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Điều kiện:$\left\{ \begin{array}{l}
1 \ne 2x - 1 > 0\\
1 \ne {x^2} - 3x + 2 > 0
\end{array} \right.$
$ \Leftrightarrow \left\{ \begin{array}{l}
1 \ne x > \frac{1}{2}\\
x < 1\,\,\,\,;\,x > 2\\
x \ne \frac{{3 \pm \sqrt 5 }}{2}
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
\frac{1}{2} < x < 1\\
x > 2\,\,\,\,\& x \ne \frac{{3 \pm \sqrt 5 }}{2}\,\,\,\,\,(*)
\end{array} \right.$
Khi đó ta có
$\begin{array}{l}
(1) \Leftrightarrow - \frac{1}{{{{\log }_2}\left( {2x - 1} \right)}} + \frac{1}{{{{\log }_2}\sqrt {{x^2} - 3x + 2} }} > 0\,\,\,\,\\
\,\,\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
{\log _2}\left( {2x - 1} \right).{\log _2}\sqrt {{x^2} - 3x + 2} > 0\\
{\log _2}\left( {2x - 1} \right) > {\log _2}\sqrt {{x^2} - 3x + 2}
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)
\end{array}$
Hoặc $\left\{ \begin{array}{l}
{\log _2}\left( {2x - 1} \right) < 0\\
{\log _2}\sqrt {{x^2} - 3x + 2} > 0
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(3)$
giải $(2)$:
$(2) \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ \begin{array}{l}
\sqrt {{x^2} - 3x + 2} > 1\\
2x - 1 > \sqrt {{x^2} - 3x + 2}
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(4)$
Hoặc $\left\{ \begin{array}{l}
2x - 1 < 1\\
2x - 1 > \sqrt {{x^2} - 3x + 2}
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(5)$
$(4) \Leftrightarrow \left\{ \begin{array}{l}
{x^2} - 3x + 1 > 0\\
3{x^2} - x - 1 > 0
\end{array} \right.\,\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
x < \frac{{3 - \sqrt 5 }}{2}\,\,\,;x > \frac{{3 + \sqrt 5 }}{2}\\
x < \frac{{1 - \sqrt {13} }}{6}\,;\,x > \frac{{1 + \sqrt {13} }}{6}
\end{array} \right.$
Xét các điều kiện $(*)$ ta có $x > \frac{{3 + \sqrt 5 }}{2}$
$(5) \Leftrightarrow \left\{ \begin{array}{l}
x < 1\\
3{x^2} - x - 1 > 0
\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
x < 1\\
x < \frac{{1 - \sqrt {13} }}{6}\,\,;\,\,x > \frac{{1 + \sqrt {13} }}{6}
\end{array} \right.$
Đối chiếu với điều kiện $(*)$ ta có $(5)$ vô nghiệm
giải $(3)$
$(3) \Leftrightarrow \left\{ \begin{array}{l}
0 < 2x - 1 < 1\\
{x^2} - 3x + 1 > 0
\end{array} \right.\,\,\,\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
\frac{1}{2} < x < 1\\
x < \frac{{3 - \sqrt 5 }}{2}\,\,\,\,;\,x > \frac{{3 + \sqrt 5 }}{2}
\end{array} \right.$
Hệ này vô nghiệm
Vậy nghiệm của $(1)$ là $x > \frac{{3 + \sqrt 5 }}{2}$
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