Lời giải của Tự Học 365

$({x^2} - 4)({\log _2}x + {\log _3}x + {\log _4}x + ... + {\log _{19}}x - \log _{20}^2x) = 0(*)$

Đkxđ: $x>0$

$(*) \Leftrightarrow \left[ \begin{array}{l}x = 2(tm)\\x =  - 2(ktm)\\{\log _2}x + {\log _3}x + {\log _4}x + ... + {\log _{19}}x - \log _{20}^2x = 0(**)\end{array} \right.$

$\begin{array}{l}(**) \Leftrightarrow \dfrac{{\log {\rm{x}}}}{{\log 2}} + \dfrac{{\log {\rm{x}}}}{{\log 3}} + \dfrac{{\log {\rm{x}}}}{{\log 4}} + ... + \dfrac{{\log {\rm{x}}}}{{\log 19}} - {\left( {\dfrac{{\log {\rm{x}}}}{{\log 20}}} \right)^2} = 0\\ \Leftrightarrow \log {\rm{x}}(\dfrac{1}{{\log 2}} + \dfrac{1}{{\log 3}} + \dfrac{1}{{\log 4}} + ... + \dfrac{1}{{\log 19}} - \dfrac{{\log {\rm{x}}}}{{{{\log }^2}20}}) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\log {\rm{x}} = 0\\\dfrac{1}{{\log 2}} + \dfrac{1}{{\log 3}} + \dfrac{1}{{\log 4}} + ... + \dfrac{1}{{\log 19}} - \dfrac{{\log {\rm{x}}}}{{{{\log }^2}20}} = 0\end{array} \right. \\ \Leftrightarrow \left[ \begin{array}{l}x = 1\\\dfrac{1}{{\log 2}} + \dfrac{1}{{\log 3}} + \dfrac{1}{{\log 4}} + ... + \dfrac{1}{{\log 19}} = \dfrac{{\log {\rm{x}}}}{{{{\log }^2}20}}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = 0\\(\dfrac{1}{{\log 2}} + \dfrac{1}{{\log 3}} + \dfrac{1}{{\log 4}} + ... + \dfrac{1}{{\log 19}}){\log ^2}20 = \log {\rm{x}}\end{array} \right. \\ \Leftrightarrow \left[ \begin{array}{l}x = 1(tm)\\x = {10^{(\dfrac{1}{{\log 2}} + \dfrac{1}{{\log 3}} + \dfrac{1}{{\log 4}} + ... + \dfrac{1}{{\log 19}}){{\log }^2}20}}(tm)\end{array} \right.\end{array}$ 

Phương trình (*) có $3$ nghiệm.

Đáp án cần chọn là: c