Question

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Answer 1

Ответ:

$2) \\ a) \: 3 \sqrt{2} = \sqrt{9 \times 2} = \sqrt{18} \\ 2 \sqrt{5} = \sqrt{4 \times 5} = \sqrt{20} \\ \sqrt{18} < \sqrt{20}$

$3) \\ a) \: \frac{7}{ \sqrt{7} } = \frac{7 \times \sqrt{7} }{ \sqrt{7} \times \sqrt{7} } = \frac{7 \sqrt{7} }{7}$

$b) \: \frac{2}{3 - \sqrt{2} } = \frac{2 \times (3 + \sqrt{2}) }{(3 - \sqrt{2} )(3 + \sqrt{2} )} = \frac{2(3 + \sqrt{2} )}{ {3}^{2} - { \sqrt{2} }^{2} } = \frac{2(3 + \sqrt{2} )}{9 - 2} = \frac{2(3 + \sqrt{2}) }{7}$

$4) \\ \frac{22}{2 \sqrt{3} + 1} - \frac{44}{2 \sqrt{3} - 1} + 4( \sqrt{3 } - 6) = \frac{22(2 \sqrt{3} - 1) }{(2 \sqrt{3} + 1)(2 \sqrt{3} - 1)} - \frac{44(2 \sqrt{3} + 1) }{(2 \sqrt{3} - 1)(2 \sqrt{3} + 1) } + 4( \sqrt{3 } - 6) = \frac{22(2 \sqrt{3} - 1)}{11} - \frac{44(2 \sqrt{3} + 1)}{11} + 4 \sqrt{3} - 6 = \frac{44 \sqrt{3} - 22 - 66 \sqrt{3} - 44 + 44 \sqrt{3} - 264}{11} = \frac{ - 330 - 66 \sqrt{3} }{11} = \frac{ - 66(5 + \sqrt{3}) }{11} = - 6(5 + \sqrt{3} )$

$b) \: \frac{2}{3} \sqrt{63} = \sqrt{ \frac{4}{9} \times 63 } = \sqrt{4 \times 7} = \sqrt{28} \\ \frac{1}{2} \sqrt{104} = \sqrt{ \frac{1}{4} \times 104} = \sqrt{26} \\ \sqrt{28} > \sqrt{26}$