Đáp án:

a. \(\dfrac{3}{2} \ge x\)

Giải thích các bước giải:

\(\begin{array}{l}
a.DK: - 2x + 3 \ge 0\\
 \to \dfrac{3}{2} \ge x\\
b.DK: - \dfrac{5}{{{x^2} + 6}} \ge 0\\
 \to \dfrac{5}{{{x^2} + 6}} \le 0\left( {vô lý} \right)\\
 \to x \in \emptyset \\
c.DK:4{x^2} - 12x + 9 \ne 0\\
 \to {\left( {2x - 3} \right)^2} \ne 0\\
 \to 2x - 3 \ne 0\\
 \to x \ne \dfrac{3}{2}\\
d.DK: - 5x \ge 0\\
 \to x \le 0\\
e.DK:x - 1 > 0 \to x > 1\\
f.DK:{x^2} - 8x + 15 \ge 0\\
 \to \left( {x - 5} \right)\left( {x - 3} \right) \ge 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 5 \ge 0\\
x - 3 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 5 \le 0\\
x - 3 \le 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
x \ge 5\\
x \le 3
\end{array} \right.\\
g.DK:x \ge 0\\
h.DK:{x^2} \ne 0\\
 \to x \ne 0\\
k.DK:\left\{ \begin{array}{l}
x - 2 \ge 0\\
x - 5 \ne 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x \ge 2\\
x \ne 5
\end{array} \right.\\
l.DK:{x^2} + 1 \ge 0\left( {ld} \right)\forall x \in R\\
 \to DK:\forall x\\
m.DK:{x^2} - 2x + 1 \ge 0\\
 \to {\left( {x - 1} \right)^2} \ge 0\left( {ld} \right)\forall x \in R\\
 \to DK:\forall x\\
n.DK:\left\{ \begin{array}{l}
\dfrac{{x + 2}}{{5 - x}} \ge 0\\
5 - x \ne 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 2 \ge 0\\
5 - x > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 2 \le 0\\
5 - x < 0
\end{array} \right.
\end{array} \right.\\
x \ne 5
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge  - 2\\
x < 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le  - 2\\
x > 5
\end{array} \right.\left( l \right)
\end{array} \right.\\
x \ne 5
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x \ge  - 2\\
x < 5
\end{array} \right.\\
o.DK:x + 3 > 0 \to x >  - 3\\
p.DK: - {x^2} - 2x - 1 \ge 0\\
 \to {x^2} + 2x + 1 \le 0\\
 \to {\left( {x + 1} \right)^2} \le 0\\
 \Leftrightarrow x + 1 = 0\\
 \Leftrightarrow x = 1\\
q.DK:\left\{ \begin{array}{l}
\dfrac{{x - 1}}{{x + 2}} \ge 0\\
x + 2 \ne 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x \ne  - 2\\
\left[ \begin{array}{l}
x \ge 1\\
x <  - 2
\end{array} \right.
\end{array} \right.
\end{array}\)