Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne 1;x \ne 2\\
\frac{{\left| {2x - 1} \right|}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} \le \frac{1}{2}\\
\to \left[ \begin{array}{l}
\frac{{2x - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} \le \frac{1}{2}\\
\frac{{2x - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} \ge - \frac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
\frac{{4x - 2 - {x^2} + 3x - 2}}{{2\left( {x - 1} \right)\left( {x - 2} \right)}} \le 0\\
\frac{{4x - 2 + {x^2} - 3x + 2}}{{2\left( {x - 1} \right)\left( {x - 2} \right)}} \ge 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\frac{{ - {x^2} + 7x - 4}}{{2\left( {x - 1} \right)\left( {x - 2} \right)}} \le 0\left( 1 \right)\\
\frac{{{x^2} + x}}{{2\left( {x - 1} \right)\left( {x - 2} \right)}} \ge 0\left( 2 \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x \in ( - \infty ;\left. {\frac{{7 - \sqrt {33} }}{2}} \right] \cup \left( {1;2} \right) \cup \left[ {\frac{{7 + \sqrt {33} }}{2}; + \infty )} \right.\\
x \in ( - \infty ; - \left. 1 \right] \cup \left[ {0;1) \cup \left( {2; + \infty } \right)} \right.
\end{array} \right.
\end{array}\)
