Cho -x+3 ______ x-2 Tìm tích để A>=1 A<1/5 A>1/6

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1. Đáp án:

   c. \(\left\{ \begin{array}{l}
   x < \dfrac{{20}}{7}\\
   x > 2
   \end{array} \right.\)

   Giải thích các bước giải:

   \(\begin{array}{l}
   DK:x \ne 2\\
   A = \dfrac{{ – x + 3}}{{x – 2}}\\
   A \ge 1\\
    \to \dfrac{{ – x + 3}}{{x – 2}} \ge 1\\
    \to \dfrac{{ – x + 3 – x + 2}}{{x – 2}} \ge 0\\
    \to \dfrac{{ – 2x + 5}}{{x – 2}} \ge 0\\
    \to \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
    – 2x + 5 \ge 0\\
   x – 2 > 0
   \end{array} \right.\\
   \left\{ \begin{array}{l}
    – 2x + 5 \le 0\\
   x – 4 < 0
   \end{array} \right.
   \end{array} \right. \to \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   x \le  – \dfrac{5}{2}\\
   x > 2
   \end{array} \right.\left( l \right)\\
   \left\{ \begin{array}{l}
   x \ge  – \dfrac{5}{2}\\
   x < 4
   \end{array} \right.
   \end{array} \right.\\
    \to  – \dfrac{5}{2} \le x < 4\\
   A < \dfrac{1}{5}\\
    \to \dfrac{{ – x + 3}}{{x – 2}} < \dfrac{1}{5}\\
    \to \dfrac{{ – 5x + 15 – x + 2}}{{5\left( {x – 2} \right)}} < 0\\
    \to \dfrac{{ – 6x + 17}}{{5\left( {x – 2} \right)}} < 0\\
    \to \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
    – 6x + 17 > 0\\
   x – 2 < 0
   \end{array} \right.\\
   \left\{ \begin{array}{l}
    – 6x + 17 < 0\\
   x – 2 > 0
   \end{array} \right.
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   x < \dfrac{{17}}{6}\\
   x < 2
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x > \dfrac{{17}}{6}\\
   x > 2
   \end{array} \right.
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   x > \dfrac{{17}}{6}\\
   x < 2
   \end{array} \right.\\
   A > \dfrac{1}{6}\\
    \to \dfrac{{ – x + 3}}{{x – 2}} > \dfrac{1}{6}\\
    \to \dfrac{{ – 6x + 18 – x + 2}}{{6\left( {x – 2} \right)}} > 0\\
    \to \dfrac{{ – 7x + 20}}{{6\left( {x – 2} \right)}} > 0\\
    \to \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
    – 7x + 20 > 0\\
   x – 2 > 0
   \end{array} \right.\\
   \left\{ \begin{array}{l}
    – 7x + 20 < 0\\
   x – 2 < 0
   \end{array} \right.
   \end{array} \right. \to \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   x < \dfrac{{20}}{7}\\
   x > 2
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x > \dfrac{{20}}{7}\\
   x < 2
   \end{array} \right.\left( l \right)
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   x < \dfrac{{20}}{7}\\
   x > 2
   \end{array} \right.
   \end{array}\)

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