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Câu hỏi: Cho các số thực a, b thỏa mãn ${{\log }_{4}}a={{\log }_{6}}b={{\log }_{9}}(4a-5b)-1$. Đặt $T=\dfrac{b}{a}$. Khẳng định nào sau đây đúng?
A. $0<T<\dfrac{1}{2}$
B. $-2<T<0$
C. $1<T<2$
D. $\dfrac{1}{2}<T<\dfrac{2}{3}$
A. $0<T<\dfrac{1}{2}$
B. $-2<T<0$
C. $1<T<2$
D. $\dfrac{1}{2}<T<\dfrac{2}{3}$
Ta có ${{\log }_{4}}a={{\log }_{6}}b={{\log }_{9}}\left( 4a-5b \right)-1=t\Leftrightarrow \left\{ \begin{aligned}
& a={{4}^{t}};b={{6}^{t}} \\
& 4a-5b={{9}^{t+1}} \\
\end{aligned} \right.$
$\Rightarrow {{4.4}^{t}}-{{5.6}^{t}}={{9.9}^{t}}\Leftrightarrow 4{{\left[ {{\left( \dfrac{2}{3} \right)}^{t}} \right]}^{2}}-4{{\left( \dfrac{2}{3} \right)}^{t}}-9=0\Leftrightarrow {{\left( \dfrac{2}{3} \right)}^{t}}=\dfrac{9}{4}\Leftrightarrow t=-2$
Do đó: $\dfrac{a}{b}={{\left( \dfrac{2}{3} \right)}^{t}}=\dfrac{9}{4}\Rightarrow \dfrac{b}{a}=\dfrac{4}{9}\in \left( 0;\dfrac{1}{2} \right)$.
& a={{4}^{t}};b={{6}^{t}} \\
& 4a-5b={{9}^{t+1}} \\
\end{aligned} \right.$
$\Rightarrow {{4.4}^{t}}-{{5.6}^{t}}={{9.9}^{t}}\Leftrightarrow 4{{\left[ {{\left( \dfrac{2}{3} \right)}^{t}} \right]}^{2}}-4{{\left( \dfrac{2}{3} \right)}^{t}}-9=0\Leftrightarrow {{\left( \dfrac{2}{3} \right)}^{t}}=\dfrac{9}{4}\Leftrightarrow t=-2$
Do đó: $\dfrac{a}{b}={{\left( \dfrac{2}{3} \right)}^{t}}=\dfrac{9}{4}\Rightarrow \dfrac{b}{a}=\dfrac{4}{9}\in \left( 0;\dfrac{1}{2} \right)$.
Đáp án A.