Алгебра: 11 класс. Умоляю гуманитария

11 класс. Умоляю гуманитария

Ответ:

drrrrrr2

25.11.2021 08:38

$1) \: \sqrt[4]{81 \times 0.0001} = \\ = \sqrt[4]{81} \times \sqrt[4]{0.0001} = \\ = \sqrt[4]{ {3}^{4} } \times \sqrt[4]{(0.1) ^{4} } = \\ = 3 \times 0.1 = 0.3$

$\\ \\ 2) \sqrt[3]{0.9 } \times \sqrt[3]{ - 0.03} = \\ = \sqrt[3]{0.9 \times ( - 0.03)} = \\ = \sqrt[3]{ - 0.027} = \sqrt[3]{( - {0.3)}^{3} } = - 0.3$

$\\ \\ 3) \sqrt[4]{54} \times \sqrt[4]{24} = \\ = \sqrt[4]{6 \times 9 \times 4 \times 6} = \\ = \sqrt[4]{3 \times 2 \times 3 \times 3 \times 2 \times 2 \times 2 \times 3} = \\ = \sqrt[4]{ {3}^{4} \times {2}^{4} } = \sqrt[4]{( {3 \times 2)}^{4} } = 3 \times 2 = 6$

$\\ \\ 4) \: 3 \times \sqrt[3]{ - 4 \frac{17}{27} } = 3 \times \sqrt[3]{ - \frac{ 125}{27} } = \\ = \sqrt[3]{ - \frac{ {5}^{3} }{ {3}^{3} } } = 3 \times \sqrt[3]{( - \frac{ {5} }{3} )^{3} } = \\ = - 3 \times \frac{5}{3} = - 5$

$\\ \\ 5) \: ( - \sqrt[6]{17} )^{6} = 17$

в 5 примере в ответе число положительное потому что показатель степени чётный

$6) \: ( - 3 \times \sqrt[5]{ \frac{1}{9} } )^{5} = ( { - \sqrt[5]{ {3}^{5} \times \frac{1}{9} } })^{5} = \\ = ( - \sqrt[5]{ \frac{ {3}^{5} }{ {3}^{2} } } ) ^{5} = ( - { \sqrt[5]{ {3}^{5 - 2} } })^{5} = \\ = ( - \sqrt[5]{ {3}^{3} } ) ^{5} = - {3}^{3} = - 27$

$\\ \\ 7) \: \sqrt[5]{81 \times 96} = \sqrt[5]{9 \times 9 \times 8 \times 12} = \\ = \sqrt[5]{3 \times 3 \times 3 \times 3 \times 4 \times 2 \times 3 \times 4} = \\ = \sqrt[5]{ {3}^{5} \times {2}^{5} } = \sqrt[5]{( {3 \times 2)}^{5} } = 3 \times 2 = 6$

$\\ \\ 8) \: \sqrt[3]{ - 20 \times 25 \times 128 } = \\ = \sqrt[3]{ - 4 \times 5 \times 5 \times 5 \times 4 \times 32} = \\ = \sqrt[3]{ - {5}^{3} \times 4 \times 4 \times 8 \times 4} = \\ = \sqrt[3]{ - {5}^{3} \times {4}^{3} \times {2}^{3} } = \\ = \sqrt[3]{ - ( {5 \times 4 \times 2)}^{3} } = \\ = - 5 \times 4 \times 2 = - 40$

$\\ \\ 9) \: \frac{ \sqrt[3]{128} }{ \sqrt[3]{2} } = \sqrt[3]{ \frac{128}{2} } = \sqrt[3]{64} = \sqrt[3]{ {4}^{3} } = 4 \\$

$\\ \\ 10) \: \sqrt[3]{121} \times \sqrt[3]{ - 11} = \\ = \sqrt[3]{ - 11 \times 121} = \sqrt[3]{ - 11 \times 11 \times 11} = \\ = \sqrt[3]{ -11^{3} } = - 11$

$11) \: \sqrt[3]{16} \sqrt[6]{16} = \sqrt[3]{ {2}^{4} } \times \sqrt[6]{ {2}^{4} } = \\ = {2}^{ \frac{4}{3} } \times 2^{ \frac{4}{6} } = 2^ { \frac{4}{3} + \frac{4}{6} } = {2}^{ 2} = 4$