Đáp án đúng: C
Giải chi tiết: 
ĐK: \(\left\{ \begin{array}{l}{x^2} - 3x + 2 > 0\\x - m > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \in \left( { - \infty ;1} \right) \cup \left( {2; + \infty } \right)\\x > m\end{array} \right.\)
\(\begin{array}{l}{\log _2}\sqrt {{x^2} - 3x + 2}  + {\log _{\frac{1}{2}}}\left( {x - m} \right) = x - m - \sqrt {{x^2} - 3x + 2} \\ \Leftrightarrow {\log _2}\sqrt {{x^2} - 3x + 2}  - {\log _2}\left( {x - m} \right) = x - m - \sqrt {{x^2} - 3x + 2} \\ \Leftrightarrow {\log _2}\sqrt {{x^2} - 3x + 2}  + \sqrt {{x^2} - 3x + 2}  = {\log _2}\left( {x - m} \right) + x - m\end{array}\)
Xét hàm số \(f\left( t \right) = {\log _2}t + t\,\,\left( {t > 0} \right)\) ta có \(f'\left( t \right) = \frac{1}{{t\ln 2}} + 1 > 0\,\,\forall t > 0 \Rightarrow \) Hàm số đồng biến trên \(\left( {0; + \infty } \right)\).
\(\begin{array}{l} \Rightarrow \sqrt {{x^2} - 3x + 2}  = x - m\\ \Leftrightarrow {x^2} - 3x + 2 = {x^2} - 2mx + {m^2}\\ \Leftrightarrow \left( {2m - 3} \right)x + 2 - {m^2} = 0\,\,\left( * \right)\end{array}\)
TH1: \(2m - 3 = 0 \Leftrightarrow m = \frac{3}{2} \Rightarrow \) Phương trình \(\left( * \right) \Leftrightarrow 0.x - \frac{1}{4} = 0\) (vô nghiệm)
TH2: \(m \ne \frac{3}{2} \Rightarrow x = \frac{{{m^2} - 2}}{{2m - 3}}\)
Để phương trình có nghiệm \( \Rightarrow \left\{ \begin{array}{l}x \in \left( { - \infty ;1} \right) \cup \left( {2; + \infty } \right)\\x > m\end{array} \right.\)
\(\begin{array}{l} \Rightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}\frac{{{m^2} - 2}}{{2m - 3}} < 1\\\frac{{{m^2} - 2}}{{2m - 3}} > 2\end{array} \right.\\\frac{{{m^2} - 2}}{{2m - 3}} > m\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}\frac{{{m^2} - 2}}{{2m - 3}} - 1 < 0\\\frac{{{m^2} - 2}}{{2m - 3}} - 2 > 0\end{array} \right.\\\frac{{{m^2} - 2}}{{2m - 3}} - m > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}\frac{{{m^2} - 2 - 2m + 3}}{{2m - 3}} < 0\\\frac{{{m^2} - 2 - 4m + 6}}{{2m - 3}} > 0\end{array} \right.\\\frac{{{m^2} - 2 - 2{m^2} + 3m}}{{2m - 3}} > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}\frac{{{m^2} - 2m + 1}}{{2m - 3}} < 0\\\frac{{{m^2} - 4m + 4}}{{2m - 3}} > 0\end{array} \right.\\\frac{{ - {m^2} + 3m - 2}}{{2m - 3}} > 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}m \in \left( { - \infty ;\frac{3}{2}} \right)\backslash \left\{ 1 \right\}\\m \in \left( {\frac{3}{2}; + \infty } \right)\backslash \left\{ 2 \right\}\end{array} \right.\\m \in \left( { - \infty ;1} \right) \cup \left( {\frac{3}{2};2} \right)\end{array} \right. \Rightarrow m \in \left( { - \infty ;1} \right) \cup \left( {\frac{3}{2};2} \right)\end{array}\) 
Chọn C.