Bài tập 11 trang 78 SGK Toán 12 NC

a)Ta có: \({\left( {\sqrt 3 } \right)^{ - \frac{5}{6}}} = {3^{ - \frac{5}{{12}}}}\) và 

\(\begin{array}{l}
\sqrt[3]{{{3^{ - 1}}\sqrt[4]{{\frac{1}{3}}}}} = \sqrt[3]{{{3^{ - 1}}\frac{1}{{{3^{\frac{1}{4}}}}}}}\\
 = \sqrt[3]{{{3^{ - 1}}{3^{\frac{{ - 1}}{4}}}}} = \sqrt[3]{{{3^{ - \frac{5}{4}}}}} = {3^{ - \frac{5}{{12}}}}
\end{array}\)

Vậy \({\left( {\sqrt 3 } \right)^{ - \frac{5}{6}}}\) = \(\sqrt[3]{{{3^{ - 1}}\sqrt[4]{{\frac{1}{3}}}}}\)

b) Ta có: \({3^{600}} = {({3^3})^{200}} = {27^{200}}\)

Và \({5^{400}} = {({5^2})^{200}} = {25^{200}}\)

Vậy 3⁶⁰⁰ > 5⁴⁰⁰

c) Ta có: \({\left( {\frac{1}{2}} \right)^{ - \frac{5}{7}}} = {2^{\frac{5}{7}}}\)

Và \(\sqrt 2 {.2^{\frac{3}{{14}}}} = {2^{\frac{1}{2}}}{.2^{\frac{3}{{14}}}} = {2^{\frac{1}{2} + \frac{3}{{14}}}} = {2^{\frac{5}{7}}}\)

Vậy \({\left( {\frac{1}{2}} \right)^{ - \frac{5}{7}}}\) = \(\sqrt 2 {.2^{\frac{3}{{14}}}}\)

d) Ta có: \({7^{30}} = {\left( {{7^3}} \right)^{10}} = {343^{10}}\)

\({4^{40}} = {\left( {{4^4}} \right)^{10}} = {256^{10}}\)

Vậy 7³⁰ > 4⁴⁰.

-- Mod Toán 12 HỌC247