In each of the following numbers rationalise | vedclass

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Question

$\frac{n^{2}}{\sqrt{m^{2}+n^{2}+m}}=\frac{n^{2}}{\sqrt{m^{2}+n^{2}+m}} 	imes \frac{\sqrt{m^{2}+n^{2}}-m}{\sqrt{m^{2}+n^{2}}-m}$

$=\frac{n^{2}\left

Vedclass · Accepted Answer

Vedclass · Answer

$\frac{n^{2}}{\sqrt{m^{2}+n^{2}+m}}=\frac{n^{2}}{\sqrt{m^{2}+n^{2}+m}} 	imes \frac{\sqrt{m^{2}+n^{2}}-m}{\sqrt{m^{2}+n^{2}}-m}$

$=\frac{n^{2}\left(\sqrt{m^{2}+n^{2}}-might)}{\left(\sqrt{m^{2}+n^{2}}ight)^{2}-(m)^{2}}$

$=\frac{n^{2}\left(\sqrt{m^{2}+n^{2}}-might)}{m^{2}+n^{2}-m^{2}}$

$=\frac{n^{2}\left(\sqrt{m^{2}+n^{2}}-might)}{n^{2}}$

$=\sqrt{m^{2}+n^{2}}-m$

In each of the following numbers rationalise the denominator

$\frac{n^{2}}{\sqrt{m^{2}+n^{2}}+m}$

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