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Câu hỏi: Cho $0<x\ne 1,0<a\ne 1$ và $M=\dfrac{1}{{{\log }_{a}}x}+\dfrac{1}{{{\log }_{{{a}^{3}}}}x}+\dfrac{1}{{{\log }_{{{a}^{5}}}}x}+...+\dfrac{1}{{{\log }_{{{a}^{2019}}}}x}.$ Khẳng định nào sau đây là đúng?
A. $M=\dfrac{{{2020}^{2}}}{{{\log }_{a}}x}.$
B. $M=\dfrac{2018.1010}{{{\log }_{a}}x}.$
C. $M=\dfrac{2020.1010}{{{\log }_{a}}x}.$
D. $M=\dfrac{{{1010}^{2}}}{{{\log }_{a}}x}.$
A. $M=\dfrac{{{2020}^{2}}}{{{\log }_{a}}x}.$
B. $M=\dfrac{2018.1010}{{{\log }_{a}}x}.$
C. $M=\dfrac{2020.1010}{{{\log }_{a}}x}.$
D. $M=\dfrac{{{1010}^{2}}}{{{\log }_{a}}x}.$
Với điều kiện $0<x\ne 1,0<a\ne 1.$
Ta có: $M={{\log }_{x}}a+{{\log }_{x}}{{a}^{3}}+{{\log }_{x}}{{a}^{5}}+...+{{\log }_{x}}{{a}^{2019}}$
$={{\log }_{x}}\left( a.{{a}^{3}}.{{a}^{5}}...{{a}^{2019}} \right)=\left( 1+3+5+...+2019 \right){{\log }_{x}}a\left( * \right)$
$={{1010}^{2}}.{{\log }_{x}}a=\dfrac{{{1010}^{2}}}{{{\log }_{a}}x}.$
Ta có: $M={{\log }_{x}}a+{{\log }_{x}}{{a}^{3}}+{{\log }_{x}}{{a}^{5}}+...+{{\log }_{x}}{{a}^{2019}}$
$={{\log }_{x}}\left( a.{{a}^{3}}.{{a}^{5}}...{{a}^{2019}} \right)=\left( 1+3+5+...+2019 \right){{\log }_{x}}a\left( * \right)$
$={{1010}^{2}}.{{\log }_{x}}a=\dfrac{{{1010}^{2}}}{{{\log }_{a}}x}.$
| ${{\log }_{a}}\left( {{y}_{1}}.{{y}_{2}}.....{{y}_{n}} \right)={{\log }_{a}}{{y}_{1}}+{{\log }_{a}}{{y}_{2}}+...+{{\log }_{a}}{{y}_{n}}$ (với $a,{{y}_{1}},{{y}_{2}},...,{{y}_{n}}>0;a\ne 1$ ) Công thức: $1+3+5+...+\left( 2n-1 \right)={{n}^{2}}.$ Chứng minh dựa vào tính tổng của một CSC với ${{u}_{1}}=1,{{u}_{n}}=2n-1,d=2.$ Khi đó tổng $S=\dfrac{n}{2}\left( {{u}_{1}}+{{u}_{n}} \right)=\dfrac{n}{2}\left( 1+2n-1 \right)={{n}^{2}}$ |
Đáp án D.