If x = 3 - sqrt{5},then frac{sqrt{x}}{sqrt{2} | vedclass

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(D) Given $x = 3 - \sqrt{5}$.First,calculate $\sqrt{x} = \sqrt{3 - \sqrt{5}} = \frac{1}{\sqrt{2}} \sqrt{6 - 2\sqrt{5}} = \frac{1}{\sqrt{2}}(\sqrt{5} -

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$1/\sqrt{5}$

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(D) Given $x = 3 - \sqrt{5}$.First,calculate $\sqrt{x} = \sqrt{3 - \sqrt{5}} = \frac{1}{\sqrt{2}} \sqrt{6 - 2\sqrt{5}} = \frac{1}{\sqrt{2}}(\sqrt{5} - 1)$.Next,calculate $3x - 2 = 3(3 - \sqrt{5}) - 2 = 9 - 3\sqrt{5} - 2 = 7 - 3\sqrt{5}$.To simplify $\sqrt{3x - 2}$,we note that $7 - 3\sqrt{5} = \frac{14 - 6\sqrt{5}}{2} = \frac{(\sqrt{5} - 3)^2}{2}$.Thus,$\sqrt{3x - 2} = \frac{\sqrt{5} - 3}{\sqrt{2}}$ is incorrect; let us re-evaluate: $7 - 3\sqrt{5} = \frac{14 - 6\sqrt{5}}{2} = \frac{3^2 + (\sqrt{5})^2 - 2(3)(\sqrt{5})}{2} = \frac{(3 - \sqrt{5})^2}{2}$.So,$\sqrt{3x - 2} = \frac{3 - \sqrt{5}}{\sqrt{2}}$.Now,$\sqrt{2} + \sqrt{3x - 2} = \sqrt{2} + \frac{3 - \sqrt{5}}{\sqrt{2}} = \frac{2 + 3 - \sqrt{5}}{\sqrt{2}} = \frac{5 - \sqrt{5}}{\sqrt{2}} = \frac{\sqrt{5}(\sqrt{5} - 1)}{\sqrt{2}}$.Finally,$\frac{\sqrt{x}}{\sqrt{2} + \sqrt{3x - 2}} = \frac{\frac{1}{\sqrt{2}}(\sqrt{5} - 1)}{\frac{\sqrt{5}}{\sqrt{2}}(\sqrt{5} - 1)} = \frac{1}{\sqrt{5}}$.

If $x = 3 - \sqrt{5}$,then $\frac{\sqrt{x}}{\sqrt{2} + \sqrt{3x - 2}} = $

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