Question

rationalise the denominator​

Answer 1

Answer:

This is the answer for the above rationalisation question

hope it helps you....

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Answer 2

Answer:

$(ii) \frac{1}{ \sqrt{7} - \sqrt{6} } \\ = \frac{1}{ \sqrt{7} - \sqrt{6} } \times \frac{ \sqrt{7} + \sqrt{6} }{ \sqrt{7} + \sqrt{6} } \\ = \frac{1( \sqrt{7} + \sqrt{6}) }{( \sqrt{7} - \sqrt{6} )( \sqrt{7} + \sqrt{6} )} \\ = \frac{ \sqrt{7} + \sqrt{6} }{ { (\sqrt{7} )}^{2} - { (\sqrt{6} )}^{2} } \\ = \frac{ \sqrt{7} + \sqrt{6} }{7 - 6} \\ = \frac{ \sqrt{7} + \sqrt{6} }{1} \\ = \sqrt{7} + \sqrt{6}$

$(iii)\frac{3 \sqrt{2} - 2 \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} } \\ = \frac{3 \sqrt{2} - 2 \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} } \times \frac{3 \sqrt{2} - 2 \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } \\ = \frac{ {(3 \sqrt{2} - 2 \sqrt{3}) }^{2} }{ {(3 \sqrt{2} )}^{2} - {(2 \sqrt{3})}^{2}} \\ = \frac{18 + 12 - 12 \sqrt{6} }{18 - 12} \\ = \frac{30 - 12 \sqrt{6} }{6} \\ = \frac{6(5 - 2 \sqrt{6} )}{6} \\ = 5 - 2 \sqrt{6}$