Đáp án:

\(\begin{array}{l}
1)\dfrac{7}{3} \ge x\\
2)x \ne \dfrac{3}{2}\\
3)4 > x \ge 2\\
4)\left[ \begin{array}{l}
x >  - \dfrac{2}{5}\\
x \le  - 1
\end{array} \right.\\
5)x > 2\\
6)3 \ge x >  - 1\\
7)3 \ge x \ge  - 2\\
8)\left[ \begin{array}{l}
x \ge 6\\
x \le 0
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
1)DK: - 3x + 7 \ge 0\\
 \to \dfrac{7}{3} \ge x\\
2)DK:9 - 12x + 4{x^2} > 0\\
 \to {\left( {3 - 2x} \right)^2} > 0\\
 \to x \ne \dfrac{3}{2}\\
3)DK:\left\{ \begin{array}{l}
\dfrac{{x + 2}}{{4 - x}} \ge 0\\
x \ne 4
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 2 \ge 0\\
4 - x > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 2 \le 0\\
4 - x < 0
\end{array} \right.
\end{array} \right.\\
x \ne 4
\end{array} \right. \to \left\{ {\left[ \begin{array}{l}
4 > x \ge 2\\
\left\{ \begin{array}{l}
x \le  - 2\\
4 < x
\end{array} \right.\left( l \right)
\end{array} \right.} \right.\\
 \to 4 > x \ge 2\\
4)DK:\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 \ge 0\\
5x + 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 \le 0\\
5x + 2 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
x >  - \dfrac{2}{5}\\
x \le  - 1
\end{array} \right.\\
5)DK:\left\{ \begin{array}{l}
\left( {x - 2} \right)\left( {x + 2} \right) \ge 0\\
x - 2 > 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x > 2\\
x + 2 \ge 0
\end{array} \right.\\
 \to x > 2\\
6)DK:\left\{ \begin{array}{l}
5 - x \ge 0\\
x + 1 > 0\\
9 - {x^2} \ge 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
5 \ge x\\
x >  - 1\\
3 \ge x >  - 3
\end{array} \right.\\
 \to 3 \ge x >  - 1\\
7)DK:\left( {3 - x} \right)\left( {x + 2} \right) \ge 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
3 - x \ge 0\\
x + 2 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
3 - x \le 0\\
x + 2 \le 0
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
3 \ge x \ge  - 2\\
\left\{ \begin{array}{l}
3 \le x\\
x \le  - 2
\end{array} \right.\left( l \right)
\end{array} \right.\\
8)DK:x\left( {x - 6} \right) \ge 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 0\\
x - 6 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 0\\
x - 6 \le 0
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x \ge 6\\
x \le 0
\end{array} \right.
\end{array}\)