\(\begin{array}{l}
{\log _{\frac{1}{3}}}({x^2} - 2x) \le  - 1\\
dk:{x^2} - 2x > 0 \Leftrightarrow x(x - 2) > 0 \Leftrightarrow \left[ \begin{array}{l}
x > 2\\
x < 0
\end{array} \right.\\
bpt \Leftrightarrow {\log _{\frac{1}{3}}}({x^2} - 2x) \le {\log _{\frac{1}{3}}}3\\
 \Leftrightarrow {x^2} - 2x \ge 3\\
 \Leftrightarrow {x^2} - 2x - 3 \ge 0\\
 \Leftrightarrow {x^2} + x - 3x - 3 \ge 0\\
 \Leftrightarrow x(x + 1) - 3(x + 1) \ge 0\\
 \Leftrightarrow (x + 1)(x - 3) \ge 0\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
x + 1 \ge 0\\
x - 3 \ge 0
\end{array} \right.}\\
{\left\{ \begin{array}{l}
x + 1 \le 0\\
x - 3 \le 0
\end{array} \right.}
\end{array}} \right.\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
x \ge  - 1\\
x \ge 3
\end{array} \right.}\\
{\left\{ \begin{array}{l}
x \le  - 1\\
x \le 3
\end{array} \right.}
\end{array}} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x \ge 3\\
x \le  - 1
\end{array} \right.
\end{array}\)

Kết hợp với điều kiện xác định ta được: \(\left[ \begin{array}{l}
x \ge 3\\
x \le  - 1
\end{array} \right.\)