If x = frac{1}{2} left( sqrt{7} + frac{1}{sqr | vedclass

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(C) Given that,$x = \frac{1}{2} \left( \sqrt{7} + \frac{1}{\sqrt{7}} \right) = \frac{1}{2} \left( \frac{7+1}{\sqrt{7}} \right) = \frac{4}{\sqrt{7}}$.T

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$3$

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(C) Given that,$x = \frac{1}{2} \left( \sqrt{7} + \frac{1}{\sqrt{7}} \right) = \frac{1}{2} \left( \frac{7+1}{\sqrt{7}} \right) = \frac{4}{\sqrt{7}}$.Then,$x^2 = \frac{16}{7}$.So,$x^2 - 1 = \frac{16}{7} - 1 = \frac{9}{7}$.Thus,$\sqrt{x^2 - 1} = \sqrt{\frac{9}{7}} = \frac{3}{\sqrt{7}}$.Now,substitute these values into the expression:$\frac{\sqrt{x^2 - 1}}{x - \sqrt{x^2 - 1}} = \frac{\frac{3}{\sqrt{7}}}{\frac{4}{\sqrt{7}} - \frac{3}{\sqrt{7}}} = \frac{\frac{3}{\sqrt{7}}}{\frac{1}{\sqrt{7}}} = 3$.

If $x = \frac{1}{2} \left( \sqrt{7} + \frac{1}{\sqrt{7}} \right)$,then $\frac{\sqrt{x^2 - 1}}{x - \sqrt{x^2 - 1}}$ is equal to

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