Question

rationalise 1/(sqrt(2) + sqrt(3) + 5)​

Answer 1

Step-by-step explanation:

$\frac{1}{ \sqrt{2} + \ \sqrt{3} + 5 } \\ = \frac{1}{ \sqrt{2} + \sqrt{3} + 5 } \times \frac{ \sqrt{2} + \sqrt{3} - 5}{ \sqrt{2} + \sqrt{3} - 5}$

$= \frac{ \sqrt{2} + \sqrt{3} - 5}{ {( \sqrt{2} + \sqrt{3}) }^{2} - {(5)}^{2} } \\ = \frac{ \sqrt{2 + } \sqrt{3} - 5 }{( \sqrt{2}) ^{2} + {( \sqrt{3}) }^{2} + 2 \times \sqrt{2} \times \sqrt{3} }$

$= \frac{ \sqrt{2} + \sqrt{3} - 5 }{2 + 3 + 2 \sqrt{6} } \\ = \frac{ \sqrt{2} + \sqrt{3} - 5 }{5 + 2 \sqrt{6} } \\ \\ = \frac{ \sqrt{2} + \sqrt{3} - 5 }{5 + 2 \sqrt{6} } \times \frac{5 - 2 \sqrt{6} }{5 - 2 \sqrt{6} }$

$= \frac{ - \sqrt{2} + \sqrt{3} - 25 + 10 \sqrt{6} }{(5)^{2} - {(2 \sqrt{6}) }^{2} } \\ \\ = \frac{ - \sqrt{2} + \sqrt{3} - 25 + 10 \sqrt{6} }{25 - 24}$

$- \sqrt{2 } + \sqrt{3} - 25 + 10 \sqrt{6}$

Hope it helps you