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Đặt $t = {5^x},t > 0$ ta có:
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,{t^2} - \left( {2m + 5} \right).t + {m^2} + 5 > 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\forall t > 0$
Gọi $\,\,\,\,\,\,\,\,\,\,\,\,f(t) = \,\,{t^2} - \left( {2m + 5} \right).t + {m^2} + 5$
Ta có $a = 1 > 0,\,\,\,\Delta = {\left( {2m + 5} \right)^2} - 4\left( {{m^2} + 5} \right) = 25$
$f(t)$có $2$ nghiệm ${t_1}<{t_2}$và $f(t) > 0$ có nghiệm là : $\left[ \begin{array}{l}
t < {t_1}\\
t > {t_2}
\end{array} \right.$
$\begin{array}{l}
Để f(t) > 0,\forall t > 0\,\,\,\, \Leftrightarrow \,\,\,{t_1} < {t_2} \le 0\\
\Leftrightarrow \left\{ \begin{array}{l}
{\rm P} \ge 0\\
S < 0
\end{array} \right.\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
{m^2} + 5 \ge 0\\
2m + 5 < 0
\end{array} \right.\,\,\,\,\,\,\,\, \Leftrightarrow m < - \frac52
\end{array}$
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Đăng bài 26-04-12 01:42 PM
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