3/2} = \frac{(\sqrt{3}+1)^3 + (\sqrt{3}-1)^3}{2\sqrt{2}} = \frac{6\sqrt{3} + 6\sqrt{3}}{2\sqrt{2}} = \frac{12\sqrt{3}}{2\sqrt{2}} = 3\sqrt{6}$.$S = \frac{1}{\sqrt{3}} [\sqrt{2}(-2\sqrt{3}) + 3\sqrt{6}] = \frac{1}{\sqrt{3}} [-2\sqrt{6} + 3\sqrt{6}] = \frac{\sqrt{6}}{\sqrt{3}} = \sqrt{2}$.

If $x = 2 + \sqrt{3}$ and $xy = 1$,then $\frac{x}{\sqrt{2} + \sqrt{x}} + \frac{y}{\sqrt{2} - \sqrt{y}} = $

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