Question

rationalist the denominator


1 upon root 5 +root 2​

Answer 1

Answer:

here is your answer mate

Answer 2

$\frac{1}{ \sqrt{5} + \sqrt{2} }$

To rationalise the denominator.

.

We will use the formula (a + b)(a - b) = a² - b²

.

$\frac{1}{ \sqrt{5} + \sqrt{2} } \\ \\ = > \frac{1}{ \sqrt{5} + \sqrt{2} } \times \frac{ \sqrt{5} - \sqrt{2} }{ \sqrt{5} - \sqrt{2} } \\ \\ = > \frac{1 \times ( \sqrt{5} - \sqrt{2} )}{( \sqrt{5} + \sqrt{2} )( \sqrt{5} - \sqrt{2} )}$

$\\ \\ = \frac{ \sqrt{5} - \sqrt{2} }{ {( \sqrt{5}) }^{2} - {( \sqrt{2} )}^{2} } .... {\tiny{using \: formula \: (a + b)(a - b) = {a}^{2} - {b}^{2} .here \: a = \sqrt{5} .b = \sqrt{2} }}$

$\\ = > \frac{ \sqrt{5} - \sqrt{2} }{5 - 2} \\ \\ = > \frac{ \sqrt{5} - \sqrt{2} }{ 3} \\ \\ \red{hence \: rationalised}$

hope it helps.