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$1$) ĐK : $x > 0$
$\begin{array}{l}
\,\,\,\,\,\,{3^{\log x + 2}} < {3^{\log {x^2} + 5}} - 2\\
\Leftrightarrow 9.{3^{\log x}} < {3^5}{.3^{2\log x}} - 2\\
\Leftrightarrow \left\{ \begin{array}{l}
9t < 243{t^2} - 2\\
t = {3^{\log x}},t > 0
\end{array} \right. \Leftrightarrow t > \frac{1}{9}\\
\Leftrightarrow {3^{\log x}} > \frac{1}{9} = {3^{ - 2}} \Leftrightarrow \log x > - 2\\
\Leftrightarrow x > {10^{ - 2}}\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow x > \frac{1}{{100}}\\
\end{array}$
$2$) Đổi về cơ số $10$ ta có:
$\left( {\frac{{\log 5 - \log 4}}{{\log 5.\log 4}}} \right)\log x \ge
1\,\,\\\,\,\,$$ \Leftrightarrow
$ $x \ge {10^{\frac{{\log 5.\log 4}}{{\log 5 - \log 4}}}}$
$3$) $t = {\log _4}\frac{{3x - 1}}{{x + 1}},\,\,\,t > 0$
Suy ra $0 < t \le 1\,\,\,\,\,$
$ \Leftrightarrow
$ $\left[ \begin{array}{l}
x \le - 5\\
x > 1
\end{array} \right.$
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Đăng bài 07-05-12 09:26 AM
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