Question

Rationalise the denominator of 1 / 3 root 2- 2 root 3

Answer 1

this is your answer

Answer 2

The value of $\dfrac{1}{3\sqrt{2}-2\sqrt{3} }$ is $\dfrac{3\sqrt{2}+2\sqrt{3}}{6}$.

Step-by-step explanation:

We need to rationalize the denominator of the given expression :

$\dfrac{1}{3\sqrt{2}-2\sqrt{3} }$

For rationalizing multiply and divide by $3\sqrt{2}+2\sqrt{3}$

So,

$\dfrac{1}{3\sqrt{2}-2\sqrt{3} }\times \dfrac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}$

Now using identity on denominator: $(a+b)(a-b)=a^2-b^2$

$=\dfrac{1}{3\sqrt{2}-2\sqrt{3} }\times \dfrac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}\\\\=\dfrac{3\sqrt{2}+2\sqrt{3}}{(3\sqrt{2})^2-(2\sqrt{3})^2 }\\\\=\dfrac{3\sqrt{2}+2\sqrt{3}}{6}$

So, the value of  $\dfrac{1}{3\sqrt{2}-2\sqrt{3} }$  is $\dfrac{3\sqrt{2}+2\sqrt{3}}{6}$.