a)|x+2|-3=-7/5 b) |8-3x|=1>0 c) |3x+1|=5 d) |x+4|=2x -6 e) |2x+5|=8 f) | x+2|=6-2x g) | 2x-3|+1=5x

0 bình luận về “a)|x+2|-3=-7/5 b) |8-3x|=1>0 c) |3x+1|=5 d) |x+4|=2x -6 e) |2x+5|=8 f) | x+2|=6-2x g) | 2x-3|+1=5x”

1. Đáp án:

   e) \(x = \dfrac{4}{7}\)

   Giải thích các bước giải:

   \(\begin{array}{l}
   a)\left| {x + 2} \right| = \dfrac{8}{5}\\
    \to \left[ \begin{array}{l}
   x + 2 = \dfrac{8}{5}\\
   x + 2 =  – \dfrac{8}{5}
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   x =  – \dfrac{2}{5}\\
   x =  – 6
   \end{array} \right.\\
   b)\left| {8 – 3x} \right| = 1\\
    \to \left[ \begin{array}{l}
   8 – 3x = 1\left( {DK:\dfrac{8}{3} \ge x} \right)\\
   8 – 3x =  – 1\left( {DK:\dfrac{8}{3} < x} \right)
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   x = \dfrac{7}{3}\\
   x = 3
   \end{array} \right.\left( {TM} \right)\\
   c)\left| {3x + 1} \right| = 5\\
    \to \left[ \begin{array}{l}
   3x + 1 = 5\\
   3x + 1 =  – 5
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   x = \dfrac{4}{3}\\
   x =  – 2
   \end{array} \right.\\
   d)\left| {x + 4} \right| = 2x – 6\\
    \to \left[ \begin{array}{l}
   x + 4 = 2x – 6\\
   x + 4 =  – 2x + 6
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   x = 10\\
   3x = 2
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   x = 10\\
   x = \dfrac{2}{3}
   \end{array} \right.\\
   e)\left| {2x + 5} \right| = 8\\
    \to \left[ \begin{array}{l}
   2x + 5 = 8\\
   2x + 5 =  – 8
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   x = \dfrac{3}{2}\\
   x =  – \dfrac{{13}}{2}
   \end{array} \right.\\
   f)\left| {x + 2} \right| = 6 – 2x\\
    \to \left[ \begin{array}{l}
   x + 2 = 6 – 2x\left( {DK:x \ge  – 2} \right)\\
   x + 2 =  – 6 + 2x\left( {DK:x <  – 2} \right)
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   3x = 4\\
   x = 8\left( l \right)
   \end{array} \right.\\
    \to x = \dfrac{4}{3}\\
   g)\left| {2x – 3} \right| = 5x – 1\\
    \to \left[ \begin{array}{l}
   2x – 3 = 5x – 1\left( {DK:x \ge \dfrac{3}{2}} \right)\\
   2x – 3 =  – 5x + 1\left( {DK:x < \dfrac{3}{2}} \right)
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   3x =  – 2\\
   7x = 4
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   x =  – \dfrac{2}{3}\left( l \right)\\
   x = \dfrac{4}{7}
   \end{array} \right.
   \end{array}\)

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