Алгебра: Решите 10 вариант корни

Решите 10 вариант корни

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Ответ:

looney2017

04.01.2021 09:26

$\frac{ {(12 \sqrt{3}) }^{2} }{9} - 2 = \frac{144 \times 3}{9} - 2 = \frac{432}{9} - 2 = \frac{432 - 18}{9} = \frac{414}{9} = 46$

$\sqrt{ \frac{4}{81} } - \sqrt{ \frac{1}{36} } = \sqrt{ {( \frac{2}{9} )}^{2} } - \sqrt{ {( \frac{1}{6} )}^{2} } = \frac{2}{9} - \frac{1}{6} = \frac{12 - 9}{36} = \frac{3}{36} = \frac{1}{12}$

$(7 - \sqrt{3} )(7 + \sqrt{3} ) = {7}^{2} - { \sqrt{3} }^{2} = 49 - 3 = 46$

$( \sqrt{5} - \sqrt{3} )^{2} = 25 - 2 \sqrt{15} + 3 = 28 - 2 \sqrt{15} = 2(14 - \sqrt{15} )$

$5 \sqrt{0.04} - \frac{1}{16} \sqrt{100} = 5 \sqrt{ \frac{1}{25} } - \frac{1}{16} \sqrt{ {10}^{2} } = 5 \times \frac{1}{5} - \frac{1}{16} \times 10 = 1 - \frac{5}{8} = \frac{8 - 5}{8} = \frac{3}{8}$

$8 \sqrt{2} - 3 \sqrt{50} + 2 \sqrt{32} = 8 \sqrt{2} - 3 \times 5 \sqrt{2} + 2 \times 4 \sqrt{2} = 8 \sqrt{2} - 15 \sqrt{2} + 8 \sqrt{2} = \sqrt{2}$

$\frac{ \sqrt{125} }{ \sqrt{5} } = \frac{ \sqrt{ {5}^{2} \times 5 } }{ \sqrt{5} } = \frac{5 \sqrt{5} }{ \sqrt{5} } = 5$

$\sqrt{ {8}^{0} \times {6}^{6} } = \sqrt{1 \times {( {6}^{3} )}^{2} } = {6}^{3}$

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Ответ:

vmalina2005

04.01.2021 09:26

$1)\frac{(12\sqrt{3)}^{2} }{9} -2= \frac{144*3}{9} -2= \frac{432}{9} -2= 48-2=46\\4) (\sqrt{5} -\sqrt{3} )^{2} = 5-3=2\\5) 5\sqrt{0,04} -\frac{1}{16} \sqrt{100} =5*0,2-\frac{10}{16} =1-\frac{10}{16}=\frac{16}{16} -\frac{10}{16} =\frac{6}{16} =0,375\\7) \frac{\sqrt{125} }{\sqrt{5} } =\sqrt{25} =5\\8)\sqrt{2^{0} *6^{6} } =6^{3} =216$