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Câu hỏi: Cho hai số thực dương $a,b$ thỏa mãn ${{\log }_{4}}a={{\log }_{6}}b={{\log }_{9}}\left( a+b \right)$. Tính $\dfrac{a}{b}$.
A. $\dfrac{1}{2}$
B. $\dfrac{-1+\sqrt{5}}{2}$
C. $\dfrac{-1-\sqrt{5}}{2}$
D. $\dfrac{1+\sqrt{5}}{2}$
A. $\dfrac{1}{2}$
B. $\dfrac{-1+\sqrt{5}}{2}$
C. $\dfrac{-1-\sqrt{5}}{2}$
D. $\dfrac{1+\sqrt{5}}{2}$
Đặt ${{\log }_{4}}a={{\log }_{6}}b={{\log }_{9}}\left( a+b \right)=t\Rightarrow \left\{ \begin{aligned}
& a={{4}^{t}} \\
& b={{6}^{t}} \\
& a+b={{9}^{t}} \\
\end{aligned} \right.$
$\Rightarrow {{4}^{t}}+{{6}^{t}}={{9}^{t}}\Rightarrow {{\left( \dfrac{4}{9} \right)}^{t}}+{{\left( \dfrac{6}{9} \right)}^{t}}=1\Rightarrow {{\left[ {{\left( \dfrac{2}{3} \right)}^{t}} \right]}^{2}}+{{\left( \dfrac{2}{3} \right)}^{t}}-1=0\Rightarrow {{\left( \dfrac{2}{3} \right)}^{t}}=\dfrac{\sqrt{5}-1}{2}.$
Ta có $\dfrac{a}{b}={{\left( \dfrac{4}{6} \right)}^{t}}={{\left( \dfrac{2}{3} \right)}^{t}}=\dfrac{\sqrt{5}-1}{2}.$
& a={{4}^{t}} \\
& b={{6}^{t}} \\
& a+b={{9}^{t}} \\
\end{aligned} \right.$
$\Rightarrow {{4}^{t}}+{{6}^{t}}={{9}^{t}}\Rightarrow {{\left( \dfrac{4}{9} \right)}^{t}}+{{\left( \dfrac{6}{9} \right)}^{t}}=1\Rightarrow {{\left[ {{\left( \dfrac{2}{3} \right)}^{t}} \right]}^{2}}+{{\left( \dfrac{2}{3} \right)}^{t}}-1=0\Rightarrow {{\left( \dfrac{2}{3} \right)}^{t}}=\dfrac{\sqrt{5}-1}{2}.$
Ta có $\dfrac{a}{b}={{\left( \dfrac{4}{6} \right)}^{t}}={{\left( \dfrac{2}{3} \right)}^{t}}=\dfrac{\sqrt{5}-1}{2}.$
Đáp án B.