`NHANH` `VS` `Ạ` `giải` `bất` `phương` `trình` `a)` `{3+x}/2` `-` `{2-x}/6` `<` `0` `b)` `{11x+1}/12` `>` `{3x+2}/6` `<` `{5x-2}/2` `+` `{1-x}/2` `c)`

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1. Đáp án:

   \(a.x <  – \frac{7}{4}\)

   Giải thích các bước giải:

   \(\begin{array}{l}
   a.\frac{{3\left( {3 + x} \right) – 2 + x}}{6} < 0\\
    \to 9 + 3x – 2 + x < 0\\
    \to 4x <  – 7\\
    \to x <  – \frac{7}{4}\\
   b.\left\{ \begin{array}{l}
   \frac{{11x + 1 – 2\left( {3x + 2} \right)}}{{12}} > 0\\
   \frac{{3x + 2}}{6} < \frac{{5x – 2 + 1 – x}}{2}
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   11x + 1 – 6x – 4 > 0\\
   \frac{{3x + 2}}{6} < \frac{{4x – 1}}{2}
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   5x > 3\\
   \frac{{3x + 2 – 3\left( {4x – 1} \right)}}{6} < 0
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   x > \frac{3}{5}\\
   3x + 2 – 12x + 3 < 0
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   x > \frac{3}{5}\\
    – 9x + 5 < 0
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   x > \frac{3}{5}\\
   x > \frac{5}{9}
   \end{array} \right.\\
   KL:x > \frac{3}{5}\\
   c.\frac{{2x – 7}}{{15}} – \frac{{3x + 2}}{6} < \frac{{5x – 2 + 1 – x}}{2}\\
    \to \frac{{2x – 7}}{{15}} – \frac{{3x + 2}}{6} < \frac{{4x – 1}}{2}\\
    \to \frac{{2\left( {2x – 7} \right) – 5\left( {3x + 2} \right) – 15\left( {4x – 1} \right)}}{{30}} < 0\\
    \to 4x – 14 – 15x – 10 – 60x + 15 < 0\\
    \to 71x >  – 9\\
    \to x >  – \frac{9}{{71}}
   \end{array}\)

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