Đề bài

Tính: a) \(A = \frac{{{{25}^{{{\log }_5}6}} + {{49}^{{{\log }_7}8}} - 3}}{{{3^{1 + {{\log }_9}4}} + {4^{2 - {{\log }_2}3}} + {5^{{{\log }_{125}}27}}}};\)        b) \(\frac{{{{36}^{{{\log }_6}5}} + {{10}^{1 - \log 2}} - 3{}^{{{\log }_9}36}}}{{{{\log }_2}\left( {{{\log }_2}\sqrt {\sqrt[4]{2}} } \right)}};\)   c) \(C = {\log _{\frac{1}{4}}}\left( {{{\log }_3}4.{{\log }_2}3} \right);\)                  d) \(D = {\log _4}2.{\log _6}4.{\log _8}6.\)

Lời giải chi tiết

a) \(A = \frac{{{{25}^{{{\log }_5}6}} + {{49}^{{{\log }_7}8}} - 3}}{{{3^{1 + {{\log }_9}4}} + {4^{2 - {{\log }_2}3}} + {5^{{{\log }_{125}}27}}}} = \frac{{{{\left( {{5^{{{\log }_5}6}}} \right)}^2} + {{\left( {{7^{{{\log }_7}8}}} \right)}^2} - 3}}{{{{3.3}^{{{\log }_{{3^2}}}{2^2}}} + {4^2}.{{\left( {{2^{{{\log }_2}3}}} \right)}^{ - 2}} + {5^{{{\log }_{{5^3}}}{3^3}}}}}\)       \( = \frac{{{6^2} + {8^2} - 3}}{{{{3.3}^{{{\log }_3}2}} + {4^2}{{.3}^{ - 2}} + {5^{{{\log }_5}3}}}} = \frac{{97}}{{3.2 + \frac{{16}}{9} + 3}} = \frac{{97}}{{\frac{{97}}{9}}} = 9.\)b) \(\frac{{{{36}^{{{\log }_6}5}} + {{10}^{1 - \log 2}} - 3{}^{{{\log }_9}36}}}{{{{\log }_2}\left( {{{\log }_2}\sqrt {\sqrt[4]{2}} } \right)}} = \frac{{{{\left( {{6^{{{\log }_6}5}}} \right)}^2} + 10.{{\left( {{{10}^{\log 2}}} \right)}^{ - 1}} - {3^{{{\log }_{{3^2}}}{6^2}}}}}{{{{\log }_2}\left( {{{\log }_2}{2^{\frac{1}{8}}}} \right)}}\)\( = \frac{{{5^2} + {{10.2}^{ - 1}} - {3^{{{\log }_3}6}}}}{{{{\log }_2}\frac{1}{8}}} = \frac{{25 + 5 - 6}}{{{{\log }_2}{2^{ - 3}}}} = \frac{{24}}{{ - 3}} =  - 8.\) c) \(C = {\log _{\frac{1}{4}}}\left( {{{\log }_3}4.{{\log }_2}3} \right) = {\log _{{2^{ - 2}}}}\left( {{{\log }_3}{2^2}.{{\log }_2}3} \right) =  - \frac{1}{2}{\log _2}\left( {2{{\log }_3}2.{{\log }_2}3} \right)\)        \( =  - \frac{1}{2}{\log _2}2 =  - \frac{1}{2}.\)        d) \(D = {\log _4}2.{\log _6}4.{\log _8}6 = {\log _{{2^2}}}2.\frac{{{{\log }_2}4}}{{{{\log }_2}6}}.{\log _{{2^3}}}6\)\( = \frac{1}{2}{\log _2}2.\frac{{{{\log }_2}{2^2}}}{{{{\log }_2}6}}.\frac{1}{3}{\log _2}6 = \frac{1}{2}.1.2.\frac{1}{3} = \frac{1}{3}.\)