Simply by rationalising the denominator

$\displaystyle \frac {\sqrt 3}

{\sqrt 6}$

thanks!

Printable View

- Mar 16th 2009, 08:06 PMwaven[SOLVED] Fractions with surds in the denominator
Simply by rationalising the denominator

$\displaystyle \frac {\sqrt 3}

{\sqrt 6}$

thanks! - Mar 16th 2009, 08:26 PMmathaddict
- Mar 16th 2009, 08:35 PMwaven
- Mar 16th 2009, 10:11 PMProve It