\(\begin{array}{l}
\frac{{2{x^2} - x - 1}}{{\left| {x + 1} \right| - 2x}} \le  - 2{x^2} + x + 1\\
 \Leftrightarrow \left[ \begin{array}{l}
\frac{{2{x^2} - x - 1}}{{x + 1 - 2x}} \le  - 2{x^2} + x + 1\\
\frac{{2{x^2} - x - 1}}{{ - \left( {x + 1} \right) - 2x}} \le  - 2{x^2} + x + 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2{x^2} + x \le 0\\
\frac{{3x\left( {2{x^2} - x - 1} \right)}}{{ - 3x - 1}} \ge 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
 - \frac{1}{2} \le x \le 0\\
 - \frac{1}{3} < x \le \frac{{ - 1}}{2} \vee 0 \le x \le 1
\end{array} \right.
\end{array}\)

Do \(x \in Z \Rightarrow x = 0,x = 1\)

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