Giải các hệ bất phương trình sau:
a) \(\left\{ \begin{array}{l}
{x^2} \ge 4\\
{\left( {2x - 1} \right)^2} \le 9
\end{array} \right.\)
b) \(\left\{ \begin{array}{l}
2x - 3 \le \left( {x + 1} \right)\left( {x - 2} \right)\\
{x^2} - x < 6
\end{array} \right.\)
a) \(\left\{ \begin{array}{l}
{x^2} \ge 4\\
{\left( {2x - 1} \right)^2} \le 9
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^2} - 4 \ge 0\\
- 3 < 2x - 1 < 3
\end{array} \right.\)
\(\left\{ \begin{array}{l}
{x^2} \ge 4\\
{\left( {2x - 1} \right)^2} \le 9
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^2} - 4 \ge 0\\
- 3 < 2x - 1 < 3
\end{array} \right.\)
b) \(\left\{ \begin{array}{l}
2x - 3 \le \left( {x + 1} \right)\left( {x - 2} \right)\\
{x^2} - x < 6
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^2} - 3x + 1 \ge 0\\
{x^2} - x - 6 < 0
\end{array} \right.\)
\(\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
x \in \left( { - \infty ;\frac{{3 - \sqrt 5 }}{2}} \right] \cup \left[ {\frac{{3 + \sqrt 5 }}{2}; + \infty } \right)\\
- 2 < x < 3
\end{array} \right.\\
\Leftrightarrow x \in \left( { - 2;\frac{{3 - \sqrt 5 }}{2}} \right] \cup \left[ {\frac{{3 + \sqrt 5 }}{2};3} \right)
\end{array}\)
-- Mod Toán 10