If x = 3 - sqrt{5},then find the value of fra | vedclass

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(B) Given $x = 3 - \sqrt{5}$.First,calculate $3x - 2$:$3x - 2 = 3(3 - \sqrt{5}) - 2 = 9 - 3\sqrt{5} - 2 = 7 - 3\sqrt{5}$.To simplify $\sqrt{7 - 3\sqrt

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$1/\sqrt{5}$

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(B) Given $x = 3 - \sqrt{5}$.First,calculate $3x - 2$:$3x - 2 = 3(3 - \sqrt{5}) - 2 = 9 - 3\sqrt{5} - 2 = 7 - 3\sqrt{5}$.To simplify $\sqrt{7 - 3\sqrt{5}}$,multiply and divide by $\sqrt{2}$:$\sqrt{7 - 3\sqrt{5}} = \sqrt{\frac{14 - 6\sqrt{5}}{2}} = \frac{\sqrt{14 - 2\sqrt{45}}}{\sqrt{2}}$.Since $9 + 5 = 14$ and $9 	imes 5 = 45$,we have $\sqrt{14 - 2\sqrt{45}} = \sqrt{9} - \sqrt{5} = 3 - \sqrt{5}$.So,$\sqrt{3x - 2} = \frac{3 - \sqrt{5}}{\sqrt{2}}$.Now substitute this into the expression:$\frac{\sqrt{x}}{\sqrt{2} + \frac{3 - \sqrt{5}}{\sqrt{2}}} = \frac{\sqrt{x}}{\frac{2 + 3 - \sqrt{5}}{\sqrt{2}}} = \frac{\sqrt{2} \cdot \sqrt{x}}{5 - \sqrt{5}}$.Since $x = 3 - \sqrt{5}$,$\sqrt{2} \cdot \sqrt{x} = \sqrt{6 - 2\sqrt{5}} = \sqrt{(\sqrt{5} - 1)^2} = \sqrt{5} - 1$.Substituting back: $\frac{\sqrt{5} - 1}{5 - \sqrt{5}} = \frac{\sqrt{5} - 1}{\sqrt{5}(\sqrt{5} - 1)} = \frac{1}{\sqrt{5}}$.

If $x = 3 - \sqrt{5}$,then find the value of $\frac{\sqrt{x}}{\sqrt{2} + \sqrt{3x - 2}}$.

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