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Câu hỏi: Cho hai số thực $x$, $y$ thỏa $1<x<y$ và ${{\log }_{x}}\left( {{y}^{4}} \right)+{{\log }_{y}}\left( {{x}^{5}} \right)=9$. Tính ${{\log }_{xy}}\dfrac{{{x}^{5}}+{{y}^{4}}}{2}$.
A. $0$.
B. $1$.
C. $\dfrac{20}{9}$.
D. $\dfrac{45}{4}$.
A. $0$.
B. $1$.
C. $\dfrac{20}{9}$.
D. $\dfrac{45}{4}$.
Đặt $t={{\log }_{x}}y\left( t>1 \right)\Leftrightarrow y={{x}^{t}}$, khi đó ${{\log }_{x}}\left( {{x}^{4t}} \right)+{{\log }_{{{x}^{t}}}}\left( {{x}^{5}} \right)=9\Leftrightarrow 4t+\dfrac{5}{t}=9\Rightarrow t=\dfrac{5}{4}$.
$\Rightarrow y={{x}^{\dfrac{5}{4}}}\Rightarrow \left\{ \begin{matrix}
{{y}^{4}}={{x}^{5}} \\
xy={{x}^{\dfrac{9}{4}}} \\
\end{matrix} \right.\Rightarrow {{\log }_{xy}}\dfrac{{{x}^{5}}+{{y}^{4}}}{2}={{\log }_{{{x}^{\dfrac{9}{4}}}}}\dfrac{2{{x}^{5}}}{2}=\dfrac{20}{9}$.
$\Rightarrow y={{x}^{\dfrac{5}{4}}}\Rightarrow \left\{ \begin{matrix}
{{y}^{4}}={{x}^{5}} \\
xy={{x}^{\dfrac{9}{4}}} \\
\end{matrix} \right.\Rightarrow {{\log }_{xy}}\dfrac{{{x}^{5}}+{{y}^{4}}}{2}={{\log }_{{{x}^{\dfrac{9}{4}}}}}\dfrac{2{{x}^{5}}}{2}=\dfrac{20}{9}$.
Đáp án C.