Question

помогите с алгеброй пж-
${27}^{ \frac{1}{3} }$
${81}^{ \frac{3}{4} }$
${9}^ -{1.5}$

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Answer 1

${27}^{ \frac{1}{3} } = \sqrt[3]{27} = \sqrt[3]{ { 3}^{3} } = 3$

${81}^{ \frac{3}{4} } = \sqrt[4]{ {81}^{ 3} } = \sqrt[4]{ {3}^{3} \times {3}^{3} \times {3}^{3} \times {3}^{3} } = \sqrt[4]{27 \times 27 \times 27 \times 27 } = \sqrt[4]{ {27}^{4} } = 27$

${9}^{ - 1.5} = {9}^{ - \frac{3}{2} } = \sqrt[2]{ {9}^{ - 3} } = \sqrt{ \frac{1}{ {9}^{3} } } = \sqrt{ \frac{1}{729 } } = \sqrt{ { (\frac{1}{27}) }^{2} } = \frac{1}{27}$