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Câu hỏi: Cho ${{\log }_{2}}5=m$, ${{\log }_{3}}5=n$. Khi đó ${{\log }_{6}}5$ tính theo $m , n$ là
A. ${{m}^{2}}+{{n}^{2}}$
B. $\dfrac{mn}{m+n}$
C. $\dfrac{1}{m+n}$
D. $m+n$
A. ${{m}^{2}}+{{n}^{2}}$
B. $\dfrac{mn}{m+n}$
C. $\dfrac{1}{m+n}$
D. $m+n$
Ta có: ${{\log }_{6}}5=\dfrac{1}{{{\log }_{5}}6}=\dfrac{1}{{{\log }_{5}}2+{{\log }_{5}}3}=\dfrac{1}{\dfrac{1}{{{\log }_{2}}5}+\dfrac{1}{{{\log }_{2}}3}}=\dfrac{mn}{m+n}$.
Đáp án B.