Đáp án:
$\begin{array}{l}
a)x\left( {2x - 3} \right) \le - 3x\left( {x - 1} \right) - 1\\
\Rightarrow 2{x^2} - 3x + 3{x^2} - 3x + 1 \le 0\\
\Rightarrow 5{x^2} - 6x + 1 \le 0\\
\Rightarrow \left( {5x - 1} \right)\left( {x - 1} \right) \le 0\\
\Rightarrow \dfrac{1}{5} \le x \le 1\\
Vậy\,\dfrac{1}{5} \le x \le 1\\
b)\dfrac{1}{{2x - 1}} \ge \dfrac{4}{{x - 3}}\\
\Rightarrow \dfrac{{x - 3 - 4\left( {2x - 1} \right)}}{{\left( {2x - 1} \right)\left( {x - 3} \right)}} \ge 0\\
\Rightarrow \dfrac{{ - 7x + 1}}{{\left( {2x - 1} \right)\left( {x - 3} \right)}} \ge 0\\
\Rightarrow \dfrac{{7x - 1}}{{\left( {2x - 1} \right)\left( {x - 3} \right)}} \le 0\\
\Rightarrow \left[ \begin{array}{l}
x \le \dfrac{1}{7}\\
\dfrac{1}{2} \le x \le 3
\end{array} \right.\\
c)Dkxd:{x^2} - 2x - 3 \ge 0\\
\Rightarrow \left( {x - 3} \right)\left( {x + 1} \right) \ge 0\\
\Rightarrow \left[ \begin{array}{l}
x \ge 3\\
x \le - 1
\end{array} \right.\\
\sqrt {{x^2} - 2x - 3} > 2x - 3\\
\Rightarrow {x^2} - 2x - 3 > 4{x^2} - 12x + 9\\
\Rightarrow 3{x^2} - 10x + 12 < 0\left( {ktm} \right)
\end{array}$
Vậy bất phương trình vô nghiệm
$\begin{array}{l}
d) - x + 2 > 0\\
\Rightarrow x < 2\\
\left| {{x^2} + 3x + 2} \right| < - x + 2\\
\Rightarrow x - 2 < {x^2} + 3x + 2 < - x + 2\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} + 2x + 4 > 0\left( {tm} \right)\\
{x^2} + 4x < 0
\end{array} \right.\\
\Rightarrow x\left( {x + 4} \right) < 0\\
\Rightarrow - 4 < x < 0\\
Vậy\, - 4 < x < 0
\end{array}$