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Câu hỏi: Với ${{\log }_{5}}3=a$ thì ${{\log }_{15}}45$ bằng
A. $\dfrac{1+{{a}^{2}}}{1+a}$.
B. $\dfrac{1+2a}{1+a}$.
C. $\dfrac{2+a}{1+a}$.
D. $\dfrac{2}{a}$.
Ta có ${{\log }_{15}}45=\dfrac{{{\log }_{5}}\left( {{3}^{2}}.5 \right)}{{{\log }_{5}}\left( 3.5 \right)}=\dfrac{2{{\log }_{5}}3+1}{{{\log }_{5}}3+1}=\dfrac{1+2a}{1+a}$.
A. $\dfrac{1+{{a}^{2}}}{1+a}$.
B. $\dfrac{1+2a}{1+a}$.
C. $\dfrac{2+a}{1+a}$.
D. $\dfrac{2}{a}$.
Ta có ${{\log }_{15}}45=\dfrac{{{\log }_{5}}\left( {{3}^{2}}.5 \right)}{{{\log }_{5}}\left( 3.5 \right)}=\dfrac{2{{\log }_{5}}3+1}{{{\log }_{5}}3+1}=\dfrac{1+2a}{1+a}$.
Đáp án B.