Đáp án:
$\begin{array}{l}
a)\left( {x - 3} \right)\left( {\sqrt 2 - x} \right) > 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 3 > 0\\
\sqrt 2 - x > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 3 < 0\\
\sqrt 2 - x < 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 3\\
x < \sqrt 2
\end{array} \right.\left( {ktm} \right)\\
\left\{ \begin{array}{l}
x < 3\\
x > \sqrt 2
\end{array} \right.
\end{array} \right.\\
Vậy\,\sqrt 2 < x < 3\\
b)\left( {x + 2} \right)\left( {{x^2} - 4} \right) \le 0\\
\Leftrightarrow \left( {x + 2} \right)\left( {x + 2} \right)\left( {x - 2} \right) \le 0\\
\Leftrightarrow {\left( {x + 2} \right)^2}\left( {x - 2} \right) \le 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 \le 0\\
x + 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x \le 2\\
x = - 2
\end{array} \right.\\
\Leftrightarrow x \le 2\\
Vậy\,x \le 2\\
c){x^2} + 4x + 3 \le 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {x + 3} \right) \le 0\\
\Leftrightarrow - 3 \le x \le - 1\\
Vậy\, - 3 \le x \le - 1\\
d)\left( {1 - 3x} \right)\left( { - 6{x^2} + 5x + 1} \right) \ge 0\\
\Leftrightarrow \left( {3x - 1} \right)\left( {6{x^2} - 5x - 1} \right) \ge 0\\
\Leftrightarrow \left( {3x - 1} \right)\left( {6x + 1} \right)\left( {x - 1} \right) \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge 1\\
- \dfrac{1}{6} \le x \le \dfrac{1}{3}
\end{array} \right.\\
Vậy\, - \dfrac{1}{6} \le x \le \dfrac{1}{3}\,hoặc\,x \ge 1\\
e)9{x^2} - 4x \le 0\\
\Leftrightarrow x\left( {9x - 4} \right) \le 0\\
\Leftrightarrow 0 \le x \le \dfrac{4}{9}\\
Vậy\,0 \le x \le \dfrac{4}{9}\\
f) - 6{x^2} + x + 1 \ge 0\\
\Leftrightarrow 6{x^2} - x - 1 \ge 0\\
\Leftrightarrow \left( {2x - 1} \right)\left( {3x + 1} \right) \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge \dfrac{1}{2}\\
x \le - \dfrac{1}{3}
\end{array} \right.\\
Vậy\,x \le - \dfrac{1}{3}\,hoặc\,x \ge \dfrac{1}{2}
\end{array}$
