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Câu hỏi: Biết ${{\log }_{6}}2=a$, ${{\log }_{6}}5=b$. Tính $I={{\log }_{3}}5$ theo $a$, $b$.
A. $I=\dfrac{b}{1+a}$.
B. $I=\dfrac{b}{1-a}$.
C. $I=\dfrac{b}{a-1}$.
D. $I=\dfrac{b}{a}$.
A. $I=\dfrac{b}{1+a}$.
B. $I=\dfrac{b}{1-a}$.
C. $I=\dfrac{b}{a-1}$.
D. $I=\dfrac{b}{a}$.
Ta có ${{\log }_{3}}5=\dfrac{{{\log }_{6}}5}{{{\log }_{6}}3}=\dfrac{{{\log }_{6}}5}{{{\log }_{6}}6-{{\log }_{6}}2}=\dfrac{b}{1-a}$.
Đáp án B.