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Câu hỏi: Cho a, b là các số dương thỏa mãn ${{\log }_{9}}a={{\log }_{16}}b={{\log }_{12}}\dfrac{5b-a}{2}$. Giá trị của $\dfrac{a}{b}$ bằng:
A. $\dfrac{a}{b}=-1+\sqrt{6}$.
B. $\dfrac{a}{b}=\dfrac{7+2\sqrt{6}}{5}$.
C. $\dfrac{a}{b}=\dfrac{1+\sqrt{6}}{5}$.
D. $\dfrac{a}{b}=-7-2\sqrt{6}$.
A. $\dfrac{a}{b}=-1+\sqrt{6}$.
B. $\dfrac{a}{b}=\dfrac{7+2\sqrt{6}}{5}$.
C. $\dfrac{a}{b}=\dfrac{1+\sqrt{6}}{5}$.
D. $\dfrac{a}{b}=-7-2\sqrt{6}$.
Đặt ${{\log }_{9}}a={{\log }_{16}}b={{\log }_{12}}\dfrac{5b-a}{2}=t$
Khi đó $\left\{ \begin{aligned}
& a={{9}^{t}} \\
& b={{16}^{t}} \\
& \dfrac{5b-a}{2}={{12}^{t}} \\
\end{aligned} \right.\Rightarrow \dfrac{a}{b}={{\left( \dfrac{3}{4} \right)}^{2t}}$
Ta có ${{5.16}^{t}}-{{9}^{t}}={{2.12}^{t}}\Leftrightarrow 5-{{\left( \dfrac{9}{16} \right)}^{t}}=2.{{\left( \dfrac{12}{16} \right)}^{t}}\Leftrightarrow -{{\left( \dfrac{3}{4} \right)}^{2t}}-2.{{\left( \dfrac{3}{4} \right)}^{t}}+5=0$
Suy ra ${{\left( \dfrac{3}{4} \right)}^{t}}=-1+\sqrt{6}\Rightarrow \dfrac{a}{b}={{\left( \dfrac{3}{4} \right)}^{2t}}=7-2\sqrt{6}$.
Khi đó $\left\{ \begin{aligned}
& a={{9}^{t}} \\
& b={{16}^{t}} \\
& \dfrac{5b-a}{2}={{12}^{t}} \\
\end{aligned} \right.\Rightarrow \dfrac{a}{b}={{\left( \dfrac{3}{4} \right)}^{2t}}$
Ta có ${{5.16}^{t}}-{{9}^{t}}={{2.12}^{t}}\Leftrightarrow 5-{{\left( \dfrac{9}{16} \right)}^{t}}=2.{{\left( \dfrac{12}{16} \right)}^{t}}\Leftrightarrow -{{\left( \dfrac{3}{4} \right)}^{2t}}-2.{{\left( \dfrac{3}{4} \right)}^{t}}+5=0$
Suy ra ${{\left( \dfrac{3}{4} \right)}^{t}}=-1+\sqrt{6}\Rightarrow \dfrac{a}{b}={{\left( \dfrac{3}{4} \right)}^{2t}}=7-2\sqrt{6}$.
Đáp án D.