Bài 104274

Question

Giải các bất phương trình sau:
$a) log_3\left( {{lo}{{g}_{\frac{{1}}{{2}}}}{x}} \right) \ge
{0}$
$b) {lo}{{g}_{3}}\frac{{\left|
{{{x}^{2}}{ - 4x}} \right|{ + 3}}}{{{{x}^{2}}\left|
{{x - 5}} \right|}} \ge {0}$
$c) {lo}{{g}_{\frac{{{x + 6}}}{{3}}}}\left(
{{lo}{{g}_{2}}\frac{{{x - 1}}}{{{x + 2}}}} \right){ > 0} (*)$

score 0 · Answer 1

$a$) Điều kiện $x > 0$
$log_3\left( {{lo}{{g}_{\frac{{1}}{{2}}}}{x}} \right) \ge {0}
\Leftrightarrow {lo}{{g}_{\frac{{1}}{{2}}}}{x} \ge {1}
\Leftrightarrow {lo}{{g}_{2}}{x} \le { - 1}$
$ \Leftrightarrow {0} \le {x} \le \frac{{1}}{{2}}$
$b) {lo}{{g}_{3}}\frac{{\left| {{{x}^{2}}{ - 4x}}
\right|{ + 3}}}{{{{x}^{2}}\left| {{x - 5}} \right|}} \ge {0}
\Leftrightarrow \frac{{\left| {{{x}^{2}}{ - 4x}} \right|{ +
3}}}{{{{x}^{2}}\left| {{x - 5}} \right|}} \ge {1}$
   $ \Leftrightarrow {x} \le { -
}\frac{{2}}{{3}}{ } \vee { }\frac{{1}}{{2}} \le {x} \le {2}$
$c) (*) \Leftrightarrow {lo}{{g}_{\frac{{{x + 6}}}{{3}}}}\left(
{{lo}{{g}_{2}}\frac{{{x - 1}}}{{{x + 2}}}} \right){ >
lo}{{g}_{\frac{{{x + 6}}}{{3}}}}{1}$
    $\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
\frac{{{x + 6}}}{{3}}{ > 0}\\
\left( {\frac{{{x + 6}}}{{3}}{ - 1}} \right)\left[
{{lo}{{g}_{2}}\frac{{{x - 1}}}{{{x + 2}}}{ - 1}}
\right]{ > 0}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x > - 6}\\
{(x + 3)}\left[ {{lo}{{g}_{2}}\frac{{{x - 1}}}{{{x +
2}}}{ - 1}} \right]{ > 0}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x > - 6}\\
{x + 3 > 0 } \vee { lo}{{g}_{2}}\frac{{{x - 1}}}{{{x +
2}}}{ - 1 > 0}
\end{array} \right. \vee \left\{ \begin{array}{l}
{x > - 6}\\
{x + 3 < 0}\\
{lo}{{g}_{2}}\frac{{{x - 1}}}{{{x + 2}}}{ - 1 < 0}
\end{array} \right.\\
\Leftrightarrow { - 3 < x < - 2 } \vee { - 6 < x < - 5}
\end{array}$

Bài 104274

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Bài 104274

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