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Câu hỏi: Cho hai số thực dương $a,b$ thỏa ${{\log }_{9}}a={{\log }_{15}}b={{\log }_{25}}\left( a+b \right)$. Tính $\dfrac{a}{b}.$
A. $\dfrac{1}{2}\cdot $
B. $\dfrac{-1-\sqrt{5}}{2}\cdot $
C. $\dfrac{1+\sqrt{5}}{2}\cdot $
D. $\dfrac{-1+\sqrt{5}}{2}\cdot $
A. $\dfrac{1}{2}\cdot $
B. $\dfrac{-1-\sqrt{5}}{2}\cdot $
C. $\dfrac{1+\sqrt{5}}{2}\cdot $
D. $\dfrac{-1+\sqrt{5}}{2}\cdot $
Đặt ${{\log }_{9}}a={{\log }_{15}}b={{\log }_{25}}\left( a+b \right)=t\Rightarrow \left\{ \begin{aligned}
& a={{9}^{t}} \\
& b={{15}^{t}} \\
& a+b={{25}^{t}} \\
\end{aligned} \right.\Rightarrow {{9}^{t}}+{{15}^{t}}={{25}^{t}}\begin{matrix}
{} & \left( * \right) \\
\end{matrix}$.
Chia cả hai vế của (*) cho ${{25}^{t}}$ ta được:
${{\left( \dfrac{9}{25} \right)}^{t}}+{{\left( \dfrac{3}{5} \right)}^{t}}=1\Leftrightarrow {{\left[ {{\left( \dfrac{3}{5} \right)}^{t}} \right]}^{2}}+{{\left( \dfrac{3}{5} \right)}^{t}}-1=0\Leftrightarrow \left[ \begin{aligned}
& {{\left( \dfrac{3}{5} \right)}^{t}}=\dfrac{1+\sqrt{5}}{2} \\
& {{\left( \dfrac{3}{5} \right)}^{t}}=\dfrac{1-\sqrt{5}}{2}<0 \\
\end{aligned} \right.$
Ta có $\dfrac{a}{b}=\dfrac{{{9}^{t}}}{{{15}^{t}}}={{\left( \dfrac{3}{5} \right)}^{t}}=\dfrac{1+\sqrt{5}}{2}$.
& a={{9}^{t}} \\
& b={{15}^{t}} \\
& a+b={{25}^{t}} \\
\end{aligned} \right.\Rightarrow {{9}^{t}}+{{15}^{t}}={{25}^{t}}\begin{matrix}
{} & \left( * \right) \\
\end{matrix}$.
Chia cả hai vế của (*) cho ${{25}^{t}}$ ta được:
${{\left( \dfrac{9}{25} \right)}^{t}}+{{\left( \dfrac{3}{5} \right)}^{t}}=1\Leftrightarrow {{\left[ {{\left( \dfrac{3}{5} \right)}^{t}} \right]}^{2}}+{{\left( \dfrac{3}{5} \right)}^{t}}-1=0\Leftrightarrow \left[ \begin{aligned}
& {{\left( \dfrac{3}{5} \right)}^{t}}=\dfrac{1+\sqrt{5}}{2} \\
& {{\left( \dfrac{3}{5} \right)}^{t}}=\dfrac{1-\sqrt{5}}{2}<0 \\
\end{aligned} \right.$
Ta có $\dfrac{a}{b}=\dfrac{{{9}^{t}}}{{{15}^{t}}}={{\left( \dfrac{3}{5} \right)}^{t}}=\dfrac{1+\sqrt{5}}{2}$.
Đáp án C.