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(D) To evaluate $\frac{\sqrt{2}}{2+\sqrt{2}}$,we first rationalize the denominator by multiplying the numerator and the denominator by the conjugate $

Vedclass · Accepted Answer

$0.414$

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(D) To evaluate $\frac{\sqrt{2}}{2+\sqrt{2}}$,we first rationalize the denominator by multiplying the numerator and the denominator by the conjugate $(2-\sqrt{2})$:$\frac{\sqrt{2}}{2+\sqrt{2}} 	imes \frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{2\sqrt{2} - (\sqrt{2})^2}{2^2 - (\sqrt{2})^2}$$= \frac{2\sqrt{2} - 2}{4 - 2} = \frac{2(\sqrt{2} - 1)}{2} = \sqrt{2} - 1$Given $\sqrt{2} = 1.414$,we substitute this value:$1.414 - 1 = 0.414$Thus,the correct value is $0.414$.

Find the value of the following expression correct to three decimal places,rationalizing the denominator if needed. Take $\sqrt{2} = 1.414$,$\sqrt{3} = 1.732$,and $\sqrt{5} = 2.236$.
$\frac{\sqrt{2}}{2+\sqrt{2}}$

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Find the value of the following expression correct to three decimal places,rationalizing the denominator if needed. Take $\sqrt{2} = 1.414$,$\sqrt{3} = 1.732$,and $\sqrt{5} = 2.236$.$\frac{\sqrt{2}}{2+\sqrt{2}}$