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Câu hỏi: Cho a, b, c là các số thực khác 0 thỏa mãn ${{6}^{a}}={{9}^{b}}={{24}^{c}}$. Tính $T=\dfrac{a}{b}+\dfrac{a}{c}$.
A. $-3$
B. 3
C. 2
D. $\dfrac{11}{12}$
A. $-3$
B. 3
C. 2
D. $\dfrac{11}{12}$
Đặt $t={{6}^{a}}={{9}^{b}}={{24}^{c}},\left( 0<t\ne 1 \right)$.
$\Rightarrow \left\{ \begin{aligned}
& a={{\log }_{6}}t \\
& b={{\log }_{9}}t \\
& c={{\log }_{24}}t \\
\end{aligned} \right.$.
$\Rightarrow T=\dfrac{{{\log }_{6}}t}{{{\log }_{9}}t}+\dfrac{{{\log }_{6}}t}{{{\log }_{24}}t}=\dfrac{{{\log }_{t}}9}{{{\log }_{t}}6}+\dfrac{{{\log }_{t}}24}{{{\log }_{t}}6}={{\log }_{6}}9+{{\log }_{6}}24=3$
$\Rightarrow \left\{ \begin{aligned}
& a={{\log }_{6}}t \\
& b={{\log }_{9}}t \\
& c={{\log }_{24}}t \\
\end{aligned} \right.$.
$\Rightarrow T=\dfrac{{{\log }_{6}}t}{{{\log }_{9}}t}+\dfrac{{{\log }_{6}}t}{{{\log }_{24}}t}=\dfrac{{{\log }_{t}}9}{{{\log }_{t}}6}+\dfrac{{{\log }_{t}}24}{{{\log }_{t}}6}={{\log }_{6}}9+{{\log }_{6}}24=3$
Đáp án B.