Đáp án:
\[B\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\dfrac{1}{x} \le 1\,\,\,\,\,\left( {x \ne 0} \right)\\
\Leftrightarrow \dfrac{1}{x} - 1 \le 0 \Leftrightarrow \dfrac{{1 - x}}{x} \le 0 \Leftrightarrow \left\{ \begin{array}{l}
x\left( {x - 1} \right) \ge 0\\
x \ne 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \ge 1\\
x < 0
\end{array} \right.\\
b,\\
2x - 3 - \dfrac{1}{x} < x - 4 - \dfrac{1}{x}\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ne 0\\
2x - 3 < x - 4
\end{array} \right. \Leftrightarrow 2x - 3 < x - 4\\
c,\\
\sqrt {1 - x} \le x\\
\Leftrightarrow \left\{ \begin{array}{l}
1 - x \ge 0\\
x \ge 0\\
1 - x \le {x^2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 1\\
x \ge 0\\
1 - x \le {x^2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
0 \le x \le 1\\
1 - x \le {x^2}
\end{array} \right.\\
d,\\
{x^2} \ge x \Leftrightarrow {x^2} - x \ge 0 \Leftrightarrow x\left( {x - 1} \right) \ge 0 \Leftrightarrow \left[ \begin{array}{l}
x \ge 1\\
x \le 0
\end{array} \right.
\end{array}\)
Vậy đáp án đúng là \(B\)
