Rationalise the denominator in each of the fo | vedclass

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(B) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{3}-\sqrt{2})$.$\fra

Vedclass · Accepted Answer

$0.318$

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(B) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{3}-\sqrt{2})$.$\frac{1}{\sqrt{3}+\sqrt{2}} = \frac{1}{\sqrt{3}+\sqrt{2}} 	imes \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$Using the identity $(a+b)(a-b) = a^2 - b^2$ in the denominator:$= \frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2}$$= \frac{\sqrt{3}-\sqrt{2}}{3-2}$$= \frac{\sqrt{3}-\sqrt{2}}{1} = \sqrt{3}-\sqrt{2}$Now,substitute the given values $\sqrt{3} = 1.732$ and $\sqrt{2} = 1.414$:$= 1.732 - 1.414$$= 0.318$

Rationalise the denominator in each of the following and hence evaluate by taking $\sqrt{2}=1.414, \sqrt{3}=1.732$ and $\sqrt{5}=2.236,$ up to three decimal places.
$\frac{1}{\sqrt{3}+\sqrt{2}}$

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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 $\sqrt{5}=2.236,$ up to three decimal places.$\frac{1}{\sqrt{3}+\sqrt{2}}$