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, $\frac{1}{3+\sqrt{2}}=\frac{3-\sqrt{2}}{7}$.