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Câu hỏi: Cho ${{\log }_{9}}x={{\log }_{12}}y={{\log }_{16}}\left( x+y \right)$. Giá trị của tỷ số $\dfrac{x}{y}$ là.
A. 2
B. $\dfrac{1-\sqrt{5}}{2}$
C. 1
D. $\dfrac{-1+\sqrt{5}}{2}$
A. 2
B. $\dfrac{1-\sqrt{5}}{2}$
C. 1
D. $\dfrac{-1+\sqrt{5}}{2}$
${{\log }_{9}}x={{\log }_{12}}y={{\log }_{16}}\left( x+y \right)$.
Đặt $t={{\log }_{9}}x\Leftrightarrow x={{9}^{t}}$. Ta được :
$t={{\log }_{12}}y={{\log }_{16}}\left( x+y \right)$.
$\Leftrightarrow \left\{ \begin{aligned}
& y={{12}^{t}} \\
& x+y={{16}^{t}} \\
\end{aligned} \right. $ hay $ {{9}^{t}}+{{12}^{t}}={{16}^{t}}\Leftrightarrow {{\left( \dfrac{3}{4} \right)}^{2t}}+{{\left( \dfrac{3}{4} \right)}^{t}}-1=0 $ $ \Leftrightarrow \left[ \begin{aligned}
& {{\left( \dfrac{3}{4} \right)}^{t}}=\dfrac{-1+\sqrt{5}}{2} \\
& {{\left( \dfrac{3}{4} \right)}^{t}}=\dfrac{-1-\sqrt{5}}{2}\left( loai \right) \\
\end{aligned} \right.$.
Khi đó: $\dfrac{x}{y}={{\left( \dfrac{3}{4} \right)}^{t}}=\dfrac{-1+\sqrt{5}}{2}$.
Đặt $t={{\log }_{9}}x\Leftrightarrow x={{9}^{t}}$. Ta được :
$t={{\log }_{12}}y={{\log }_{16}}\left( x+y \right)$.
$\Leftrightarrow \left\{ \begin{aligned}
& y={{12}^{t}} \\
& x+y={{16}^{t}} \\
\end{aligned} \right. $ hay $ {{9}^{t}}+{{12}^{t}}={{16}^{t}}\Leftrightarrow {{\left( \dfrac{3}{4} \right)}^{2t}}+{{\left( \dfrac{3}{4} \right)}^{t}}-1=0 $ $ \Leftrightarrow \left[ \begin{aligned}
& {{\left( \dfrac{3}{4} \right)}^{t}}=\dfrac{-1+\sqrt{5}}{2} \\
& {{\left( \dfrac{3}{4} \right)}^{t}}=\dfrac{-1-\sqrt{5}}{2}\left( loai \right) \\
\end{aligned} \right.$.
Khi đó: $\dfrac{x}{y}={{\left( \dfrac{3}{4} \right)}^{t}}=\dfrac{-1+\sqrt{5}}{2}$.
Đáp án D.