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Câu hỏi: Với các số thực dương $x,y$ tùy ý, đặt ${{\log }_{3}}x=\alpha ,{{\log }_{3}}y=\beta $. Mệnh đề nào dưới đây đúng?
A. ${{\log }_{27}}{{\left( \dfrac{\sqrt{x}}{y} \right)}^{3}}=9\left( \dfrac{\alpha }{2}-\beta \right)$
B. ${{\log }_{27}}{{\left( \dfrac{\sqrt{x}}{y} \right)}^{3}}=\dfrac{\alpha }{2}+\beta $
C. ${{\log }_{27}}{{\left( \dfrac{\sqrt{x}}{y} \right)}^{3}}=9\left( \dfrac{\alpha }{2}+\beta \right)$
D. ${{\log }_{27}}{{\left( \dfrac{\sqrt{x}}{y} \right)}^{3}}=\dfrac{\alpha }{2}-\beta $
A. ${{\log }_{27}}{{\left( \dfrac{\sqrt{x}}{y} \right)}^{3}}=9\left( \dfrac{\alpha }{2}-\beta \right)$
B. ${{\log }_{27}}{{\left( \dfrac{\sqrt{x}}{y} \right)}^{3}}=\dfrac{\alpha }{2}+\beta $
C. ${{\log }_{27}}{{\left( \dfrac{\sqrt{x}}{y} \right)}^{3}}=9\left( \dfrac{\alpha }{2}+\beta \right)$
D. ${{\log }_{27}}{{\left( \dfrac{\sqrt{x}}{y} \right)}^{3}}=\dfrac{\alpha }{2}-\beta $
${{\log }_{27}}{{\left( \dfrac{\sqrt{x}}{y} \right)}^{3}}=\dfrac{3}{2}{{\log }_{27}}x-3{{\log }_{27}}y=\dfrac{1}{2}{{\log }_{3}}x-{{\log }_{3}}y=\dfrac{\alpha }{2}-\beta $
Đáp án D.