Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
11,\\
\left\{ \begin{array}{l}
{x^2} - 5x + 6 \ge 0\\
- 6x + 24 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left( {x - 2} \right)\left( {x - 3} \right) \ge 0\\
6x - 24 < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 3\\
x \le 2
\end{array} \right.\\
x < 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x \le 2\\
3 \le x < 4
\end{array} \right.\\
\Rightarrow S = \left( { - \infty ;2} \right] \cup \left[ {3;4} \right)\\
12,
\end{array}\)
Bất phương trình bậc 2 một ẩn là: \({x^2} - 3x - 4 = 0\)
\(\begin{array}{l}
13,\\
f\left( x \right) > 0,\,\,\,\,\forall x \in R\\
\Leftrightarrow a\,{x^2} + bx + c > 0,\,\,\,\,\forall x \in R\\
\Leftrightarrow \left\{ \begin{array}{l}
a > 0\\
\Delta ' < 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a > 0\\
{b^2} - 4ac < 0
\end{array} \right.
\end{array}\)
