Đáp án đúng: D
Giải chi tiết:ĐKXĐ: \(x > 2,\,\,x \ne 4\)
Ta có:
\(\begin{array}{l}{\log _{\sqrt 3 }}\left( {x - 2} \right) + {\log _3}{\left( {x - 4} \right)^2} = 0\\ \Leftrightarrow {\log _3}{\left( {x - 2} \right)^2} + {\log _3}{\left( {x - 4} \right)^2} = 0\end{array}\)
\( \Leftrightarrow {\log _3}{\left[ {\left( {x - 2} \right)\left( {x - 4} \right)} \right]^2} = 0\)
\(\begin{array}{l} \Leftrightarrow {\left[ {\left( {x - 2} \right)\left( {x - 4} \right)} \right]^2} = 1\\ \Leftrightarrow \left[ \begin{array}{l}\left( {x - 2} \right)\left( {x - 4} \right) = 1\\\left( {x - 2} \right)\left( {x - 4} \right) = - 1\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}{x^2} - 6x + 7 = 0\\{x^2} - 6x + 9 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 3 + \sqrt 2 \,\,\,\,\left( {tm} \right)\\x = 3 - \sqrt 2 \,\,\,\left( {ktm} \right)\\x = 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {tm} \right)\end{array} \right.\end{array}\)
Tổng các nghiệm đó là: \(S = 6 + \sqrt 2 \Rightarrow a = 6,\,b = 1 \Rightarrow Q = a.b = 6\).
Chọn D.
