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Câu hỏi: Đặt $a={{\log }_{2}}5,b={{\log }_{5}}3$. Mệnh đề nào dưới đây đúng?
A. ${{\log }_{48}}45=\dfrac{a+2b}{4+ab}$
B. ${{\log }_{48}}45=\dfrac{a+2\text{a}b}{4+ab}$
C. ${{\log }_{48}}45=\dfrac{1+2b}{4\text{a}+b}$
D. ${{\log }_{\dfrac{1}{27}}}\left( \dfrac{x}{{{y}^{3}}} \right)=\dfrac{1}{3}a+b$
A. ${{\log }_{48}}45=\dfrac{a+2b}{4+ab}$
B. ${{\log }_{48}}45=\dfrac{a+2\text{a}b}{4+ab}$
C. ${{\log }_{48}}45=\dfrac{1+2b}{4\text{a}+b}$
D. ${{\log }_{\dfrac{1}{27}}}\left( \dfrac{x}{{{y}^{3}}} \right)=\dfrac{1}{3}a+b$
Cách 1: Ta có ${{\log }_{2}}3={{\log }_{2}}5.{{\log }_{5}}3=ab$
${{\log }_{48}}45=\dfrac{{{\log }_{2}}45}{{{\log }_{2}}48}=\dfrac{{{\log }_{2}}\left( {{3}^{2}}.5 \right)}{{{\log }_{2}}\left( {{2}^{4}}.3 \right)}=\dfrac{2{{\log }_{2}}3+{{\log }_{2}}5}{4+{{\log }_{2}}3}=\dfrac{a+2\text{a}b}{4+ab}$.
Cách 2:
Lưu biến nhớ ${{\log }_{2}}5\to A,{{\log }_{5}}3\to B$
Bấm ${{\log }_{48}}45-\dfrac{A+2\text{A}B}{4+AB}=0$ nên đáp án B đúng.
${{\log }_{48}}45=\dfrac{{{\log }_{2}}45}{{{\log }_{2}}48}=\dfrac{{{\log }_{2}}\left( {{3}^{2}}.5 \right)}{{{\log }_{2}}\left( {{2}^{4}}.3 \right)}=\dfrac{2{{\log }_{2}}3+{{\log }_{2}}5}{4+{{\log }_{2}}3}=\dfrac{a+2\text{a}b}{4+ab}$.
Cách 2:
Lưu biến nhớ ${{\log }_{2}}5\to A,{{\log }_{5}}3\to B$
Bấm ${{\log }_{48}}45-\dfrac{A+2\text{A}B}{4+AB}=0$ nên đáp án B đúng.
Đáp án B.