If x=sqrt{frac{sqrt{5}+1}{sqrt{5}-1}}, then t | vedclass

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(C) Given $x=\sqrt{\frac{\sqrt{5}+1}{\sqrt{5}-1}}$.Rationalizing the denominator inside the square root:$x=\sqrt{\frac{(\sqrt{5}+1)(\sqrt{5}+1)}{(\sqr

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$4$

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(C) Given $x=\sqrt{\frac{\sqrt{5}+1}{\sqrt{5}-1}}$.Rationalizing the denominator inside the square root:$x=\sqrt{\frac{(\sqrt{5}+1)(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}} = \sqrt{\frac{(\sqrt{5}+1)^{2}}{5-1}} = \sqrt{\frac{(\sqrt{5}+1)^{2}}{4}} = \frac{\sqrt{5}+1}{2}$.Now,substitute $x = \frac{\sqrt{5}+1}{2}$ into the expression $5x^{2}-5x-1$:$5\left(\frac{\sqrt{5}+1}{2}\right)^{2} - 5\left(\frac{\sqrt{5}+1}{2}\right) - 1$$= 5\left(\frac{5+1+2\sqrt{5}}{4}\right) - \frac{5\sqrt{5}+5}{2} - 1$$= 5\left(\frac{6+2\sqrt{5}}{4}\right) - \frac{5\sqrt{5}+5}{2} - 1$$= 5\left(\frac{3+\sqrt{5}}{2}\right) - \frac{5\sqrt{5}+5}{2} - 1$$= \frac{15+5\sqrt{5}-5\sqrt{5}-5-2}{2} = \frac{8}{2} = 4$.

If $x=\sqrt{\frac{\sqrt{5}+1}{\sqrt{5}-1}},$ then the value of $5x^{2}-5x-1$ is

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