If x=(sqrt{2}-1)^{-1/2}, then the value of le | vedclass

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(A) Given that $x=(\sqrt{2}-1)^{-1/2}$.Squaring both sides,we get $x^2 = ((\sqrt{2}-1)^{-1/2})^2 = (\sqrt{2}-1)^{-1} = \frac{1}{\sqrt{2}-1}$.Rationali

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$2$

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(A) Given that $x=(\sqrt{2}-1)^{-1/2}$.Squaring both sides,we get $x^2 = ((\sqrt{2}-1)^{-1/2})^2 = (\sqrt{2}-1)^{-1} = \frac{1}{\sqrt{2}-1}$.Rationalizing the denominator for $x^2$: $x^2 = \frac{1}{\sqrt{2}-1} 	imes \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{\sqrt{2}+1}{2-1} = \sqrt{2}+1$.Now,find $\frac{1}{x^2}$: $\frac{1}{x^2} = \frac{1}{\sqrt{2}+1} = \frac{1}{\sqrt{2}+1} 	imes \frac{\sqrt{2}-1}{\sqrt{2}-1} = \frac{\sqrt{2}-1}{2-1} = \sqrt{2}-1$.Finally,calculate $x^2 - \frac{1}{x^2} = (\sqrt{2}+1) - (\sqrt{2}-1) = \sqrt{2} + 1 - \sqrt{2} + 1 = 2$.

If $x=(\sqrt{2}-1)^{-1/2},$ then the value of $\left(x^{2}-\frac{1}{x^{2}}\right)$ is

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