Question

Rationalize the denominator
1÷√5+√2-1

Answer 1

Step-by-step explanation:

To rationalize the denominator:

$\frac{1}{ \sqrt{5} + \sqrt{2} - 1} \times \frac{ \sqrt{5} + \sqrt{2} + 1 }{ \sqrt{5} + \sqrt{2} + 1} \\ \\ = > \frac{ \sqrt{5} + \sqrt{2} + 1}{ {( \sqrt{5} + \sqrt{2} )}^{2} - ( {1)}^{2} } \\ \\ \\ = > \frac{\sqrt{5} + \sqrt{2} + 1}{( { \sqrt{5}) }^{2} + {( \sqrt{2} )}^{2} + 2 \sqrt{5} \sqrt{2} - 1 } \\ \\ \\ = > \frac{ \sqrt{5} + \sqrt{2} + 1}{5 + 2 - 1 + 2 \sqrt{10} } \\ \\ \\ = > \frac{ \sqrt{5} + \sqrt{2} + 1 }{6 + 2 \sqrt{10} }$

Now again there is a irrational number in the denominator,so rationalized again

$\frac{ \sqrt{5} + \sqrt{2} + 1}{6 + 2 \sqrt{10} } \times \frac{6 - 2 \sqrt{10} }{6 - 2 \sqrt{10} } \\ \\ = > \frac{(\sqrt{5} + \sqrt{2} + 1)(6 - 2 \sqrt{10}) }{36 - 40} \\ \\ = > \frac{(\sqrt{5} + \sqrt{2} + 1).2.(3- \sqrt{10}) }{ - 4} \\ \\ = > \frac{ - (\sqrt{5} + \sqrt{2} + 1)(3- \sqrt{10}) }{ 2} \\ \\ = > \frac{ - 3 \sqrt{5} - 3 \sqrt{2} - 3 + \sqrt{50} + \sqrt{20} + \sqrt{10} }{2} \\ \\ = > \frac{ - 3 \sqrt{5} - 3 \sqrt{2} - 3 + 5\sqrt{2} + 2\sqrt{5} + \sqrt{10} }{2} \\ \\ = > \frac{2 \sqrt{2} - \sqrt{5} + \sqrt{10} - 3}{2}$

Hope it helps you