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Câu hỏi: Biết rằng ${{\log }_{3}}4=a$ và $T={{\log }_{12}}18$. Phát biểu nào sau đây là đúng?
A. $T=\dfrac{a+2}{2a+2}$.
B. $T=\dfrac{a+4}{2a+2}$.
C. $T=\dfrac{\sqrt{a}+2}{a+1}$.
D. $T=\dfrac{\sqrt{a}-2}{a+1}$.
A. $T=\dfrac{a+2}{2a+2}$.
B. $T=\dfrac{a+4}{2a+2}$.
C. $T=\dfrac{\sqrt{a}+2}{a+1}$.
D. $T=\dfrac{\sqrt{a}-2}{a+1}$.
Ta có $T={{\log }_{12}}18=\dfrac{{{\log }_{3}}18}{{{\log }_{3}}12}=\dfrac{{{\log }_{3}}\left( {{2.3}^{2}} \right)}{{{\log }_{3}}\left( {{2}^{2}}.3 \right)}=\dfrac{{{\log }_{3}}2+2}{2{{\log }_{3}}2+1}$.
Mà ${{\log }_{3}}4=a\Rightarrow 2{{\log }_{3}}2=a\Leftrightarrow {{\log }_{3}}2=\dfrac{a}{2}$.
Nên $T={{\log }_{12}}18==\dfrac{{{\log }_{3}}2+2}{2{{\log }_{3}}2+1}=\dfrac{\dfrac{a}{2}+2}{a+1}=\dfrac{a+4}{2a+2}$.
Mà ${{\log }_{3}}4=a\Rightarrow 2{{\log }_{3}}2=a\Leftrightarrow {{\log }_{3}}2=\dfrac{a}{2}$.
Nên $T={{\log }_{12}}18==\dfrac{{{\log }_{3}}2+2}{2{{\log }_{3}}2+1}=\dfrac{\dfrac{a}{2}+2}{a+1}=\dfrac{a+4}{2a+2}$.
Đáp án B.