Bài tập 2.15 trang 109 SBT Toán 12

a)

\(\begin{array}{l}
\frac{1}{2}{\log _7}36 - {\log _7}14 - 3{\log _7}\sqrt[3]{{21}}\\
 = {\log _7}\sqrt {36}  - {\log _7}14 - 3.\frac{1}{3}{\log _7}21\\
 = {\log _7}6 - {\log _7}14 - {\log _7}21\\
 = {\log _7}\frac{6}{{14.21}} = {\log _7}\frac{1}{{49}} =  - 2
\end{array}\)

b)

\(\frac{{{{\log }_2}24 - {{\log }_2}\sqrt {72} }}{{{{\log }_3}18 - {{\log }_3}\sqrt[3]{{72}}}} = \frac{{{{\log }_2}\frac{{24}}{{\sqrt {72} }}}}{{{{\log }_3}\frac{{18}}{{\sqrt[3]{{72}}}}}} = \frac{{{{\log }_2}\frac{{24}}{{3\sqrt 8 }}}}{{{{\log }_3}\frac{{18}}{{2\sqrt[3]{9}}}}} = \frac{{{{\log }_2}\frac{8}{{\sqrt 8 }}}}{{{{\log }_3}\frac{9}{{\sqrt[3]{9}}}}} = \frac{{{{\log }_2}{2^{3 - \frac{3}{2}}}}}{{{{\log }_3}{3^{2 - \frac{2}{3}}}}} = \frac{{\frac{3}{2}}}{{\frac{4}{3}}} = \frac{9}{8}\)

c)

\(\begin{array}{l}
\frac{{{{\log }_2}4 + {{\log }_2}\sqrt {10} }}{{{{\log }_2}20 + 3{{\log }_2}2}} = \frac{{{{\log }_2}\left( {4.\sqrt {10} } \right)}}{{{{\log }_2}\left( {{{20.2}^3}} \right)}} = \frac{{{{\log }_2}\left( {{2^2}{{.2}^{\frac{1}{2}}}{{.5}^{\frac{1}{2}}}} \right)}}{{{{\log }_2}\left( {{2^2}{{.5.2}^3}} \right)}}\\
 = \frac{{{{\log }_2}{2^{\frac{5}{2}}} + {{\log }_2}{5^{\frac{1}{2}}}}}{{{{\log }_2}{2^5} + {{\log }_2}5}} = \frac{{\frac{1}{2}\left( {{{\log }_2}{2^5} + {{\log }_2}5} \right)}}{{{{\log }_2}{2^5} + {{\log }_2}5}} = \frac{1}{2}
\end{array}\)

-- Mod Toán 12 HỌC247