If sqrt{2} = 1.4142,then sqrt{frac{sqrt{2}-1} | vedclass

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Question

(A) To simplify the expression $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$,we rationalize the denominator:$\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}} = \sqrt{\fr

Vedclass · Accepted Answer

$0.4142$

Vedclass · Answer

(A) To simplify the expression $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$,we rationalize the denominator:$\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}} = \sqrt{\frac{(\sqrt{2}-1) 	imes (\sqrt{2}-1)}{(\sqrt{2}+1) 	imes (\sqrt{2}-1)}}$$= \sqrt{\frac{(\sqrt{2}-1)^2}{(\sqrt{2})^2 - 1^2}}$$= \sqrt{\frac{(\sqrt{2}-1)^2}{2-1}}$$= \sqrt{(\sqrt{2}-1)^2}$$= \sqrt{2}-1$Given $\sqrt{2} = 1.4142$,we have:$1.4142 - 1 = 0.4142$Hence,the correct option is $(A)$.

If $\sqrt{2} = 1.4142$,then $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$ is equal to

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