Question

СРОЧНО!!!! ПОМОГИТЕ МНЕ ПОЖАЛУЙСТА с алгеброй. ​Заранее спасибо!!!

Answer 1

Объяснение:

№1

а) $\frac{2}{3} \sqrt{12.96} + \frac{1}{7} \sqrt{4.41} = \frac{2}{3} \times \frac{36}{10} + \frac{1}{7} \times \frac{21}{10} = \frac{12}{5} + \frac{3}{10} = \frac{27}{10} = 2.7$

б) ${( - \frac{2}{ \sqrt{6} })}^{2} = \frac{ {2}^{2} }{ { \sqrt{6} }^{2} } = \frac{4}{6} = \frac{2}{3}$

в) $\sqrt{1 \frac{40}{81} \times \frac{4}{49} } - \sqrt{1} = \sqrt{ \frac{121}{81} \times \frac{4}{49} } - \sqrt{1} = \frac{11}{9} \times \frac{2}{7} - 1 = \frac{22}{63} - 1 = - \frac{41}{63}$

Г) $\frac{3}{4} \sqrt{10.24} + \frac{1}{6} \sqrt{5.76} = \frac{3}{4} \times \frac{32}{10} + \frac{1}{6} \times \frac{24}{10} = \frac{24}{10} + \frac{4}{10} = \frac{28}{10} = 2.8$

д) ${( - \frac{ \sqrt{15} }{3} )}^{2} = \frac{ { \sqrt{15} }^{2} }{ {3}^{2} } = \frac{15}{9} = \frac{5}{3} = 1 \frac{2}{3}$

е) $\sqrt{2 \frac{14}{121} \times \frac{4}{25} } - \sqrt{1} = \sqrt{ \frac{256}{121} \times \frac{4}{25} } - \sqrt{1} = \frac{16}{11} \times \frac{2}{5} - 1 = \frac{32}{55} - 1 = - \frac{23}{55}$

№2

а) $\sqrt{21} \times \sqrt{3 \frac{6}{7} } = \sqrt{ \frac{21}{1} \times \frac{27}{7} } = \sqrt{ \frac{3 \times 27}{1} } = \sqrt{81} = 9$

б) $\frac{ \sqrt{147} }{ \sqrt{48} } = \sqrt{ \frac{147}{48} } = \sqrt{ \frac{49}{16} } = \frac{7}{4} = 1.75$

в) $\sqrt{{( - 2) }^{6} \times {9}^{3} } = \sqrt{ {4}^{3} \times {9}^{3} } = {2}^{3} \times {3}^{3} = 8 \times 27 = 216$

Г) $\sqrt{15} \times \sqrt{6 \frac{2}{3} } = \sqrt{ \frac{15}{1} \times \frac{20}{3} } = \sqrt{ \frac{5 \times 20}{1} } = \sqrt{100} = 10$

д) $\frac{ \sqrt{75} }{ \sqrt{108} } = \sqrt{ \frac{75}{108} } = \sqrt{ \frac{25}{36} } = \frac{5}{6}$

е) $\sqrt{{( - 5)}^{4} \times {4}^{5} } = \sqrt{ {25}^{2} \times {4}^{5} } = {5}^{2} \times {2}^{5} = 25 \times 32 = 800$