Tập xác định: \[D = \left( {0; + \infty } \right)\].

Ta có \({\log _4}2x + {\log _6}2x \ge 1 + {\log _4}2x.{\log _6}2x\)

\( \Leftrightarrow {\log _4}2x - 1 + {\log _6}2x - {\log _4}2x.{\log _6}2x \ge 0\)

\( \Leftrightarrow {\log _4}2x - 1 + {\log _6}2x\left( {1 - {{\log }_4}2x} \right) \ge 0\)

\(\begin{array}{l} \Leftrightarrow \left( {{{\log }_4}2x - 1} \right)\left( {{{\log }_6}2x - 1} \right) \le 0\\ \Leftrightarrow {\log _4}\frac{{2x}}{4} \cdot {\log _6}\frac{{2x}}{6} \le 0\end{array}\)

\( \Leftrightarrow {\log _4}\frac{x}{2} \cdot {\log _6}\frac{x}{3} \le 0\) \( \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}{\log _4}\frac{x}{2} \ge 0\\{\log _6}\frac{x}{3} \le 0\end{array} \right.\\\left\{ \begin{array}{l}\log { _4}\frac{x}{2} \le 0\\\log { _6}\frac{x}{3} \ge 0\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}\frac{x}{2} \ge 1\\\frac{x}{3} \le 1\end{array} \right.\\\left\{ \begin{array}{l}\frac{x}{2} \le 1\\\frac{x}{3} \ge 1\end{array} \right.\end{array} \right.\)\( \Leftrightarrow \left[ \begin{array}{l}2 \le x \le 3\\\left\{ \begin{array}{l}x \le 2\\x \ge 3\end{array} \right.\,\end{array} \right.\)\( \Leftrightarrow 2 \le x \le 3\).

Vậy có 2 nghiệm nguyên của bất phương trình. Chọn A.