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Câu hỏi: Cho $\text{lo}{{\text{g}}_{2}}5=a; \text{lo}{{\text{g}}_{3}}5=b$. Khi đó ${{\log }_{5}}6$ tính theo $a$ và $b$ là:
A. $a+b$.
B. $\dfrac{ab}{a+b}$.
C. $\dfrac{a+b}{ab}$.
D. ${{a}^{2}}+{{b}^{2}}.$
Ta có ${{\log }_{5}}6={{\log }_{5}}2+{{\log }_{5}}3=\dfrac{1}{{{\log }_{2}}5}+\dfrac{1}{{{\log }_{3}}5}=\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{a+b}{a.b}$.
A. $a+b$.
B. $\dfrac{ab}{a+b}$.
C. $\dfrac{a+b}{ab}$.
D. ${{a}^{2}}+{{b}^{2}}.$
Ta có ${{\log }_{5}}6={{\log }_{5}}2+{{\log }_{5}}3=\dfrac{1}{{{\log }_{2}}5}+\dfrac{1}{{{\log }_{3}}5}=\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{a+b}{a.b}$.
Đáp án C.