Tìm x nguyên dương ( x ∈ Z ) thoả mãn: $\frac{-7}{3}$ < |$\frac{2}{7}$ - x| - $\frac{5}{2}$ < $\frac{-7}{4}$

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1. Đáp án:

   $\begin{array}{l}
    – \dfrac{7}{3} < \left| {\dfrac{2}{7} – x} \right| – \dfrac{5}{2} < \dfrac{{ – 7}}{4}\\
    \Rightarrow \dfrac{{ – 7}}{3} + \dfrac{5}{2} < \left| {\dfrac{2}{7} – x} \right| < \dfrac{{ – 7}}{4} + \dfrac{5}{2}\\
    \Rightarrow \dfrac{{ – 14 + 15}}{6} < \left| {\dfrac{2}{7} – x} \right| < \dfrac{{ – 7 + 10}}{4}\\
    \Rightarrow \dfrac{1}{6} < \left| {\dfrac{2}{7} – x} \right| < \dfrac{3}{4}\\
    \Rightarrow \left[ \begin{array}{l}
   \dfrac{{ – 3}}{4} < x – \dfrac{2}{7} < \dfrac{{ – 1}}{6}\\
   \dfrac{1}{6} < \dfrac{2}{7} – x < \dfrac{3}{4}
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
   \dfrac{{ – 3}}{4} + \dfrac{2}{7} < x < \dfrac{{ – 1}}{6} + \dfrac{2}{7}\\
   \dfrac{1}{6} – \dfrac{2}{7} <  – x < \dfrac{3}{4} – \dfrac{2}{7}
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
   \dfrac{{ – 13}}{{28}} < x < \dfrac{5}{{42}}\\
   \dfrac{{ – 5}}{{42}} <  – x < \dfrac{{13}}{{28}}
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
    – \dfrac{{13}}{{28}} < x < \dfrac{5}{{42}}\\
   \dfrac{{ – 13}}{{28}} < x < \dfrac{5}{{42}}
   \end{array} \right.\\
    \Rightarrow \dfrac{{ – 13}}{{28}} < x < \dfrac{5}{{42}}\\
   Do:x \in N*
   \end{array}$

   => ko có giá trị của x thỏa mãn.

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