Đáp án:
\(\begin{array}{l}
a)A \in \left( { - \dfrac{5}{2};\dfrac{7}{2}} \right)\\
b)B \in \left( { - 2;0} \right] \cup \left[ {2;4} \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{3}{{\left| {2x - 1} \right|}} > \dfrac{1}{2}\\
\to \dfrac{{6 - \left| {2x - 1} \right|}}{{2\left| {2x - 1} \right|}} > 0\\
\to 6 - \left| {2x - 1} \right| > 0\left( {do:2\left| {2x - 1} \right| > 0\forall x \ne \dfrac{1}{2}} \right)\\
\to 6 > \left| {2x - 1} \right|\\
\to \left\{ \begin{array}{l}
6 > 2x - 1\\
2x - 1 > - 6
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{7}{2} > x\\
x > - \dfrac{5}{2}
\end{array} \right.\\
\to \dfrac{7}{2} > x > - \dfrac{5}{2}\\
\to A \in \left( { - \dfrac{5}{2};\dfrac{7}{2}} \right)\\
b)1 \le \left| {1 - x} \right| < 3\\
\to \left\{ \begin{array}{l}
1 \le \left| {1 - x} \right|\\
\left| {1 - x} \right| < 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
1 \le 1 - x\\
- 1 \ge 1 - x
\end{array} \right.\\
- 3 < 1 - x < 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \le 0\\
x \ge 2
\end{array} \right.\\
- 4 < - x < 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \le 0\\
x \ge 2
\end{array} \right.\\
4 > x > - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
4 > x \ge 2\\
0 \ge x > - 2
\end{array} \right.\\
\to B \in \left( { - 2;0} \right] \cup \left[ {2;4} \right)
\end{array}\)
