Đáp án:
\(\frac{7}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
P = 2{\log _{ab}}b + \frac{1}{2}{\log _{ab}}c = \frac{2}{{{{\log }_b}ab}} + \frac{1}{{2.{{\log }_c}ab}} = \frac{2}{{{{\log }_b}a + 1}} + \frac{1}{{2.{{\log }_c}a + 2.{{\log }_c}b}}\\
= \frac{2}{{{{\log }_b}a + 1}} + \frac{1}{{2.{{\log }_c}b.{{\log }_b}a + 2.{{\log }_c}b}} = \frac{2}{{\frac{1}{2} + 1}} + \frac{1}{{2.\frac{1}{3}.\frac{1}{2} + 2.\frac{1}{3}}} = \frac{7}{3}
\end{array}\)