Question

rationalise the denominators 2root 6+root2​

Answer 1

Answer:

hope this will help you

Answer 2

Answer:

$= \frac{1}{2 \sqrt{6} + \sqrt{2} } \\ \\ = \frac{1}{2 \sqrt{6} + \sqrt{2} } \times \frac{2 \sqrt{6} - \sqrt{2} }{2 \sqrt{6} - \sqrt{2} } \\ \\ the \: denominator \: is \: in \: the \: form \\ \\ (x + y)(x - y) = {x}^{2} - {y}^{2} \\ \\ here \: x = 2 \sqrt{6} \\ y = \sqrt{2} \\ \\ = \frac{2 \sqrt{6} - \sqrt{2} }{(2 \sqrt{6) } ^{2} - \sqrt{2} ^{2} } \\ \\ = \frac{2 \sqrt{6} - \sqrt{2} }{(2 \times 6) - 2} \\ \\ = \frac{2 \sqrt{6} - \sqrt{2} }{12 - 2} \\ \\ = \frac{2 \sqrt{6} - \sqrt{2} }{10} \\ \\ this \: is \: the \: correct \: answer$