Đáp án:
\(\left\{ \begin{array}{l}
x > \frac{1}{2}\\
x \ne 1
\end{array} \right.\)
Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
2x - 1 \ne 0\\
x - 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne \frac{1}{2}\\
x \ne 1
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
\frac{{ - x + 1}}{{\left( {2x - 1} \right)\left( {x - 1} \right)}} \le 0\\
\Leftrightarrow \frac{{ - \left( {x - 1} \right)}}{{\left( {2x - 1} \right)\left( {x - 1} \right)}} \le 0\\
\Leftrightarrow \frac{{ - 1}}{{2x - 1}} \le 0\\
\Leftrightarrow 2x - 1 > 0\\
\Leftrightarrow x > \frac{1}{2}
\end{array}\)
Vậy \(\left\{ \begin{array}{l}
x > \frac{1}{2}\\
x \ne 1
\end{array} \right.\)
