|
|
Điều kiện : $x > 1,\,\,\,$ta có $\left| {1 - x} \right| = x - 1$
$(1)
\Leftrightarrow \,\,\,\,\,\,\,\,\,\,\,6{\log _3}\left( {x - 1} \right) +
\log _3^2\left( {x - 1} \right) + 5 \ge
0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$
Đặt $t = {\log _3}\left( {x - 1} \right)$ ta có ${t^2} + 6t + 5 \ge 0$
$\begin{array}{l}
\Leftrightarrow \left[ \begin{array}{l}
t \le - 5\\
t \ge - 1
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \left[ \begin{array}{l}
{\log _3}\left( {x - 1} \right) \le - 5\\
{\log _3}\left( {x - 1} \right) \ge - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
0 \le x - 1 \le \frac{1}{{{3^5}}}\\
x - 1 \ge \frac{1}{3}\,\,\,\,\,\,\,\,\,\,
\end{array} \right.\,\,\,\,\,\,\, \Leftrightarrow \left[ \begin{array}{l}
1 < x \le 1 + \frac{1}{{{3^5}}}\\
x \ge \frac{4}{3}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
1 < x \le \frac{{244}}{{243}}\\
x \ge \frac{4}{3}
\end{array} \right.
\end{array}$
|
|
|
Trả lời 10-07-12 11:29 AM
|
|