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Câu hỏi: Cho x, y, z là các số thực dương tùy ý khác 1 và xyz khác 1. Đặt $a={{\log }_{x}}y,b={{\log }_{z}}y.$ Mệnh đề nào sau đây đúng?
A. ${{\log }_{xyz}}\left( {{y}^{3}}{{z}^{2}} \right)=\dfrac{3ab+2a}{a+b+1}.$
B. ${{\log }_{xyz}}\left( {{y}^{3}}{{z}^{2}} \right)=\dfrac{3ab+2b}{a+b+1}.$
C. ${{\log }_{xyz}}\left( {{y}^{3}}{{z}^{2}} \right)=\dfrac{3ab+2a}{ab+a+b}.$
D. ${{\log }_{xyz}}\left( {{y}^{3}}{{z}^{2}} \right)=\dfrac{3ab+2b}{ab+a+b}.$
A. ${{\log }_{xyz}}\left( {{y}^{3}}{{z}^{2}} \right)=\dfrac{3ab+2a}{a+b+1}.$
B. ${{\log }_{xyz}}\left( {{y}^{3}}{{z}^{2}} \right)=\dfrac{3ab+2b}{a+b+1}.$
C. ${{\log }_{xyz}}\left( {{y}^{3}}{{z}^{2}} \right)=\dfrac{3ab+2a}{ab+a+b}.$
D. ${{\log }_{xyz}}\left( {{y}^{3}}{{z}^{2}} \right)=\dfrac{3ab+2b}{ab+a+b}.$
Ta có ${{\log }_{xyz}}\left( {{y}^{3}}{{z}^{2}} \right)=\dfrac{{{\log }_{y}}\left( {{y}^{3}}{{z}^{2}} \right)}{{{\log }_{y}}\left( xyz \right)}=\dfrac{3+2{{\log }_{y}}z}{{{\log }_{y}}x+1+{{\log }_{y}}z}=\dfrac{3+\dfrac{2}{b}}{\dfrac{1}{a}+1+\dfrac{1}{b}}=\dfrac{3ab+2a}{b+ab+a}.$
Đáp án C.