Question

please answer 5th question
its very urgent

Answer 1

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$5. \: a) \\ \\ x = 3 - 2 \sqrt{2} \\ \\ \frac{1}{x} = \frac{1}{3 - 2 \sqrt{2} } \times \frac{3 + 2 \sqrt{2} }{3 + 2 \sqrt{2} } \\ \\ \frac{1}{x} = \frac{3 + 2 \sqrt{2} }{ {(3)}^{2} - {(2 \sqrt{2}) }^{2} } \\ \\ \frac{1}{x} = \frac{3 + 2 \sqrt{2} }{9 - 8} \\ \\ \frac{1}{x} = 3 + 2 \sqrt{2} \\ \\ now. \\ \\ = {(x)}^{2} + { (\frac{1}{x} )}^{2} \\ \\ = {(x + \frac{1}{x} )}^{2} - 2 \times x \times \frac{1}{x} \\ \\ = {(3 - 2 \sqrt{2} + 3 + 2 \sqrt{2} ) }^{2} - 2 \\ \\ = {6}^{2} - 2 \\ \\ = 36 - 2 \\ \\ = 34$
$\\ \\ 5. \: b) \\ \\ = \frac{1}{ \sqrt{8} - \sqrt{7} } + \frac{1}{ \sqrt{7} - \sqrt{6} } - \frac{1}{ \sqrt{6} - \sqrt{5} } \\ \\ = \frac{1}{ \sqrt{8} - \sqrt{7} } \times \frac{ \sqrt{8} + \sqrt{7} }{ \sqrt{8} + \sqrt{7} } + \frac{1}{ \sqrt{7} - \sqrt{6} } \times \frac{ \sqrt{7} + \sqrt{6} }{ \sqrt{7} + \sqrt{6} } - \frac{1}{ \sqrt{6} - \sqrt{5} } \times \frac{ \sqrt{6} + \sqrt{5} }{ \sqrt{6} + \sqrt{5} } \\ \\ = \frac{ \sqrt{8} + \sqrt{7} }{8 - 7} + \frac{ \sqrt{7} + \sqrt{6} }{7 - 6} - \frac{ \sqrt{6} + \sqrt{5} }{6 - 5} \\ \\ \frac{ \sqrt{8} + \sqrt{7} }{1} + \frac{ \sqrt{7} + \sqrt{6} }{1} - \frac{ \sqrt{6} + \sqrt{5} }{1} \\ \\ = \sqrt{8} + \sqrt{7} + \sqrt{7} + \sqrt{6} - \sqrt{6} - \sqrt{5} \\ \\ = \sqrt{8} + 2 \sqrt{7} - \sqrt{5} \\ \\ = 2 \sqrt{2} + 2 \sqrt{7} - \sqrt{5}$
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Answer 2

Answer:

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