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Câu hỏi: Cho $a$, $b$, $c$ là các số thực khác $0$ thỏa mãn ${{4}^{a}}={{9}^{b}}={{6}^{c}}$. Khi đó $\frac{c}{a}+\frac{c}{b}$ bằng
A. $\frac{1}{2}$.
B. $\frac{1}{6}$.
C. $\sqrt{6}$.
D. $2$.
A. $\frac{1}{2}$.
B. $\frac{1}{6}$.
C. $\sqrt{6}$.
D. $2$.
Đặt ${{4}^{a}}={{9}^{b}}={{6}^{c}}=t \left( 0<t\ne 1 \right)$ ta được : $\left\{ \begin{aligned}
& a={{\log }_{4}}t \\
& b={{\log }_{9}}t \\
& c={{\log }_{6}}t \\
\end{aligned} \right.$$\Rightarrow \frac{c}{a}+\frac{c}{b}=c\left( \frac{1}{a}+\frac{1}{b} \right) $ $ ={{\log }_{6}}t\left( \frac{1}{{{\log }_{4}}t}+\frac{1}{{{\log }_{9}}t} \right) $ $ ={{\log }_{6}}t\left( {{\log }_{t}}4+{{\log }_{t}}9 \right) $ $ ={{\log }_{6}}t . {{\log }_{t}}36 $ $ ={{\log }_{6}}36=2$.
& a={{\log }_{4}}t \\
& b={{\log }_{9}}t \\
& c={{\log }_{6}}t \\
\end{aligned} \right.$$\Rightarrow \frac{c}{a}+\frac{c}{b}=c\left( \frac{1}{a}+\frac{1}{b} \right) $ $ ={{\log }_{6}}t\left( \frac{1}{{{\log }_{4}}t}+\frac{1}{{{\log }_{9}}t} \right) $ $ ={{\log }_{6}}t\left( {{\log }_{t}}4+{{\log }_{t}}9 \right) $ $ ={{\log }_{6}}t . {{\log }_{t}}36 $ $ ={{\log }_{6}}36=2$.
Đáp án D.