Write the rationalising factor of the denominator in

Question:

Write the rationalising factor of the denominator in $\frac{1}{\sqrt{2}+\sqrt{3}}$.

 

Solution:

$\frac{1}{\sqrt{2}+\sqrt{3}}$

$=\frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

$=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}$

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

$=\frac{\sqrt{3}-\sqrt{2}}{1}$

Here, the denominator i.e. 1 is a rational number. Thus, the rationalising factor of the denominator in $\frac{1}{\sqrt{2}+\sqrt{3}}$ is $\sqrt{3}-\sqrt{2}$.

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