Cho \({{\log }_{9}}x={{\log }_{12}}y={{\log }_{16}}\left( x+y \right)\). Tính giá trị tỷ số \(\frac{x}{y}\) ?

* Hướng dẫn giải

ĐK: \(x>0;y>0\).

Đặt \({{\log }_{9}}x={{\log }_{12}}y={{\log }_{16}}\left( x+y \right)=t\)

\(\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
x = {9^t}\\
y = {12^t}\\
x + y = {16^t}
\end{array} \right. \Rightarrow {9^t} + {12^t} = {16^t} \Leftrightarrow {\left( {\frac{9}{{16}}} \right)^t} + {\left( {\frac{3}{4}} \right)^t} = 1\\
\Leftrightarrow {\left( {\frac{3}{4}} \right)^{2t}} + {\left( {\frac{3}{4}} \right)^t} - 1 = 0 \Leftrightarrow {\left( {\frac{3}{4}} \right)^t} = \frac{{ - 1 + \sqrt 5 }}{2}\\
\Rightarrow \frac{x}{y} = \frac{{{9^t}}}{{{{12}^t}}} = {\left( {\frac{3}{4}} \right)^t} = \frac{{ - 1 + \sqrt 5 }}{2}
\end{array}\)

Chọn C.