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Câu hỏi: Biết ${{4}^{{{x}_{1}}}}=5$, ${{5}^{{{x}_{2}}}}=6$, ${{6}^{{{x}_{3}}}}=7$, …, ${{6}^{{{x}_{60}}}}=64$, khi đó ${{x}_{1}}{{x}_{2}}.{{x}_{2}}...{{x}_{60}}$ bằng
A. $4$.
B. $3$.
C. $\dfrac{3}{2}$.
D. $\dfrac{5}{2}$.
A. $4$.
B. $3$.
C. $\dfrac{3}{2}$.
D. $\dfrac{5}{2}$.
Ta có
$\left\{ \begin{aligned}
& {{4}^{{{x}_{1}}}}=5 \\
& {{5}^{{{x}_{2}}}}=6 \\
& {{6}^{{{x}_{3}}}}=7 \\
& ... \\
& {{63}^{{{x}_{60}}}}=64 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{x}_{1}}={{\log }_{4}}5 \\
& {{x}_{2}}={{\log }_{5}}6 \\
& {{x}_{3}}={{\log }_{6}}7 \\
& ... \\
& {{x}_{60}}={{\log }_{63}}64 \\
\end{aligned} \right.\Rightarrow {{x}_{1}}{{x}_{2}}.{{x}_{3}}...{{x}_{60}}={{\log }_{4}}5.{{\log }_{5}}6.{{\log }_{6}}7. ... .{{\log }_{63}}64={{\log }_{4}}64=3$.
$\left\{ \begin{aligned}
& {{4}^{{{x}_{1}}}}=5 \\
& {{5}^{{{x}_{2}}}}=6 \\
& {{6}^{{{x}_{3}}}}=7 \\
& ... \\
& {{63}^{{{x}_{60}}}}=64 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{x}_{1}}={{\log }_{4}}5 \\
& {{x}_{2}}={{\log }_{5}}6 \\
& {{x}_{3}}={{\log }_{6}}7 \\
& ... \\
& {{x}_{60}}={{\log }_{63}}64 \\
\end{aligned} \right.\Rightarrow {{x}_{1}}{{x}_{2}}.{{x}_{3}}...{{x}_{60}}={{\log }_{4}}5.{{\log }_{5}}6.{{\log }_{6}}7. ... .{{\log }_{63}}64={{\log }_{4}}64=3$.
Đáp án B.