Đáp án đúng: B
Giải chi tiết:Ta có: \({\log _3}5 = a;\,\,\,\,{\log _5}7 = b \Rightarrow {\log _3}7 = ab.\)
\(\begin{array}{l} \Rightarrow P = {\log _{45}}175 = {\log _{45}}\left( {{5^2}.7} \right)\\\,\,\,\,\,\,\,\,\,\,\,\, = {\log _{45}}{5^2} + {\log _{45}}7\\\,\,\,\,\,\,\,\,\,\,\,\, = 2{\log _{45}}5 + {\log _{45}}7\\\,\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{{{{\log }_5}45}} + \frac{1}{{{{\log }_7}45}}\\\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{{{{\log }_5}\left( {{{5.3}^2}} \right)}} + \frac{1}{{{{\log }_7}\left( {{{5.3}^2}} \right)}}\\\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{{{{\log }_5}5 + {{\log }_5}{3^2}}} + \frac{1}{{{{\log }_7}5 + {{\log }_7}{3^2}}}\\\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{{1 + 2{{\log }_5}3}} + \frac{1}{{{{\log }_7}5 + 2{{\log }_7}3}}\\\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{{1 + \frac{2}{a}}} + \frac{1}{{\frac{1}{b} + \frac{2}{{ab}}}} = \frac{2}{{\frac{{a + 2}}{a}}} + \frac{1}{{\frac{{a + 2}}{{ab}}}}\\\,\,\,\,\,\,\,\,\,\, = \frac{{2a}}{{a + 2}} + \frac{{ab}}{{a + 2}} = \frac{{a\left( {b + 2} \right)}}{{a + 2}}.\end{array}\)
Chọn B.
