Bài tập 5 trang 8 SBT Toán 11 Tập 2 Chân trời sáng tạo - CTST

a) Ta có: $8^{-\frac{2}{3}}={\left(2^3\right)}^{-\frac{2}{3}}=2^{-2}=\frac{1}{2^2}=\frac{1}{4}$.

b) Ta có: ${32}^{-\frac{2}{5}}={\left(\frac{1}{32}\right)}^{\frac{2}{5}}=\frac{1}{\sqrt[5]{{32}^2}}=\frac{1}{\sqrt[5]{{\left(2^5\right)}^2}}=\frac{1}{2^2}=\frac{1}{4}$.

c) Ta có: ${81}^{1,25}={81}^{\frac{5}{4}}=\sqrt[4]{{81}^5}=\sqrt[4]{{\left(3^4\right)}^5}=3^5=243$.

d) Ta có: ${1000}^{-\frac{5}{3}}={\left(\frac{1}{1000}\right)}^{\frac{5}{3}}=\frac{1}{\sqrt[3]{{1000}^5}}=\frac{1}{\sqrt[3]{{\left({10}^3\right)}^5}}=\frac{1}{{10}^5}={10}^{-5}$.

e) Ta có: ${\left(\frac{16}{81}\right)}^{-\frac{1}{4}}={\left(\frac{81}{16}\right)}^{\frac{1}{4}}=\sqrt[4]{\frac{81}{16}}=\sqrt[4]{\frac{3^4}{2^4}}=\sqrt[4]{{\left(\frac{3}{2}\right)}^4}=\frac{3}{2}$.

g) Ta có: ${\left(\frac{8}{27}\right)}^{-\frac{2}{3}}={\left(\frac{27}{8}\right)}^{\frac{2}{3}}=\sqrt[3]{{\left(\frac{27}{8}\right)}^2}$

= $\sqrt[3]{{\left[{\left(\frac{3}{2}\right)}^3\right]}^2}={\left(\frac{3}{2}\right)}^2=\frac{9}{4}$.

-- Mod Toán 11 HỌC247