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

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.