Rationalise the denominator of the following | vedclass

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Question

(A) To rationalise the denominator of $\frac{\sqrt{2}}{2+\sqrt{2}}$,we multiply the numerator and the denominator by the conjugate of the denominator,

Vedclass · Accepted Answer

$0.414$

Vedclass · Answer

(A) To rationalise the denominator of $\frac{\sqrt{2}}{2+\sqrt{2}}$,we multiply the numerator and the denominator by the conjugate of the denominator,which is $(2-\sqrt{2})$.$\frac{\sqrt{2}}{2+\sqrt{2}} = \frac{\sqrt{2}}{2+\sqrt{2}} 	imes \frac{2-\sqrt{2}}{2-\sqrt{2}}$Using the identity $(a+b)(a-b) = a^2 - b^2$ in the denominator:$= \frac{\sqrt{2}(2-\sqrt{2})}{(2)^2 - (\sqrt{2})^2}$$= \frac{2\sqrt{2} - 2}{4 - 2}$$= \frac{2\sqrt{2} - 2}{2}$$= \frac{2(\sqrt{2} - 1)}{2}$$= \sqrt{2} - 1$Given $\sqrt{2} = 1.414$,we substitute the value:$= 1.414 - 1 = 0.414$

Rationalise the denominator of the following expression and evaluate it by taking $\sqrt{2}=1.414, \sqrt{3}=1.732$,and $\sqrt{5}=2.236$ up to three decimal places:
$\frac{\sqrt{2}}{2+\sqrt{2}}$

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Rationalise the denominator of the following expression and evaluate it by taking $\sqrt{2}=1.414, \sqrt{3}=1.732$,and $\sqrt{5}=2.236$ up to three decimal places:$\frac{\sqrt{2}}{2+\sqrt{2}}$