Question

Rationalise 1/ root 5 + root 2 - 1

Answer 1

Multipley root 5 in both numerator and denominator, your ans will be

Root 7 by 2

Answer 2

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