Giải các bất phương trình :
$\begin{array}{l}
1)\log _x2.\log _{2x}2 > \log _{4x}2                         (1)\\
2)\log _{\left| x \right|}\left( {x^2 - \frac{1}{2}x} \right) > 1                            (2)
\end{array}$
1) Điều kiện:
$\,\,\,x > 0,\,\,\,x \ne 1,\,\,x \ne \frac{1}{2},\,\,x \ne \frac{1}{4}$
Khi đó ta có
$(1) \Leftrightarrow \frac{1}{{{{\log }_2}x}}.\frac{1}{{1 + {{\log }_2}x}} > \frac{1}{{2 + {{\log }_2}x}}$  
Đặt $u = {\log _2}x,$ ta có: $\begin{array}{l}
\frac{1}{u}.\frac{1}{{1 + u}} > \frac{1}{{2 + u}}\\

\end{array}$
$ \Leftrightarrow \frac{{2 - {u^2}}}{{u\left( {1 + u} \right)\left( {2 + u} \right)}} > 0\,\,\,\,\,\,\,\,(1');$
Xét dấu vế trái, đặt $f(u)$, bằng phương pháp khoảng

Nghiệm của $(1')$là: $\left[ \begin{array}{l}
u <  - 2\\
- \sqrt 2  < u <  - 1\\
0 < u < \sqrt 2
\end{array} \right.$
Suy ra $\left[ \begin{array}{l}
{\log _2}x <  - 2\\
 - \sqrt 2  < {\log _2}x <  - 1\\
0 < {\log _2}x < \sqrt 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
0 < x < \frac{1}{4}\\
\frac{1}{{{2^{\sqrt 2 }}}} < x < \frac{1}{2}\\
1 < x < {2^{\sqrt 2 }}
\end{array} \right.$
Tập nghiệm của BPT là:
$S=\left (0;\frac{1}{4} \right )\cup \left (\frac{1}{2^\sqrt2};\frac{1}{2} \right )\cup \left (1;2^\sqrt2\right )$

$2)$
Điều kiện:$\left\{ \begin{array}{l}
x \ne 0\\
x \ne  \pm 1\\
x < 0\,\,\,\,;\,\,\,\,x > \frac{1}{2}
\end{array} \right.$
Xét $2$ trường hợp :
$a)\left\{ \begin{array}{l}
\left| x \right| < 1\\
x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
 - 1 < x < 1\\
x \ne 0
\end{array} \right.$  khi đó $(2) \Leftrightarrow 0 < {x^2} - \frac{1}{2}x < \left| x \right|$
Suy ra      $\left[ \begin{array}{l}
 - \frac{1}{2} < x < 0\\
\frac{1}{2} < x < 1
\end{array} \right.$
$b)$  $\left| x \right| > 1 \Leftrightarrow \left[ \begin{array}{l}
x > 1\\
x <  - 1
\end{array} \right.$
với
$\begin{array}{l}
x > 1:\,\,(2) \Leftrightarrow {x^2} - \frac{1}{2}x > x\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow {x^2} - \frac{3}{2}x \Leftrightarrow x > \frac{3}{2}
\end{array}$
Với $x <  - 1:(2) \Leftrightarrow {x^2} - \frac{1}{2}x >  - x$
                       $ \Leftrightarrow {x^2} + \frac{1}{2}x > 0\,\,\,\,\, \Leftrightarrow x <  - 1$
Tập nghiệm của $(2)$ là:
$\begin{array}{l}
S = ( - \infty , - 1) \cup ( - \frac{1}{2},0) \cup (\frac{1}{2},1) \cup (\frac{3}{2}, + \infty )\\

\end{array}$

Thẻ

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