Đáp án đúng: D
Giải chi tiết:\(\begin{array}{l}{\log _5}\dfrac{{4\sqrt 2 }}{{\sqrt {15} }} = {\log _5}4\sqrt 2 - {\log _5}\sqrt {15} = {\log _5}\sqrt {32} - {\log _5}\sqrt {15} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{2}{\log _5}32 - \dfrac{1}{2}{\log _5}15 = \dfrac{1}{2}{\log _5}{2^5} - \dfrac{1}{2}\left( {{{\log }_5}3 + {{\log }_5}5} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{5}{2}{\log _5}2 - \dfrac{1}{2}\left( {{{\log }_5}3 + 1} \right) = \dfrac{5}{2}a - \dfrac{1}{2}\left( {b + 1} \right) = \dfrac{{5a - b - 1}}{2}.\end{array}\)
Chọn D.