Question

1/ root2+root9 rationalize denominator

Answer 1

We have,

To rationalize:-

$\frac{1}{ \sqrt{2} \: + \: \sqrt{9} } = \frac{1}{ \sqrt{2} \: + \: 3}$

Dividing numerator and denominator by

$\sqrt{2} - 3$

$\frac{ \sqrt{2} - 3}{( { \sqrt{2} )}^{2} - ( {3}^{2}) } \\ = > \frac{ \sqrt{2} - 3}{2 - 9} \\ = > \frac{ \sqrt{2} - 3}{ - 7} \\ = > \frac{3 - \sqrt{2} }{7} ans$

Regards

Kshitij

Answer 2

Step-by-step explanation:

$1 \div \sqrt{2} + \sqrt{9}$

multiply numerator and denominator by

$\sqrt{2} - \sqrt{9}$

so you will get

$(\sqrt{2} - \sqrt{9} ) \div (\sqrt{2} + \sqrt{9} )( \sqrt{2} - \sqrt{9} )$

so

$- (\sqrt{2} - \sqrt{9} ) \div 7$