Rationalise the denominator in each of the fo | vedclass

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Question

$\frac{\sqrt{2}}{2+\sqrt{2}}=\frac{\sqrt{2}}{2+\sqrt{2}} 	imes \frac{2-\sqrt{2}}{2-\sqrt{2}}=\frac{\sqrt{2}(2-\sqrt{2})}{(2)^{2}-(\sqrt{2})^{2}}$

Vedclass · Accepted Answer

$0.414$

Vedclass · Answer

$\frac{\sqrt{2}}{2+\sqrt{2}}=\frac{\sqrt{2}}{2+\sqrt{2}} 	imes \frac{2-\sqrt{2}}{2-\sqrt{2}}=\frac{\sqrt{2}(2-\sqrt{2})}{(2)^{2}-(\sqrt{2})^{2}}$

$=\frac{\sqrt{2}(2-\sqrt{2})}{4-2}$

$=\frac{\sqrt{2}(2-\sqrt{2})}{2}$

$=\frac{2 \sqrt{2}-2}{2}$

$=\sqrt{2}-1=1.414-1$

$=0.414$

Rationalise the denominator in each of the following and hence evaluate by taking $\sqrt{2}=1.414, \sqrt{3}=1.732$ and $\sqrt{5}=2.236,$ upto three places of decimal.

$\frac{\sqrt{2}}{2+\sqrt{2}}$

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Rationalise the denominator in each of the following and hence evaluate by taking $\sqrt{2}=1.414, \sqrt{3}=1.732$ and $\sqrt{5}=2.236,$ upto three places of decimal. $\frac{\sqrt{2}}{2+\sqrt{2}}$