Đáp án:

\[S = \left( { - \infty ;2} \right]\]

Giải thích các bước giải:

 ĐKXĐ: \(\left\{ \begin{array}{l}
x \le 2\\
x \ne 0
\end{array} \right.\)

Ta có:

\(\begin{array}{l}
\frac{{\sqrt {2 - x}  + 4x - 3}}{x} \ge 2\\
 \Leftrightarrow \frac{{\sqrt {2 - x}  + 4x - 3}}{x} - 2 \ge 0\\
 \Leftrightarrow \frac{{\sqrt {2 - x}  + 4x - 3 - 2x}}{x} \ge 0\\
 \Leftrightarrow \frac{{\sqrt {2 - x}  + 2x - 3}}{x} \ge 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\
TH1:\,\,\,\,0 < x \le 2\\
\left( 1 \right) \Leftrightarrow \sqrt {2 - x}  + 2x - 3 \ge 0\\
 \Leftrightarrow \sqrt {2 - x}  \ge 3 - 2x\\
 \Leftrightarrow \left[ \begin{array}{l}
3 - 2x \le 0\\
\left\{ \begin{array}{l}
3 - 2x \ge 0\\
2 - x \ge {\left( {3 - 2x} \right)^2}
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \ge \frac{3}{2}\\
\left\{ \begin{array}{l}
x \le \frac{3}{2}\\
2 - x \ge 4{x^2} - 12x + 9
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x \ge \frac{3}{2}\\
\left\{ \begin{array}{l}
x \le \frac{3}{2}\\
4{x^2} - 11x + 7 \le 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \ge \frac{3}{2}\\
\left\{ \begin{array}{l}
x \le \frac{3}{2}\\
1 \le x \le \frac{7}{4}
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \ge \frac{3}{2}\\
1 \le x \le \frac{3}{2}
\end{array} \right. \Leftrightarrow x \ge 1\\
 \Rightarrow {S_1} = \left[ {1;2} \right]\\
TH2:\,\,\,\,x < 0\\
\left( 1 \right) \Leftrightarrow \sqrt {2 - x}  + 2x - 3 \le 0\\
 \Leftrightarrow \sqrt {2 - x}  \le 3 - 2x\\
 \Leftrightarrow \left\{ \begin{array}{l}
3 - 2x \ge 0\\
2 - x \le {\left( {3 - 2x} \right)^2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le \frac{3}{2}\\
2 - x \le 4{x^2} - 12x + 9
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \le \frac{3}{2}\\
4{x^2} - 11x + 7 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le \frac{3}{2}\\
\left[ \begin{array}{l}
x \ge \frac{7}{4}\\
x \le 1
\end{array} \right.
\end{array} \right. \Leftrightarrow x \le 1\\
 \Rightarrow {S_2} = \left( { - \infty ;1} \right]\\
 \Rightarrow S = {S_1} \cup {S_2} = \left( { - \infty ;2} \right]
\end{array}\)

Vậy tập nghiệm của bất phương trình đã cho là \(S = \left( { - \infty ;2} \right]\)