Đáp án:
$\begin{array}{l}
{3.3^{2x}} - {28.3^x} + 9 \le 0\\
\Leftrightarrow 3.{\left( {{3^x}} \right)^2} - {28.3^x} + 9 \le 0\\
\Leftrightarrow 3.{\left( {{3^x}} \right)^2} - {27.3^x} - {3^x} + 9 \le 0\\
\Leftrightarrow \left( {{3^x} - 9} \right){.3.3^x} - \left( {{3^x} - 9} \right) \le 0\\
\Leftrightarrow \left( {{3^x} - 9} \right)\left( {{{3.3}^x} - 1} \right) \le 0\\
\Leftrightarrow \dfrac{1}{3} \le {3^x} \le 9\\
\Leftrightarrow - 1 \le x \le 2
\end{array}$
