Cho hai số thực dương $a,b$ thỏa mãn ${{\log }_{4}}a={{\log }_{6}}b={{\log }_{9}}\left( a+b \right).$ Tính $\dfrac{a}{b}.$

Đặt ${{\log }_{4}}a={{\log }_{6}}b={{\log }_{9}}\left( a+b \right)=t.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{4}}a=t \\
& {{\log }_{6}}b=t \\
& {{\log }_{9}}\left( a+b \right)=t \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a={{4}^{t}} \\
& b={{6}^{t}} \\
& a+b={{9}^{t}} \\
\end{aligned} \right..$
Ta có ${{4}^{t}}+{{6}^{t}}={{9}^{t}}\Leftrightarrow {{\left( \dfrac{4}{9} \right)}^{t}}+{{\left( \dfrac{2}{3} \right)}^{t}}-1=0\Leftrightarrow \left[ \begin{aligned}
& {{\left( \dfrac{2}{3} \right)}^{t}}=\dfrac{-1+\sqrt{5}}{2} \\
& {{\left( \dfrac{2}{3} \right)}^{t}}=\dfrac{-1-\sqrt{5}}{2}\left( VN \right) \\
\end{aligned} \right.\Leftrightarrow \dfrac{a}{b}=\dfrac{-1+\sqrt{5}}{2}.$