Cho \({\log _2}3 = a;{\log _3}7 = b\) Biểu diễn \(P = {\log _{21}}126\) theo a, b.

* Hướng dẫn giải

Ta có: \({\log _2}7 = {\log _2}3.{\log _3}7 = ab.\)

\(\begin{array}{l}P = {\log _{21}}126 = {\log _{21}}\left( {21.6} \right) = {\log _{21}}21 + {\log _{21}}6\\\,\,\,\,\, = 1 + {\log _{21}}2 + {\log _{21}}3 = 1 + \dfrac{1}{{{{\log }_2}21}} + \dfrac{1}{{{{\log }_3}21}}\\\,\,\,\,\, = 1 + \dfrac{1}{{{{\log }_2}3 + {{\log }_2}7}} + \dfrac{1}{{{{\log }_3}3 + {{\log }_3}7}}\\\,\,\,\,\, = 1 + \dfrac{1}{{a + ab}} + \dfrac{1}{{1 + b}} = \dfrac{{a + ab + 1 + a}}{{a + ab}}\\\,\,\,\,\, = \dfrac{{ab + 2a + 1}}{{a + ab}}.\end{array}\)

Chọn  A.