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Câu hỏi: Đặt ${{\log }_{2}}5=a$, ${{\log }_{3}}2=b$. Tính ${{\log }_{15}}20$ theo $a$ và $b$ ta được
A. ${{\log }_{15}}20=\dfrac{2b+1}{1+ab}$.
B. ${{\log }_{15}}20=\dfrac{2b+a}{1+ab}$.
C. ${{\log }_{15}}20=\dfrac{b+ab+1}{1+ab}$.
D. ${{\log }_{15}}20=\dfrac{2b+ab}{1+ab}$.
A. ${{\log }_{15}}20=\dfrac{2b+1}{1+ab}$.
B. ${{\log }_{15}}20=\dfrac{2b+a}{1+ab}$.
C. ${{\log }_{15}}20=\dfrac{b+ab+1}{1+ab}$.
D. ${{\log }_{15}}20=\dfrac{2b+ab}{1+ab}$.
Ta có: ${{\log }_{15}}20=\dfrac{{{\log }_{2}}20}{{{\log }_{2}}15}=\dfrac{2+{{\log }_{2}}5}{{{\log }_{2}}3+{{\log }_{2}}5}=\dfrac{2+a}{\dfrac{1}{b}+a}=\dfrac{2b+ab}{1+ab}$
Đáp án A.