Giải các hệ bất phương trình (ẩn m) sau:
a) \(\left\{ \begin{array}{l}
{\left( {2m - 1} \right)^2} - 4\left( {{m^2} - m} \right) \ge 0\\
\frac{1}{{{m^2} - m}} > 0\\
\frac{{2m - 1}}{{{m^2} - m}} > 0
\end{array} \right.\)
b) \(\left\{ \begin{array}{l}
{\left( {m - 2} \right)^2} - \left( {m + 3} \right)\left( {m - 1} \right) \ge 0\\
\frac{{m - 2}}{{m + 3}} > 0\\
\frac{{m - 1}}{{m + 3}} > 0
\end{array} \right.\)
a) \(\left\{ \begin{array}{l}
{\left( {2m - 1} \right)^2} - 4\left( {{m^2} - m} \right) \ge 0\\
\frac{1}{{{m^2} - m}} > 0\\
\frac{{2m - 1}}{{{m^2} - m}} > 0
\end{array} \right.\)
\( \Leftrightarrow \left\{ \begin{array}{l}
1 \ge 0\\
{m^2} - m > 0\\
2m - 1 > 0
\end{array} \right. \Leftrightarrow m > 1\)
b) \(\left\{ \begin{array}{l}
{\left( {m - 2} \right)^2} - \left( {m + 3} \right)\left( {m - 1} \right) \ge 0\\
\frac{{m - 2}}{{m + 3}} > 0\\
\frac{{m - 1}}{{m + 3}} > 0
\end{array} \right.\)
\( \Leftrightarrow \left\{ \begin{array}{l}
- 6m + 7 \ge 0\\
\left( {m - 2} \right)\left( {m + 3} \right) < 0\\
\left( {m - 1} \right)\left( {m + 3} \right) > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m \le \frac{7}{6}\\
- 3 < m < 2\\
\left[ \begin{array}{l}
m > 1\\
m < - 3
\end{array} \right.
\end{array} \right. \Leftrightarrow 1 < m \le \frac{7}{6}\)
-- Mod Toán 10