Đáp án đúng: B
Giải chi tiết:\(\begin{array}{l}{\log _6}20 = {\log _6}\left( {{2^2}.5} \right) = 2{\log _6}2 + {\log _6}5\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{2}{{{{\log }_2}6}} + \dfrac{1}{{{{\log }_5}6}} = \dfrac{2}{{{{\log }_2}\left( {2.3} \right)}} + \dfrac{1}{{{{\log }_5}\left( {2.3} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{2}{{1 + {{\log }_2}3}} + \dfrac{1}{{{{\log }_5}2 + {{\log }_5}3}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{2}{{1 + \dfrac{1}{{{{\log }_3}2}}}} + \dfrac{1}{{\dfrac{{{{\log }_3}2}}{{{{\log }_3}5}} + \dfrac{1}{{{{\log }_3}5}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{2}{{1 + \dfrac{1}{a}}} + \dfrac{1}{{\dfrac{a}{b} + \dfrac{1}{b}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{2a}}{{a + 1}} + \dfrac{b}{{a + 1}} = \dfrac{{2a + b}}{{a + 1}}.\end{array}\)
Chọn B.