Đáp án:
\( \to x \in \left[ {2;\left. {2 + \sqrt 2 } \right]} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
ĐK:x \ne 0\\
\left\{ \begin{array}{l}
\frac{{{x^2} - 3x + 2 - x}}{x} \le 0\left( 1 \right)\\
\frac{{\left( {x - 2} \right)(x - 1)}}{x} > 0\left( 2 \right)
\end{array} \right.
\end{array}\)
Xét: (1)
BXD:
x -∞ 0 2-√2 2+√2 +∞
f(x) - // + 0 - 0 +
\( \to x \in \left( { - \infty ;0} \right) \cup \left[ {2 - \sqrt 2 ;2 + \sqrt 2 } \right]\)
Xét (2)
BXD:
x -∞ 0 1 2 +∞
f(x) - // + 0 - 0 +
\( \to x \in (0;\left. 1 \right] \cup \left[ {2; + \infty )} \right.\)
KL: \( \to x \in \left[ {2;\left. {2 + \sqrt 2 } \right]} \right.\)
