Bài tập 28 trang 38 SBT Toán 11 Tập 2 Cánh diều - CD

a) Ta có: \(A = \frac{{{{25}^{{{\log }_5}6}} + {{49}^{{{\log }_7}8}} - 3}}{{{3^{1 + {{\log }_9}4}} + {4^{2 - {{\log }_2}3}} + {5^{{{\log }_{125}}27}}}} \)

\(= \frac{{{{\left( {{5^{{{\log }_5}6}}} \right)}^2} + {{\left( {{7^{{{\log }_7}8}}} \right)}^2} - 3}}{{{{3.3}^{{{\log }_{{3^2}}}{2^2}}} + {4^2}.{{\left( {{2^{{{\log }_2}3}}} \right)}^{ - 2}} + {5^{{{\log }_{{5^3}}}{3^3}}}}}\)

\( = \frac{{{6^2} + {8^2} - 3}}{{{{3.3}^{{{\log }_3}2}} + {4^2}{{.3}^{ - 2}} + {5^{{{\log }_5}3}}}} = \frac{{97}}{{3.2 + \frac{{16}}{9} + 3}} \\= \frac{{97}}{{\frac{{97}}{9}}} = 9.\)

b) Ta có: \(B= \frac{{{{36}^{{{\log }_6}5}} + {{10}^{1 - \log 2}} - 3{}^{{{\log }_9}36}}}{{{{\log }_2}\left( {{{\log }_2}\sqrt {\sqrt[4]{2}} } \right)}}\)

\(= \frac{{{{\left( {{6^{{{\log }_6}5}}} \right)}^2} + 10.{{\left( {{{10}^{\log 2}}} \right)}^{ - 1}} - {3^{{{\log }_{{3^2}}}{6^2}}}}}{{{{\log }_2}\left( {{{\log }_2}{2^{\frac{1}{8}}}} \right)}}\)

\( = \frac{{{5^2} + {{10.2}^{ - 1}} - {3^{{{\log }_3}6}}}}{{{{\log }_2}\frac{1}{8}}} \\= \frac{{25 + 5 - 6}}{{{{\log }_2}{2^{ - 3}}}} = \frac{{24}}{{ - 3}} = - 8.\)

c) Ta có: \(C = {\log _{\frac{1}{4}}}\left( {{{\log }_3}4.{{\log }_2}3} \right) \)

\(= {\log _{{2^{ - 2}}}}\left( {{{\log }_3}{2^2}.{{\log }_2}3} \right) \\= - \frac{1}{2}{\log _2}\left( {2{{\log }_3}2.{{\log }_2}3} \right)\)

\( = - \frac{1}{2}{\log _2}2 = - \frac{1}{2}.\)

d) Ta có: \(D = {\log _4}2.{\log _6}4.{\log _8}6 \)

\(= {\log _{{2^2}}}2.\frac{{{{\log }_2}4}}{{{{\log }_2}6}}.{\log _{{2^3}}}6\)

\( = \frac{1}{2}{\log _2}2.\frac{{{{\log }_2}{2^2}}}{{{{\log }_2}6}}.\frac{1}{3}{\log _2}6 \\= \frac{1}{2}.1.2.\frac{1}{3} = \frac{1}{3}.\)

-- Mod Toán 11 HỌC247