If x = 2 + sqrt{3} and xy = 1,then find the v | vedclass

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(A) Given $x = 2 + \sqrt{3}$ and $xy = 1$. Since $xy = 1$,we have $y = \frac{1}{x} = \frac{1}{2 + \sqrt{3}}$. Rationalizing the denominator,$y = \frac

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$\sqrt{2}$

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(A) Given $x = 2 + \sqrt{3}$ and $xy = 1$. Since $xy = 1$,we have $y = \frac{1}{x} = \frac{1}{2 + \sqrt{3}}$. Rationalizing the denominator,$y = \frac{1(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$. Now,we need to evaluate $S = \frac{x}{\sqrt{2} + \sqrt{x}} + \frac{y}{\sqrt{2} - \sqrt{y}}$. Note that $\sqrt{x} = \sqrt{2 + \sqrt{3}} = \sqrt{\frac{4 + 2\sqrt{3}}{2}} = \frac{\sqrt{3} + 1}{\sqrt{2}}$ and $\sqrt{y} = \sqrt{2 - \sqrt{3}} = \sqrt{\frac{4 - 2\sqrt{3}}{2}} = \frac{\sqrt{3} - 1}{\sqrt{2}}$. Substituting these values: $\frac{x}{\sqrt{2} + \frac{\sqrt{3} + 1}{\sqrt{2}}} = \frac{x \sqrt{2}}{2 + \sqrt{3} + 1} = \frac{(2 + \sqrt{3}) \sqrt{2}}{3 + \sqrt{3}} = \frac{\sqrt{2}(2 + \sqrt{3})}{\sqrt{3}(\sqrt{3} + 1)}$. This simplifies to $\frac{\sqrt{2}(2 + \sqrt{3})}{\sqrt{3}(\sqrt{3} + 1)} 	imes \frac{\sqrt{3}-1}{\sqrt{3}-1} = \frac{\sqrt{2}(2\sqrt{3} - 2 + 3 - \sqrt{3})}{\sqrt{3}(3 - 1)} = \frac{\sqrt{2}(\sqrt{3} + 1)}{2\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{6}}$. Similarly,$\frac{y}{\sqrt{2} - \sqrt{y}} = \frac{2 - \sqrt{3}}{\sqrt{2} - \frac{\sqrt{3} - 1}{\sqrt{2}}} = \frac{(2 - \sqrt{3}) \sqrt{2}}{2 - \sqrt{3} + 1} = \frac{\sqrt{2}(2 - \sqrt{3})}{3 - \sqrt{3}} = \frac{\sqrt{2}(2 - \sqrt{3})}{\sqrt{3}(\sqrt{3} - 1)}$. Rationalizing: $\frac{\sqrt{2}(2 - \sqrt{3})(\sqrt{3} + 1)}{\sqrt{3}(3 - 1)} = \frac{\sqrt{2}(2\sqrt{3} + 2 - 3 - \sqrt{3})}{2\sqrt{3}} = \frac{\sqrt{2}(\sqrt{3} - 1)}{2\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{6}}$. Adding both: $\frac{\sqrt{3} + 1 + \sqrt{3} - 1}{\sqrt{6}} = \frac{2\sqrt{3}}{\sqrt{6}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

If $x = 2 + \sqrt{3}$ and $xy = 1$,then find the value of $\frac{x}{\sqrt{2} + \sqrt{x}} + \frac{y}{\sqrt{2} - \sqrt{y}}$.

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