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Câu hỏi: Cho $a,b,c$ là các số thực khác $0$ thỏa mãn ${{6}^{a}}={{9}^{b}}={{24}^{c}}$. Giá trị của biểu thức $T=\dfrac{a}{b}+\dfrac{a}{c}$ bằng
A. $\dfrac{11}{12}$.
B. $\dfrac{1}{3}$.
C. $3$.
D. $2$.
A. $\dfrac{11}{12}$.
B. $\dfrac{1}{3}$.
C. $3$.
D. $2$.
Đặt ${{6}^{a}}={{9}^{b}}={{24}^{c}}=t$, $\left( t>0,t\ne 1 \right)$.
$\Rightarrow \left\{ \begin{matrix}
a={{\log }_{6}}t=\dfrac{1}{{{\log }_{t}}6} \\
b={{\log }_{9}}t=\dfrac{1}{{{\log }_{t}}9} \\
c={{\log }_{24}}t=\dfrac{1}{{{\log }_{t}}24} \\
\end{matrix} \right.$$\Rightarrow T=\dfrac{a}{b}+\dfrac{a}{c}=a\left( \dfrac{1}{b}+\dfrac{1}{c} \right)=\dfrac{1}{{{\log }_{t}}6}\left( {{\log }_{t}}9+{{\log }_{t}}24 \right)$
$\Leftrightarrow T=\dfrac{1}{{{\log }_{t}}6}.\left( {{\log }_{t}}216 \right)=\dfrac{1}{{{\log }_{t}}6}.\left( {{\log }_{t}}{{6}^{3}} \right)=3$.
Vậy $T=3$.
$\Rightarrow \left\{ \begin{matrix}
a={{\log }_{6}}t=\dfrac{1}{{{\log }_{t}}6} \\
b={{\log }_{9}}t=\dfrac{1}{{{\log }_{t}}9} \\
c={{\log }_{24}}t=\dfrac{1}{{{\log }_{t}}24} \\
\end{matrix} \right.$$\Rightarrow T=\dfrac{a}{b}+\dfrac{a}{c}=a\left( \dfrac{1}{b}+\dfrac{1}{c} \right)=\dfrac{1}{{{\log }_{t}}6}\left( {{\log }_{t}}9+{{\log }_{t}}24 \right)$
$\Leftrightarrow T=\dfrac{1}{{{\log }_{t}}6}.\left( {{\log }_{t}}216 \right)=\dfrac{1}{{{\log }_{t}}6}.\left( {{\log }_{t}}{{6}^{3}} \right)=3$.
Vậy $T=3$.
Đáp án C.