Đáp án:
d. \(\left[ \begin{array}{l}
x > \dfrac{1}{2}\\
x < - \dfrac{2}{3}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left( {x - 1} \right)\left( {x - 2} \right) \le 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 1 \ge 0\\
x - 2 \le 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 1 \le 0\\
x - 2 \ge 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 1\\
x \le 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 1\\
x \ge 2
\end{array} \right.\left( l \right)
\end{array} \right.\\
b.\left( {x + 4} \right)\left( {x - 2} \right) \ge 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 4 \ge 0\\
x - 2 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 4 \le 0\\
x - 2 \le 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge - 4\\
x \ge 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le - 4\\
x \le 2
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
x \ge 2\\
x \le - 4
\end{array} \right.\\
c.\left( {2x - 1} \right)\left( {x - 2} \right) < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2x - 1 > 0\\
x - 2 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
2x - 1 < 0\\
x - 2 > 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > \dfrac{1}{2}\\
x < 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x < \dfrac{1}{2}\\
x > 2
\end{array} \right.\left( l \right)
\end{array} \right.\\
d.\left( {3x + 2} \right)\left( {2x - 1} \right) > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
3x + 2 > 0\\
2x - 1 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
3x + 2 < 0\\
2x - 1 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > - \dfrac{2}{3}\\
x > \dfrac{1}{2}
\end{array} \right.\\
\left\{ \begin{array}{l}
x < - \dfrac{2}{3}\\
x < \dfrac{1}{2}
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x > \dfrac{1}{2}\\
x < - \dfrac{2}{3}
\end{array} \right.
\end{array}\)
