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Câu hỏi: Cho các số thực dương x, y thỏa mãn ${{\log }_{4}}x={{\log }_{9}}y={{\log }_{6}}\left( \dfrac{xy}{4}+1 \right)$. Giá trị của biểu thức $P={{x}^{{{\log }_{4}}6}}+{{y}^{{{\log }_{9}}6}}$ bằng
A. 2
B. 5
C. 4
D. 6
A. 2
B. 5
C. 4
D. 6
Đặt ${{\log }_{4}}x={{\log }_{9}}y={{\log }_{6}}\left( \dfrac{xy}{4}+1 \right)=t\Rightarrow x={{4}^{t}},y={{9}^{t}},xy={{4.6}^{t}}-4$
$\Rightarrow {{6}^{2t}}-{{4.6}^{t}}+4=0\Rightarrow {{6}^{t}}=2\Rightarrow t={{\log }_{6}}2$.
Khi đó ${{\log }_{4}}x+{{\log }_{9}}y={{\log }_{6}}\left( \dfrac{xy}{4}+1 \right)=t={{\log }_{6}}2\Leftrightarrow x={{4}^{{{\log }_{6}}2}},y={{9}^{{{\log }_{6}}2}}$.
Do đó $P={{\left( {{4}^{{{\log }_{6}}2}} \right)}^{{{\log }_{4}}6}}+{{\left( {{9}^{{{\log }_{6}}2}} \right)}^{{{\log }_{9}}6}}={{\left( {{4}^{{{\log }_{4}}6}} \right)}^{{{\log }_{6}}2}}+{{\left( {{9}^{{{\log }_{9}}6}} \right)}^{{{\log }_{6}}2}}={{6}^{{{\log }_{6}}2}}+{{6}^{{{\log }_{6}}2}}=4$.
$\Rightarrow {{6}^{2t}}-{{4.6}^{t}}+4=0\Rightarrow {{6}^{t}}=2\Rightarrow t={{\log }_{6}}2$.
Khi đó ${{\log }_{4}}x+{{\log }_{9}}y={{\log }_{6}}\left( \dfrac{xy}{4}+1 \right)=t={{\log }_{6}}2\Leftrightarrow x={{4}^{{{\log }_{6}}2}},y={{9}^{{{\log }_{6}}2}}$.
Do đó $P={{\left( {{4}^{{{\log }_{6}}2}} \right)}^{{{\log }_{4}}6}}+{{\left( {{9}^{{{\log }_{6}}2}} \right)}^{{{\log }_{9}}6}}={{\left( {{4}^{{{\log }_{4}}6}} \right)}^{{{\log }_{6}}2}}+{{\left( {{9}^{{{\log }_{9}}6}} \right)}^{{{\log }_{6}}2}}={{6}^{{{\log }_{6}}2}}+{{6}^{{{\log }_{6}}2}}=4$.
Đáp án C.