Đáp án A

Đặt \(t = {\log _4}x = {\log _6}y = {\log _9}\left( {x + y} \right) \Rightarrow \left\{ \begin{array}{l} x = {4^t}\\ y = {6^t}\\ x + y = {9^t} \end{array} \right. \Rightarrow \dfrac{x}{y} = {\left( {\dfrac{2}{3}} \right)^t}.\)
Suy ra \({4^t} + {6^t} = {9^t} \Leftrightarrow {\left( {\dfrac{2}{3}} \right)^{2t}} + {\left( {\dfrac{2}{3}} \right)^t} - 1 = 0 \Leftrightarrow \left[ \begin{array}{l} {\left( {\dfrac{2}{3}} \right)^t} = \dfrac{{ - 1 + \sqrt 5 }}{2}\\ {\left( {\dfrac{2}{3}} \right)^t} = \dfrac{{ - 1 - \sqrt 5 }}{2} \end{array} \right.\)
\( \Rightarrow {\left( {\dfrac{2}{3}} \right)^t} = \dfrac{{ - 1 + \sqrt 5 }}{2} \Leftrightarrow \dfrac{x}{y} = \dfrac{{ - 1 + \sqrt 5 }}{2}.\) Chọn A.