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Rationalize the following :
$\frac{1}{\sqrt{3}+\sqrt{4}}$
Given :
The given term is $\frac{1}{\sqrt{3}+\sqrt{4}}$
To do :
We have to rationalize the given term.
Solution :
$\frac{1}{\sqrt{3} +\sqrt{4}}$
$=\frac{1}{\sqrt{3} +\sqrt{4}} =\frac{1}{\sqrt{3} +2}$
Rationalizing factor of the denominator $\sqrt{3} +2$ is $\sqrt{3} -2$.
Therefore,
$= \frac{1}{\sqrt{3} +\sqrt{4}} =\frac{1}{\sqrt{3} +2}$
$=\frac{1}{\sqrt{3} +2} \times \frac{\sqrt{3} -2}{\sqrt{3} -2}$
$=\frac{\sqrt{3} -2}{\left(\sqrt{3}\right)^{2} -( 2)^{2}}$
$=\frac{\sqrt{3} -2}{3-4}$
$=\frac{\sqrt{3} -2}{-1}$
$=2-\sqrt{3}$ .
The Rationalized term of $\frac{1}{\sqrt{3}+\sqrt{4}}$ is $2-\sqrt{3}$
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