Đáp án:
\(\begin{array}{l}
B1:\\
\left[ \begin{array}{l}
x \ge 3\\
x < - 2
\end{array} \right.\\
B2:\\
1 \le m < 3
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
DK:\left\{ \begin{array}{l}
{x^2} + 3x - 4 \ge 0\\
{x^2} - 4 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {x - 3} \right)\left( {x + 1} \right) \ge 0\\
\left( {x - 2} \right)\left( {x + 2} \right) > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 3\\
x \le - 1
\end{array} \right.\\
\left[ \begin{array}{l}
x > 2\\
x < - 2
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x \ge 3\\
x < - 2
\end{array} \right.\\
B2:\\
DK:\left\{ \begin{array}{l}
x + m > 0\\
- x - 2m + 6 \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x > - m\\
- 2m + 6 \ge x
\end{array} \right.\\
\to - m < x \le 6 - 2m\\
Do:TXD:D = \left( { - 1;0} \right)\\
\to - 1 < x < 0\\
\to \left\{ \begin{array}{l}
- m \le - 1\\
0 < 6 - 2m
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ge 1\\
2m < 6
\end{array} \right.\\
\to 1 \le m < 3
\end{array}\)