Question

Please solve anyone

Answer 1

here is your answer ,

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$q - 1 \: \: \: \: \: it \: must \: be \: solved \: by \: \\ double \: rationalisation \\ \\ \frac{1}{ (\sqrt{3} - \sqrt{2} ) + 1 } \times \frac{( \sqrt{3} - \sqrt{2} ) - 1 }{( \sqrt{3} - \sqrt{2} ) - 1} \\ \\ = \frac{ \sqrt{3} - \sqrt{2} + 1 } { { \sqrt{3} }^{2} + { \sqrt{2}}^{2} - 2 \sqrt{6} + 1 } \\ \\ = \frac{ \sqrt{3} - \sqrt{2} + 1 } { - 2 \sqrt{6} } \\ \\ = \frac{ \sqrt{3 } - \sqrt{2} + 1 }{ - 2 \sqrt{6} } \times \frac{ \sqrt{6} } { \sqrt{6} } \\ \\ = \frac{2 \sqrt{3} - 3 \sqrt{2} - \sqrt{6} }{12 } \\ \\ hence \: now \: the \: term \\ \: is \: rationalised$
$q - 2 \\ a) \frac{1}{ \sqrt{3} - \sqrt{2} } = \frac{1}{0.3} = \frac{10}{3} \\ \\ b) \frac{1}{3 + 2 \sqrt{2} } = \frac{1}{0.2} = 5$

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