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Câu hỏi: Cho ${{\log }_{2}}5=a$ và ${{\log }_{3}}5=b$, ${{\log }_{6}}5$ bằng
A. $\dfrac{ab}{a+b}$.
B. ${{a}^{2}}+{{b}^{2}}$.
C. $a+b$.
D. $\dfrac{1}{a+b}$.
A. $\dfrac{ab}{a+b}$.
B. ${{a}^{2}}+{{b}^{2}}$.
C. $a+b$.
D. $\dfrac{1}{a+b}$.
Ta có
${{\log }_{6}}5=\dfrac{1}{{{\log }_{5}}6}=\dfrac{1}{{{\log }_{5}}2+{{\log }_{5}}3}=\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}=\dfrac{ab}{a+b}$.
${{\log }_{6}}5=\dfrac{1}{{{\log }_{5}}6}=\dfrac{1}{{{\log }_{5}}2+{{\log }_{5}}3}=\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}=\dfrac{ab}{a+b}$.
Đáp án A.