Bài 1: a)|2x-1/3|=2/3 c)1/2-|1/3-x|=1/4 b)|1/2x+1/3|=3/2 d)1/2.|2x-1/4|=1/3 Bài 2: a)|x+1/4|=0

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1. Giải thích các bước giải:

   \(\begin{array}{l}
   1,\\
   a,\\
   \left| {2x – \dfrac{1}{3}} \right| = \dfrac{2}{3}\\
    \Leftrightarrow \left[ \begin{array}{l}
   2x – \dfrac{1}{3} = \dfrac{2}{3}\\
   2x – \dfrac{1}{3} =  – \dfrac{2}{3}
   \end{array} \right.\\
    \Leftrightarrow \left[ \begin{array}{l}
   2x = 1\\
   2x =  – \dfrac{1}{3}
   \end{array} \right.\\
    \Leftrightarrow \left[ \begin{array}{l}
   x = \dfrac{1}{2}\\
   x =  – \dfrac{1}{6}
   \end{array} \right.\\
   b,\\
   \left| {\dfrac{1}{2}x + \dfrac{1}{3}} \right| = \dfrac{3}{2}\\
    \Leftrightarrow \left[ \begin{array}{l}
   \dfrac{1}{2}x + \dfrac{1}{3} = \dfrac{3}{2}\\
   \dfrac{1}{2}x + \dfrac{1}{3} =  – \dfrac{3}{2}
   \end{array} \right.\\
    \Leftrightarrow \left[ \begin{array}{l}
   \dfrac{1}{2}x = \dfrac{7}{6}\\
   \dfrac{1}{2}x =  – \dfrac{{11}}{6}
   \end{array} \right.\\
    \Leftrightarrow \left[ \begin{array}{l}
   x = \dfrac{7}{3}\\
   x =  – \dfrac{{11}}{3}
   \end{array} \right.\\
   c,\\
   \dfrac{1}{2} – \left| {\dfrac{1}{3} – x} \right| = \dfrac{1}{4}\\
    \Leftrightarrow \left| {\dfrac{1}{3} – x} \right| = \dfrac{1}{2} – \dfrac{1}{4}\\
    \Leftrightarrow \left| {\dfrac{1}{3} – x} \right| = \dfrac{1}{4}\\
    \Leftrightarrow \left[ \begin{array}{l}
   \dfrac{1}{3} – x = \dfrac{1}{4}\\
   \dfrac{1}{3} – x =  – \dfrac{1}{4}
   \end{array} \right.\\
    \Leftrightarrow \left[ \begin{array}{l}
   x = \dfrac{1}{{12}}\\
   x = \dfrac{7}{{12}}
   \end{array} \right.\\
   d,\\
   \dfrac{1}{2}.\left| {2x – \dfrac{1}{4}} \right| = \dfrac{1}{3}\\
    \Leftrightarrow \left| {2x – \dfrac{1}{4}} \right| = \dfrac{2}{3}\\
    \Leftrightarrow \left[ \begin{array}{l}
   2x – \dfrac{1}{4} = \dfrac{2}{3}\\
   2x – \dfrac{1}{4} =  – \dfrac{2}{3}
   \end{array} \right.\\
    \Leftrightarrow \left[ \begin{array}{l}
   2x = \dfrac{{11}}{{12}}\\
   2x =  – \dfrac{5}{{12}}
   \end{array} \right.\\
    \Leftrightarrow \left[ \begin{array}{l}
   x = \dfrac{{11}}{{24}}\\
   x =  – \dfrac{5}{{24}}
   \end{array} \right.\\
   2,\\
   a,\\
   \left| {x + \dfrac{1}{4}} \right| = 0\\
    \Leftrightarrow x + \dfrac{1}{4} = 0\\
    \Leftrightarrow x =  – \dfrac{1}{4}\\
   b,\\
   \left| {x – \dfrac{2}{5}} \right| \ge 0,\,\,\,\forall x \Rightarrow \left| {x – \dfrac{2}{5}} \right| >  – \dfrac{1}{2}
   \end{array}\) 

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