Rationalise the denominator of the following: | vedclass

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Question

$\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$

$=\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}} 	imes \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}$

$=\frac{\sqrt{6}(\

Vedclass · Accepted Answer

$3 \sqrt{2}-2 \sqrt{3}$

Vedclass · Answer

$\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$

$=\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}} 	imes \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}$

$=\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{(\sqrt{2})^{2}-(\sqrt{3})^{2}}$

$=\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{2-3}$

$=\sqrt{6}(\sqrt{3}-\sqrt{2})$

$=\sqrt{6 	imes 3}-\sqrt{6 	imes 2}$

$=\sqrt{18}-\sqrt{12}$

$=\sqrt{9 	imes 2}-\sqrt{4 	imes 3}$

$=3 \sqrt{2}-2 \sqrt{3}$

Rationalise the denominator of the following:

$\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$

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Rationalise the denominator of the following: $\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$