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