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

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(C) Given that,$x = \frac{1}{2} \left( \sqrt{7} + \frac{1}{\sqrt{7}} \right)$.Squaring both sides,we get $x^2 = \frac{1}{4} \left( 7 + \frac{1}{7} + 2

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$3$

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(C) Given that,$x = \frac{1}{2} \left( \sqrt{7} + \frac{1}{\sqrt{7}} \right)$.Squaring both sides,we get $x^2 = \frac{1}{4} \left( 7 + \frac{1}{7} + 2 \right) = \frac{1}{4} \left( \frac{49 + 1 + 14}{7} \right) = \frac{1}{4} \left( \frac{64}{7} \right) = \frac{16}{7}$.Now,$x^2 - 1 = \frac{16}{7} - 1 = \frac{9}{7}$,so $\sqrt{x^2 - 1} = \frac{3}{\sqrt{7}}$.Substituting these values into the expression $\frac{\sqrt{x^2 - 1}}{x - \sqrt{x^2 - 1}}$:$= \frac{\frac{3}{\sqrt{7}}}{\frac{1}{2} \left( \sqrt{7} + \frac{1}{\sqrt{7}} \right) - \frac{3}{\sqrt{7}}}$$= \frac{\frac{3}{\sqrt{7}}}{\frac{1}{2} \left( \frac{7 + 1}{\sqrt{7}} \right) - \frac{3}{\sqrt{7}}}$$= \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 the value of $\frac{\sqrt{x^2 - 1}}{x - \sqrt{x^2 - 1}}$ is equal to

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