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Câu hỏi: Cho $x,y$ là các số thực lớn hơn $1$ thoả mãn ${{x}^{2}}+9{{y}^{2}}=6xy$. Tính $M=\dfrac{1+{{\log }_{12}}x+{{\log }_{12}}y}{2{{\log }_{12}}\left( x+3y \right)}$.
A. $M=\dfrac{1}{2}$.
B. $M=\dfrac{1}{3}$.
C. $M=\dfrac{1}{4}$.
D. $M=1$
A. $M=\dfrac{1}{2}$.
B. $M=\dfrac{1}{3}$.
C. $M=\dfrac{1}{4}$.
D. $M=1$
Ta có ${{x}^{2}}+9{{y}^{2}}=6xy\Leftrightarrow {{\left( x-3y \right)}^{2}}=0\Leftrightarrow x=3y$.
Khi đó $M=\dfrac{1+{{\log }_{12}}x+{{\log }_{12}}y}{2{{\log }_{12}}\left( x+3y \right)}=\dfrac{{{\log }_{12}}\left( 12xy \right)}{{{\log }_{12}}{{\left( x+3y \right)}^{2}}}=\dfrac{{{\log }_{12}}\left( 36{{y}^{2}} \right)}{{{\log }_{12}}\left( 36{{y}^{2}} \right)}=1$.
Khi đó $M=\dfrac{1+{{\log }_{12}}x+{{\log }_{12}}y}{2{{\log }_{12}}\left( x+3y \right)}=\dfrac{{{\log }_{12}}\left( 12xy \right)}{{{\log }_{12}}{{\left( x+3y \right)}^{2}}}=\dfrac{{{\log }_{12}}\left( 36{{y}^{2}} \right)}{{{\log }_{12}}\left( 36{{y}^{2}} \right)}=1$.
Đáp án D.