Tìm x 1) | x + 3/5 | – | x – 7/3 | = 0 2) |x – 2 |< 3 3) | 5/3 x | = |-1/6| 4) | 1/2 x | = 3x - 2 5) | x - 1 | = 3x + 2 6) |5x| = x - 12 7) | 7

0 bình luận về “Tìm x 1) | x + 3/5 | – | x – 7/3 | = 0 2) |x – 2 |< 3 3) | 5/3 x | = |-1/6| 4) | 1/2 x | = 3x - 2 5) | x - 1 | = 3x + 2 6) |5x| = x - 12 7) | 7”

1. Giải thích các bước giải:

   Ta có:

   \(\begin{array}{l}
   1,\\
   \left| {x + \dfrac{3}{5}} \right| – \left| {x – \dfrac{7}{3}} \right| = 0\\
    \Leftrightarrow \left| {x + \dfrac{3}{5}} \right| = \left| {x – \dfrac{7}{3}} \right|\\
    \Leftrightarrow \left[ \begin{array}{l}
   x + \dfrac{3}{5} = x – \dfrac{7}{3}\\
   x + \dfrac{3}{5} = \dfrac{7}{3} – x
   \end{array} \right. \Leftrightarrow x = \dfrac{{13}}{{15}}\\
   2,\\
   \left| {x – 2} \right| < 3\\
    \Leftrightarrow  – 3 < x – 2 < 3\\
    \Leftrightarrow  – 1 < x < 5\\
   3,\\
   \left| {\dfrac{5}{3}x} \right| = \left| { – \dfrac{1}{6}} \right|\\
    \Leftrightarrow \left[ \begin{array}{l}
   \dfrac{5}{3}x =  – \dfrac{1}{6}\\
   \dfrac{5}{3}x = \dfrac{1}{6}
   \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
   x =  – \dfrac{1}{{10}}\\
   x = \dfrac{1}{{10}}
   \end{array} \right.\\
   4,\\
   \left| {\dfrac{1}{2}x} \right| = 3x – 2\\
    \Leftrightarrow \left\{ \begin{array}{l}
   3x – 2 \ge 0\\
   \left[ \begin{array}{l}
   \dfrac{1}{2}x = 3x – 2\\
   \dfrac{1}{2}x =  – 3x + 2
   \end{array} \right.
   \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
   x \ge \dfrac{2}{3}\\
   \left[ \begin{array}{l}
   x = \dfrac{4}{5}\\
   x = \dfrac{4}{7}
   \end{array} \right.
   \end{array} \right. \Leftrightarrow x = \dfrac{4}{5}\\
   5,\\
   \left| {x – 1} \right| = 3x + 2\\
    \Leftrightarrow \left\{ \begin{array}{l}
   3x + 2 \ge 0\\
   \left[ \begin{array}{l}
   x – 1 = 3x + 2\\
   x – 1 =  – 3x – 2
   \end{array} \right.
   \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
   x \ge  – \dfrac{2}{3}\\
   \left[ \begin{array}{l}
   x =  – \dfrac{3}{2}\\
   x =  – \dfrac{1}{4}
   \end{array} \right.
   \end{array} \right. \Leftrightarrow x =  – \dfrac{1}{4}\\
   6,\\
   \left| {5x} \right| = x – 12\\
    \Leftrightarrow \left\{ \begin{array}{l}
   x – 12 \ge 0\\
   \left[ \begin{array}{l}
   5x = x – 12\\
   5x =  – x + 12
   \end{array} \right.
   \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
   x \ge 12\\
   \left[ \begin{array}{l}
   x =  – 3\\
   x = 2
   \end{array} \right.
   \end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {vn} \right)\\
   7,\\
   \left| {7 – x} \right| = 5x + 1\\
    \Leftrightarrow \left\{ \begin{array}{l}
   5x + 1 \ge 0\\
   \left[ \begin{array}{l}
   7 – x = 5x + 1\\
   7 – x =  – 5x – 1
   \end{array} \right.
   \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
   x \ge  – \dfrac{1}{5}\\
   \left[ \begin{array}{l}
   x = 1\\
   x =  – 2
   \end{array} \right.
   \end{array} \right. \Leftrightarrow x = 1\\
   8,\\
   \left| {\dfrac{3}{4}x – \dfrac{3}{4}} \right| – \dfrac{3}{4} = \left| { – \dfrac{3}{4}} \right|\\
    \Leftrightarrow \left| {\dfrac{3}{4}x – \dfrac{3}{4}} \right| – \dfrac{3}{4} = \dfrac{3}{4}\\
    \Leftrightarrow \left| {\dfrac{3}{4}x – \dfrac{3}{4}} \right| = \dfrac{3}{2}\\
    \Leftrightarrow \left[ \begin{array}{l}
   \dfrac{3}{4}x – \dfrac{3}{4} = \dfrac{3}{2}\\
   \dfrac{3}{4}x – \dfrac{3}{4} =  – \dfrac{3}{2}
   \end{array} \right.\\
    \Leftrightarrow \left[ \begin{array}{l}
   x = 3\\
   x =  – 1
   \end{array} \right.
   \end{array}\)

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