Question

алгебра 9 класс срочно

Answer 1

$1.$

$1) \: \: \: \: \: ( \sqrt{6} )^{2} = \sqrt{6} \times \sqrt{6} = 6$

$2) \: \: \: \: \: {( - \frac{ \sqrt{6} }{3}) }^{2} = \frac{( \sqrt{6} ) {}^{2} }{ {3}^{2} } = \frac{6}{9} = \frac{1}{3}$

$2.$

$1) \: \: \: \: \: {( \sqrt{9}) }^{2} = \sqrt{9} \times \sqrt{9} = 9$

$2) \: \: \: \: \: {( \frac{ \sqrt{3} }{2} )}^{2} = \frac{ { \sqrt{3} }^{2} }{ {2}^{2} } = \frac{3}{4}$

$3) \: \: \: \: \: \sqrt{16} + \sqrt{9} = 4 + 3 = 7$

$4) \: \: \: \: \: \sqrt{0.81} + \sqrt{0.01} = 0.9 + 0.1 = 1$

$5) \: \: \: \: \: \frac{1}{3} \sqrt{0.09} - 2 = \frac{1}{3} \times 0.3 - 2 = 0.1 - 2 = - 1.9$

$6) \: \: \: \: \: \frac{1}{6} \times ( \sqrt{18} )^{2} - ( \frac{1}{2} \sqrt{24} )^{2} = \frac{1}{6} \times 18 - ( \frac{1}{4} \times 24) = 3 - 6 = - 3$

$7) \: \: \: \: \: 50 \times ( - \frac{1}{5} \sqrt{2} )^{2} = 50 \times \frac{1}{25} \times 2 = 2 \times 2 = 4$