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Câu hỏi: Cho $a,b,c>0;a,b\ne 1$. Biểu thức $A={{\log }_{a}}\left( {{b}^{2}} \right).{{\log }_{b}}\left( \sqrt{bc} \right)-{{\log }_{a}}\left( c \right)$ bằng
A. ${{\log }_{a}}c$
B. 1
C. ${{\log }_{a}}b$
D. ${{\log }_{a}}bc$
A. ${{\log }_{a}}c$
B. 1
C. ${{\log }_{a}}b$
D. ${{\log }_{a}}bc$
Có $A={{\log }_{a}}\left( {{b}^{2}} \right).{{\log }_{b}}\left( \sqrt{bc} \right)-{{\log }_{a}}\left( c \right)=2{{\log }_{a}}b.\dfrac{1}{2}{{\log }_{b}}\left( bc \right)-{{\log }_{a}}\left( c \right)$
$=2{{\log }_{a}}b.\dfrac{1}{2}\left( {{\log }_{b}}b+{{\log }_{b}}c \right)-{{\log }_{a}}\left( c \right)={{\log }_{a}}b.\left( 1+{{\log }_{b}}c \right)-{{\log }_{a}}c$
$={{\log }_{a}}b+{{\log }_{a}}b.{{\log }_{b}}c-{{\log }_{a}}c={{\log }_{a}}b+{{\log }_{a}}c-{{\log }_{a}}c={{\log }_{a}}b$
$=2{{\log }_{a}}b.\dfrac{1}{2}\left( {{\log }_{b}}b+{{\log }_{b}}c \right)-{{\log }_{a}}\left( c \right)={{\log }_{a}}b.\left( 1+{{\log }_{b}}c \right)-{{\log }_{a}}c$
$={{\log }_{a}}b+{{\log }_{a}}b.{{\log }_{b}}c-{{\log }_{a}}c={{\log }_{a}}b+{{\log }_{a}}c-{{\log }_{a}}c={{\log }_{a}}b$
Đáp án C.