Question

rationalise the denominators 1 by3+root 2

Answer 1

$\frac{1}{3 + \sqrt{2} } \\ \\ = \frac{1}{3 + \sqrt{2} } \times \frac{3 - \sqrt{2} }{3 - \sqrt{2} } \\ \\ = \frac{3 - \sqrt{2} }{9 - 2} \\ \\ = \frac{3 - \sqrt{2} }{7}$

Answer 2

hiii!!!

here's Ur answer...

$given \: = > \: \frac{1}{3 + \sqrt{2} } \\ \\ = \frac{1}{3 + \sqrt{2} } \times \frac{3 - \sqrt{2} }{3 - \sqrt{2} } \\ \\ = \frac{3 - \sqrt{2} }{(3 + \sqrt{2} )(3 - \sqrt{2} )} \\ \\ = \frac{3 - \sqrt{2} }{( {3})^{2} - ( { \sqrt{2} )}^{2} } \\ \\ = \frac{3 - \sqrt{2} }{9 - 2} \\ \\ = \frac{3 - \sqrt{2} }{7}$

hope this helps..!!