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Câu hỏi: Đặt $a={{\log }_{2}}3, b={{\log }_{5}}3$. Hãy biểu diễn ${{\log }_{6}}45$ theo a và b?
A. ${{\log }_{6}}45=\dfrac{a+2ab}{ab}$
B. ${{\log }_{6}}45=\dfrac{2{{a}^{2}}-2ab}{ab}$
C. ${{\log }_{6}}45=\dfrac{a+2ab}{ab+b}$
D. ${{\log }_{6}}45=\dfrac{2{{a}^{2}}-2ab}{ab+b}$
A. ${{\log }_{6}}45=\dfrac{a+2ab}{ab}$
B. ${{\log }_{6}}45=\dfrac{2{{a}^{2}}-2ab}{ab}$
C. ${{\log }_{6}}45=\dfrac{a+2ab}{ab+b}$
D. ${{\log }_{6}}45=\dfrac{2{{a}^{2}}-2ab}{ab+b}$
Ta có ${{\log }_{2}}3=a\Leftrightarrow {{\log }_{3}}2=\dfrac{1}{a};{{\log }_{5}}3=b\Leftrightarrow {{\log }_{3}}5=\dfrac{1}{b}$
Vậy ${{\log }_{6}}45=\dfrac{{{\log }_{3}}45}{{{\log }_{3}}6}=\dfrac{{{\log }_{3}}9+{{\log }_{3}}5}{{{\log }_{3}}3+{{\log }_{3}}2}=\dfrac{2+{{\log }_{3}}5}{1+{{\log }_{3}}2}=\dfrac{2+\dfrac{1}{b}}{1+\dfrac{1}{a}}=\dfrac{a\left( 1+2b \right)}{b\left( 1+a \right)}=\dfrac{a+2ab}{ab+b}$
Vậy ${{\log }_{6}}45=\dfrac{{{\log }_{3}}45}{{{\log }_{3}}6}=\dfrac{{{\log }_{3}}9+{{\log }_{3}}5}{{{\log }_{3}}3+{{\log }_{3}}2}=\dfrac{2+{{\log }_{3}}5}{1+{{\log }_{3}}2}=\dfrac{2+\dfrac{1}{b}}{1+\dfrac{1}{a}}=\dfrac{a\left( 1+2b \right)}{b\left( 1+a \right)}=\dfrac{a+2ab}{ab+b}$
Đáp án C.