Question

2root6-root5/3root5-2root6

Answer 1

Given:- 2root6-root5/3root5-2root6

To find:- 2root6-root5/3root5-2root6=?

Solution:-

$\frac{2 \sqrt{6}- \sqrt{5} }{3 \sqrt{5} -2 \sqrt{6 } }$

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

$\frac{2 \sqrt{6} \times 3 \sqrt{5} - 2 \sqrt{6} \times 2 \sqrt{6} - \sqrt{5} \times 3 \sqrt{5} - \sqrt{5} \times 2 \sqrt{6} }{ {(3 \sqrt{5}) }^{2} - {(2 \sqrt{6}) }^{2} }$

$= \frac{6 \sqrt{6} + 4 \times 6 - 3 \times 5 - 2 \sqrt{30} }{ {(3 \sqrt{5} )}^{2} - {(2 \sqrt{6} )}^{2} }$

$= \frac{4 \sqrt{30 } + 9}{45 - 24}$

$= \frac{9 + 4 \sqrt{30}}{21}$

Hence , 2root6-root5/3root5-2root6= 9+4√30/21 (ans)

Answer 2

Step-by-step explanation:

Given that:

$\frac{2 \sqrt{6} - \sqrt{5} }{3 \sqrt{5} - 2 \sqrt{6} }$

To find: Simplified form by rationalization

Solution: Multiply both numerator and denominator by the RF

$\frac{2 \sqrt{6} - \sqrt{5} }{3 \sqrt{5} - 2 \sqrt{6} } \times \frac{3 \sqrt{5} + 2 \sqrt{6} }{3 \sqrt{5} + 2 \sqrt{6} } \\ \\ = \frac{(2 \sqrt{6} - \sqrt{5})(3 \sqrt{5} + 2 \sqrt{6}) }{( {3 \sqrt{5} })^{2} - ( {2 \sqrt{6} })^{2} } \\ \\ \because(x - y)(x + y) = {x}^{2} - {y}^{2} \\ \\ = \frac{6 \sqrt{30} + 4 \times 6 - 3 \times 5 - 2 \sqrt{30} }{9 \times 5 - 4 \times 6} \\ \\ =\frac{4 \sqrt{30} + 24 - 15}{45 - 24} \\ \\ = \frac{9+4 \sqrt{30} }{21}$

Thus,

$\bold{\frac{2 \sqrt{6} - \sqrt{5} }{3 \sqrt{5} - 2 \sqrt{6} } = \frac{9+4 \sqrt{30}}{21}}$

Hope it helps you.