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Câu hỏi: Với các số thực $a,b>0,a\ne 1$ tùy ý, biểu thức ${{\log }_{{{a}^{2}}}}\left( a{{b}^{2}} \right)$ bằng:
A. $2+{{\log }_{a}}b.$
B. $2+4{{\log }_{a}}b.$
C. $\dfrac{1}{2}+{{\log }_{a}}b.$
D. $\dfrac{1}{2}+4{{\log }_{a}}b.$
A. $2+{{\log }_{a}}b.$
B. $2+4{{\log }_{a}}b.$
C. $\dfrac{1}{2}+{{\log }_{a}}b.$
D. $\dfrac{1}{2}+4{{\log }_{a}}b.$
Ta có:
${{\log }_{{{a}^{2}}}}\left( a{{b}^{2}} \right)=\dfrac{1}{2}{{\log }_{a}}\left( a{{b}^{2}} \right)=\dfrac{1}{2}\left( {{\log }_{a}}a+{{\log }_{a}}{{b}^{2}} \right)=\dfrac{1}{2}\left( 1+2{{\log }_{a}}b \right)=\dfrac{1}{2}+{{\log }_{a}}b.$
${{\log }_{{{a}^{2}}}}\left( a{{b}^{2}} \right)=\dfrac{1}{2}{{\log }_{a}}\left( a{{b}^{2}} \right)=\dfrac{1}{2}\left( {{\log }_{a}}a+{{\log }_{a}}{{b}^{2}} \right)=\dfrac{1}{2}\left( 1+2{{\log }_{a}}b \right)=\dfrac{1}{2}+{{\log }_{a}}b.$
Đáp án C.