GIẢI BẤT PHƯƠNG TRÌNH SAU: $\frac{\sqrt[]{2-x}+4x-3}{x}$ $\geq$ 2

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1. 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 \(S = \left( { – \infty ;2} \right]\)

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