|
|
$\begin{array}{l}
(1) \Leftrightarrow \left\{ \begin{array}{l}
{x^2} + 1 > \frac{{{x^2} + 4x + m}}{5}\\
{x^2} + 4x + m > 0
\end{array} \right.\\
\,\,\,\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
m > - {x^2} - 4x = f(x)\\
m < 4{x^2} + 4x + 5 = g(x)
\end{array} \right.
\end{array}$
Hệ trên thỏa mãn $\forall x \in (2,3)$
$ \Leftrightarrow \left\{ \begin{array}{l}
m \ge \,\mathop {Max\,}\limits_{2 \le x \le 3} f(x) = - 12\,\,\,\,\,\,khi\,\,\,\,\,\,\,x = 2\\
m \le \mathop {\min }\limits_{2 \le x \le 3} \,g(x) = 13\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,khi\,\,\,\,\,\,\,\,x = 2
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow - 12 \le m \le 13$
|
|
|
Đăng bài 26-04-12 02:08 PM
|
|