Google
Câu hỏi: Cho hai số thực a, b thỏa mãn ${{\log }_{100}}a={{\log }_{40}}b={{\log }_{16}}\dfrac{a-4b}{12}.$ Giá trị $\dfrac{a}{b}$ bằng:
A. 4.
B. 12.
C. 6.
D. 2.
A. 4.
B. 12.
C. 6.
D. 2.
Ta có ${{\log }_{100}}a={{\log }_{40}}b={{\log }_{16}}\dfrac{a-4b}{12}=t\Leftrightarrow \left\{ \begin{aligned}
& a={{100}^{t}};b={{40}^{t}} \\
& a-4b={{12.16}^{t}} \\
\end{aligned} \right.$
Khi đó ${{100}^{t}}-{{4.40}^{t}}={{12.16}^{t}}\Leftrightarrow {{\left( {{10}^{t}} \right)}^{2}}-{{4.10}^{t}}{{.4}^{t}}-12.{{\left( {{4}^{t}} \right)}^{2}}=0\Leftrightarrow {{\left[ {{\left( \dfrac{10}{4} \right)}^{t}} \right]}^{2}}-4.{{\left( \dfrac{10}{4} \right)}^{t}}-12=0$
$\Leftrightarrow {{\left( \dfrac{10}{4} \right)}^{t}}=6$ mà $\dfrac{a}{b}=\dfrac{{{100}^{t}}}{{{40}^{t}}}=\left( \dfrac{{{100}^{t}}}{{{40}^{t}}} \right)={{\left( \dfrac{10}{4} \right)}^{t}}.$ Vậy $\dfrac{a}{b}={{\left( \dfrac{10}{4} \right)}^{t}}=6.$
& a={{100}^{t}};b={{40}^{t}} \\
& a-4b={{12.16}^{t}} \\
\end{aligned} \right.$
Khi đó ${{100}^{t}}-{{4.40}^{t}}={{12.16}^{t}}\Leftrightarrow {{\left( {{10}^{t}} \right)}^{2}}-{{4.10}^{t}}{{.4}^{t}}-12.{{\left( {{4}^{t}} \right)}^{2}}=0\Leftrightarrow {{\left[ {{\left( \dfrac{10}{4} \right)}^{t}} \right]}^{2}}-4.{{\left( \dfrac{10}{4} \right)}^{t}}-12=0$
$\Leftrightarrow {{\left( \dfrac{10}{4} \right)}^{t}}=6$ mà $\dfrac{a}{b}=\dfrac{{{100}^{t}}}{{{40}^{t}}}=\left( \dfrac{{{100}^{t}}}{{{40}^{t}}} \right)={{\left( \dfrac{10}{4} \right)}^{t}}.$ Vậy $\dfrac{a}{b}={{\left( \dfrac{10}{4} \right)}^{t}}=6.$
Đáp án C.