Question

2 root 7 + 3 root 5 upon 4 roots 5 upon 4 + 3 root 7​

Answer 1

Answer:

Identity used :

(x + y)(x - y) = {x}^{2} - {y}^{2}

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

On rationalizing the denominator we get,

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

Step-by-step explanation:

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

\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} } \\ = \frac{6 \sqrt{30} + 24 - 15 - 2 \sqrt{30} }{45 - 24} \\ = \frac{4 \sqrt{30} + 9 }{21} {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} } \\ = \frac{6 \sqrt{30} + 24 - 15 - 2 \sqrt{30} }{45 - 24} \\ = \frac{4 \sqrt{30} + 9 }{21}