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Câu hỏi: Cho $a,b,c$ là các số thực dương thỏa mãn ${{a}^{2}}=9bc$. Tính $S=2{{\log }_{3}}a-{{\log }_{3}}b-{{\log }_{3}}c$.
A. $S=2{{\log }_{3}}\left( \dfrac{a}{bc} \right)$.
B. $S=1$.
C. $S=-2{{\log }_{3}}\left( \dfrac{a}{bc} \right)$.
D. $S=2$.
A. $S=2{{\log }_{3}}\left( \dfrac{a}{bc} \right)$.
B. $S=1$.
C. $S=-2{{\log }_{3}}\left( \dfrac{a}{bc} \right)$.
D. $S=2$.
Ta có: $S=2{{\log }_{3}}a-{{\log }_{3}}b-{{\log }_{3}}c$
$\begin{aligned}
& S={{\log }_{3}}{{a}^{2}}-{{\log }_{3}}b-{{\log }_{3}}c={{\log }_{3}}9bc-{{\log }_{3}}b-{{\log }_{3}}c \\
& ={{\log }_{3}}\dfrac{9bc}{bc}={{\log }_{3}}9=2 \\
\end{aligned}$
$\begin{aligned}
& S={{\log }_{3}}{{a}^{2}}-{{\log }_{3}}b-{{\log }_{3}}c={{\log }_{3}}9bc-{{\log }_{3}}b-{{\log }_{3}}c \\
& ={{\log }_{3}}\dfrac{9bc}{bc}={{\log }_{3}}9=2 \\
\end{aligned}$
Đáp án D.