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1)
Điều kiện: ${{6^{x + 1}} - {{36}^x}}>0$.
Khi đó ta có
$\begin{array}{l}
\,\,\,\,{\log _{\frac{1}{{\sqrt 5 }}}}\left( {{6^{x + 1}} - {{36}^x}} \right) \ge - 2\\ \Leftrightarrow \left\{ \begin{array}{l}
{6^{x + 1}} - {36^x} > 0\\
{6^{x + 1}} - {36^x} \le \left (\frac{1}{\sqrt5}\right )^{-2}=5
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{6^x}\left( {6 - {6^x}} \right) > 0\\
{6.6^x} - {6^{2x}} \le 5
\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}
{6^x} < 6\\
{6^{2x}} - {6.6^x} + 5 \ge 0
\end{array} \right.\\
\,\,\, \Leftrightarrow \left[ \begin{array}{l}
{6^x} \le 1\\
5 \le {6^x} < 6
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \left[ \begin{array}{l}
x \le 0\\
{\log _6}5 \le x < 1
\end{array} \right.(TM)
\end{array}$
Vậy BPT có nghiệm $\left[ \begin{array}{l}
x \le 0\\
{\log _6}5 \le x < 1
\end{array} \right.$
$2)$
Điều kiện: $5^{x+1}-25^x>0$.
Khi đó ta có:
$\begin{array}{l}
\,\,\,\,{\log _{\frac{1}{{\sqrt 6 }}}}\left( {{5^{x
+ 1}} - {{25}^x}} \right) \ge - 2\\ \Leftrightarrow \left\{
\begin{array}{l}
{5^{x + 1}} - {25^x} > 0\\
{5^{x + 1}} - {25^x} \le \left(\frac{1}{\sqrt6}\right )^{-2}=6
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{5^x}\left( {5 - {5^x}} \right) > 0\\
{5.5^x} - {5^{2x}} \le 6
\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}
{5^x} < 5\\
{5^{2x}} - {5.5^x} + 6 \ge 0
\end{array} \right.\\
\,\,\, \Leftrightarrow \left[ \begin{array}{l}
{5^x} \le 2\\
3 \le {5^x} < 5
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \left[ \begin{array}{l}
x \le \log_52\\
{\log _5}3 \le x < 1
\end{array} \right.(TM)
\end{array}$
Vậy BPT có nghiệm $\left[ \begin{array}{l}
x \le \log_52\\
{\log _5}3 \le x < 1
\end{array} \right.$
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Đăng bài 26-04-12 10:32 AM
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