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Express each one of the following with rational denominator:
$ \frac{1}{3+\sqrt{2}} $
Given:
\( \frac{1}{3+\sqrt{2}} \)
To do:
We have to express the given fraction with rational denominator.
Solution:
We know that,
Rationalising factor of a fraction with denominator ${\sqrt{a}}$ is ${\sqrt{a}}$.
Rationalising factor of a fraction with denominator ${\sqrt{a}-\sqrt{b}}$ is ${\sqrt{a}+\sqrt{b}}$.
Rationalising factor of a fraction with denominator ${\sqrt{a}+\sqrt{b}}$ is ${\sqrt{a}-\sqrt{b}}$.
Therefore,
$\frac{1}{3+\sqrt{2}}=\frac{1(3-\sqrt{2})}{(3+\sqrt{2})(3-\sqrt{2})}$
$=\frac{3-\sqrt{2}}{(3)^{2}-(\sqrt{2})^{2}}$ [Since $(a+b)(a-b)=a^2-b^2$]
$=\frac{3-\sqrt{2}}{9-2}$
$=\frac{3-\sqrt{2}}{7}$
Hence, $\frac{1}{3+\sqrt{2}}=\frac{3-\sqrt{2}}{7}$.
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