Đáp án B

Đặt \({\log _9}x = {\log _6}y = {\log _4}\left( {2x + y} \right) = t \Leftrightarrow \left\{ \begin{array}{l} x = {9^t}\\ y = {6^t}\\ 2x + y = {4^t} \end{array} \right. \Rightarrow \left( {\dfrac{x}{y}} \right) = {\left( {\dfrac{9}{6}} \right)^t} = {\left( {\dfrac{3}{2}} \right)^t}\)
Suy ra \({2.9^t} + {6^t} = {4^t} \Leftrightarrow 2.{\left( {\dfrac{9}{4}} \right)^t} + {\left( {\dfrac{6}{4}} \right)^t} = 1 \Leftrightarrow 2.{\left( {\dfrac{3}{2}} \right)^{2t}} + {\left( {\dfrac{3}{2}} \right)^t} - 1 = 0 \Leftrightarrow \left[ \begin{array}{l} {\left( {\dfrac{3}{2}} \right)^t} = \dfrac{1}{2}\\ {\left( {\dfrac{3}{2}} \right)^t} = - 1 \end{array} \right.\)
Vậy \(\dfrac{x}{y} = {\left( {\dfrac{3}{2}} \right)^t} = \dfrac{1}{2}.\) Chọn B.