Đáp án:
$\begin{array}{l}
c)Dkxd: - {x^2} + 3x - 2 \ge 0\\
\Leftrightarrow {x^2} - 3x + 2 \le 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {x - 2} \right) \le 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 1 \le 0\\
x - 2 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 1 \ge 0\\
x - 2 \le 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \le 1\\
x \ge 2
\end{array} \right.\left( {ktm} \right)\\
\left\{ \begin{array}{l}
x \ge 1\\
x \le 2
\end{array} \right.
\end{array} \right.\\
Vậy\,1 \le x \le 2\\
d)Dkxd:\dfrac{{ - 3}}{{4{x^2} - 5x + 1}} \ge 0\\
\Leftrightarrow 4{x^2} - 5x + 1 < 0\\
\Leftrightarrow \left( {4x - 1} \right)\left( {x - 1} \right) < 0\\
\Leftrightarrow \dfrac{1}{4} < x < 1\\
Vậy\,\dfrac{1}{4} < x < 1
\end{array}$
