Cho $x,y>0$ thỏa mãn ${{\log }_{6}}x={{\log }_{9}}y={{\log }_{4}}\left( 2x+2y \right).$ Tính $\dfrac{x}{y}.$

Đặt ${{\log }_{6}}x={{\log }_{9}}y={{\log }_{4}}\left( 2x+2y \right)=t\Rightarrow \left\{ \begin{aligned}
& x={{6}^{t}} \\
& y={{9}^{t}} \\
& 2x+2y={{4}^{t}} \\
\end{aligned} \right.$
$\Rightarrow {{2.6}^{t}}+{{2.9}^{t}}={{4}^{t}}\Leftrightarrow {{\left( \dfrac{2}{3} \right)}^{2t}}-2.{{\left( \dfrac{2}{3} \right)}^{t}}-2=0\Leftrightarrow \left[ \begin{aligned}
& {{\left( \dfrac{2}{3} \right)}^{t}}=1+\sqrt{3}\left( n \right) \\
& {{\left( \dfrac{2}{3} \right)}^{t}}=1-\sqrt{3}\left( l \right) \\
\end{aligned} \right.\Rightarrow {{\left( \dfrac{2}{3} \right)}^{t}}=1+\sqrt{3}.$
Vậy $\dfrac{x}{y}=1+\sqrt{3}.$