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Câu hỏi: Rút gọn biểu thức $M=\dfrac{1}{{{\log }_{a}}x}+\dfrac{1}{{{\log }_{{{a}^{2}}}}x}+...+\dfrac{1}{{{\log }_{{{a}^{k}}}}x}$ ta được:
A. $M=\dfrac{k\left( k+1 \right)}{3{{\log }_{a}}x}.$
B. $M=\dfrac{k\left( k+1 \right)}{2{{\log }_{a}}x}.$
C. $M=\dfrac{k\left( k+1 \right)}{{{\log }_{a}}x}.$
D. $M=\dfrac{4k\left( k+1 \right)}{{{\log }_{a}}x}.$
A. $M=\dfrac{k\left( k+1 \right)}{3{{\log }_{a}}x}.$
B. $M=\dfrac{k\left( k+1 \right)}{2{{\log }_{a}}x}.$
C. $M=\dfrac{k\left( k+1 \right)}{{{\log }_{a}}x}.$
D. $M=\dfrac{4k\left( k+1 \right)}{{{\log }_{a}}x}.$
Ta có $\dfrac{1}{{{\log }_{a}}x}={{\log }_{x}}a;\dfrac{1}{{{\log }_{{{a}^{2}}}}x}=2{{\log }_{x}}a;\dfrac{1}{{{\log }_{{{a}^{k}}}}x}=k.{{\log }_{x}}a$
Khi đó $M={{\log }_{x}}a\left( 1+2+3+...+k \right)=\dfrac{k\left( k+1 \right)}{2{{\log }_{a}}x}.$
Khi đó $M={{\log }_{x}}a\left( 1+2+3+...+k \right)=\dfrac{k\left( k+1 \right)}{2{{\log }_{a}}x}.$
Đáp án B.