Giải các hệ bất phương trình (ẩn m) sau:
a) \(\left\{ \begin{array}{l}
2m - 1 > 0\\
{m^2} - \left( {m - 2} \right)\left( {2m - 1} \right) < 0
\end{array} \right.\)
b) \(\left\{ \begin{array}{l}
{m^2} - m - 2 > 0\\
{\left( {2m - 1} \right)^2} - 4\left( {{m^2} - m - 2} \right) \le 0
\end{array} \right.\)
a) \(\left\{ \begin{array}{l}
2m - 1 > 0\\
{m^2} - \left( {m - 2} \right)\left( {2m - 1} \right) < 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m > \frac{1}{2}\\
- {m^2} + 5m - 2 < 0
\end{array} \right.\)
\( \Leftrightarrow \left\{ \begin{array}{l}
m > 0,5\\
\left[ \begin{array}{l}
m > \frac{{5 + \sqrt {17} }}{2}\\
m < \frac{{5 - \sqrt {17} }}{2}
\end{array} \right.
\end{array} \right. \Leftrightarrow m > \frac{{5 + \sqrt {17} }}{2}\)
b) \(\left\{ \begin{array}{l}
{m^2} - m - 2 > 0\\
{\left( {2m - 1} \right)^2} - 4\left( {{m^2} - m - 2} \right) \le 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
- 1 < m < 2\\
9 \le 0
\end{array} \right.\)
Hệ vô nghiệm
-- Mod Toán 10