Bài 2.13 trang 108 Sách bài tập (SBT) Giải tích 12

Tính:

a) \(\displaystyle \frac{1}{2}{\log _7}36 - {\log _7}14 - 3{\log _7}\sqrt[3]{{21}}\)

b) \(\displaystyle \frac{{{{\log }_2}24 - \frac{1}{2}{{\log }_2}72}}{{{{\log }_3}18 - \frac{1}{3}{{\log }_3}72}}\)

c) \(\displaystyle \frac{{{{\log }_2}4 + {{\log }_2}\sqrt {10} }}{{{{\log }_2}20 + 3{{\log }_2}2}}\)

Hướng dẫn làm bài:

a) \(\displaystyle {\log _7}\sqrt {36}  - {\log _7}14 - {\log _7}21\) \( = {\log _7}\dfrac{1}{{49}} =  - 2\)

b) \(\displaystyle \frac{{{{\log }_2}24 - {{\log }_2}\sqrt {72} }}{{{{\log }_3}18 - {{\log }_3}\sqrt[3]{{72}}}}\) \(\displaystyle = \frac{{{{\log }_2}\frac{{24}}{{\sqrt {72} }}}}{{{{\log }_3}\frac{{18}}{{\sqrt[3]{{72}}}}}} = \frac{{{{\log }_2}\sqrt 8 }}{{{{\log }_3}\sqrt[3]{{81}}}}\) \(\displaystyle = \frac{{{{\log }_2}{2^{\frac{3}{2}}}}}{{{{\log }_3}{3^{\frac{4}{3}}}}} = \frac{3}{2}:\frac{4}{3} = \frac{9}{8}\)

c) \(\displaystyle \frac{{{{\log }_2}4 + {{\log }_2}\sqrt {10} }}{{{{\log }_2}20 + 3{{\log }_2}2}}\) \(\displaystyle = \frac{{2{{\log }_2}2 + \frac{1}{2}{{\log }_2}\left( {2.5} \right)}}{{{{\log }_2}\left( {{2^2}.5} \right) + 3}}\) \(\displaystyle = \frac{{2 + \frac{1}{2}\left( {1 + {{\log }_2}5} \right)}}{{2 + {{\log }_2}5 + 3}}\) \(\displaystyle = \frac{{\frac{5}{2} + \frac{1}{2}{{\log }_2}5}}{{5 + {{\log }_2}5}} = \frac{1}{2}\)

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