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Câu hỏi: Cho $a={{\log }_{7}}12$ và $b={{\log }_{12}}14$. Biểu diễn $c={{\log }_{84}}54$ theo a và b, ta được kết quả:
A. $c=\dfrac{2a+5\left( 1+ab \right)}{a+1}.$
B. $c=\dfrac{a+1}{3a-5\left( 1+ab \right)}.$
C. $c=\dfrac{a+1}{3a+5\left( 1+ab \right)}.$
D. $c=\dfrac{3a+5\left( 1-ab \right)}{a+1}.$
A. $c=\dfrac{2a+5\left( 1+ab \right)}{a+1}.$
B. $c=\dfrac{a+1}{3a-5\left( 1+ab \right)}.$
C. $c=\dfrac{a+1}{3a+5\left( 1+ab \right)}.$
D. $c=\dfrac{3a+5\left( 1-ab \right)}{a+1}.$
Ta có $a={{\log }_{7}}12={{\log }_{7}}\left( {{2}^{2}}.3 \right)=2{{\log }_{7}}2+{{\log }_{7}}3\ \ \ \ \left( 1 \right)$.
$b={{\log }_{12}}14=\dfrac{{{\log }_{7}}14}{{{\log }_{7}}12}=\dfrac{{{\log }_{7}}\left( 7.2 \right)}{a}=\dfrac{1+{{\log }_{7}}2}{a}\Rightarrow 1+{{\log }_{7}}2=ab\Leftrightarrow {{\log }_{7}}2=ab-1.$
Thay ${{\log }_{7}}2=ab-1$ vào (1) ta được $a=2\left( ab-1 \right)+{{\log }_{7}}3\Rightarrow {{\log }_{7}}3=a-2\left( ab-1 \right)$.
Do đó $c={{\log }_{84}}54=\dfrac{{{\log }_{7}}54}{{{\log }_{7}}84}=\dfrac{{{\log }_{7}}\left( {{2.3}^{3}} \right)}{{{\log }_{7}}\left( {{2}^{2}}.3.7 \right)}=\dfrac{{{\log }_{7}}2+3{{\log }_{7}}3}{2{{\log }_{7}}2+{{\log }_{7}}3+1}=\dfrac{3a+5\left( 1-ab \right)}{a+1}.$
$b={{\log }_{12}}14=\dfrac{{{\log }_{7}}14}{{{\log }_{7}}12}=\dfrac{{{\log }_{7}}\left( 7.2 \right)}{a}=\dfrac{1+{{\log }_{7}}2}{a}\Rightarrow 1+{{\log }_{7}}2=ab\Leftrightarrow {{\log }_{7}}2=ab-1.$
Thay ${{\log }_{7}}2=ab-1$ vào (1) ta được $a=2\left( ab-1 \right)+{{\log }_{7}}3\Rightarrow {{\log }_{7}}3=a-2\left( ab-1 \right)$.
Do đó $c={{\log }_{84}}54=\dfrac{{{\log }_{7}}54}{{{\log }_{7}}84}=\dfrac{{{\log }_{7}}\left( {{2.3}^{3}} \right)}{{{\log }_{7}}\left( {{2}^{2}}.3.7 \right)}=\dfrac{{{\log }_{7}}2+3{{\log }_{7}}3}{2{{\log }_{7}}2+{{\log }_{7}}3+1}=\dfrac{3a+5\left( 1-ab \right)}{a+1}.$
Đáp án D.