Question

Sample Pa
66. The number obtained on rationalising
denominator of
15--2V6 is
(1) 13+ 50
o
3-2
(3) (3 +52)
(13+15)
​

Answer 1

3 rd option is correct if it helps u mark me as brainliest

Answer 2

Answer:

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

Step-by-step explanation:

Given :

To rationalise the denominator of the fraction $\frac{1}{\sqrt{5-2\sqrt{6} } }$

Solution :

$\frac{1}{\sqrt{5-2\sqrt{6} } }$

$\frac{1}{\sqrt{3 + 2 - 2(\sqrt{3})(\sqrt{2})}}$

$\frac{1}{\sqrt{(\sqrt{3})^2 + (\sqrt{2})^2 - 2(\sqrt{3})(\sqrt{2})}}$

$\frac{1}{\sqrt{(\sqrt{3} - \sqrt{2})^2}}$

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

By taking conjugate & multiplying,

We get,

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

$\frac{\sqrt{3} + \sqrt{2}}{(\sqrt{3} - \sqrt{2})(\sqrt{3}+\sqrt{2})}$

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

$\frac{\sqrt{3} + \sqrt{2}}{(3 - 2)}$

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

⇒ $\sqrt{3} + \sqrt{2}$

∴ $\frac{1}{\sqrt{5+2\sqrt{6} } } = \sqrt{3} + \sqrt{2}$