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$1)$ Điều kiện: $\left\{ \begin{array}{l}
1 \ne 2 - x > 0\\
1 \ne 4 - y > 0\\
2 - y > 0\\
2x - 2 > 0
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}
1 < x < 2\\
y < 2
\end{array} \right.$
$1 < x < 2\,\,\, \Rightarrow \,\,0 < 2 - x < 1\,\,$
$y < 2\,\,\,\, \Rightarrow \,\,\,4 - y > 1$ do đó :
$\begin{array}{l}
\,\,\,\,\,\,\,\left\{ \begin{array}{l}
{\log _{2 - x}}\left( {2 - y} \right) > 0\\
{\log _{4 - y}}\left( {2x - 2} \right) > 0
\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\left\{ \begin{array}{l}
0 < 2 - y < 1\\
2x - 2 > 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x > \frac{3}{2}\\
1 < y < 2
\end{array} \right.
\end{array}$
Vậy ĐS: $\left\{ \begin{array}{l}
\frac{3}{2} < x < 2\\
1 < y < 2
\end{array} \right.$
$2)$ Điều kiện: $\left\{ \begin{array}{l} 1\neq x-1>0\\ 1\neq 2-y>0\\5-y>0\\4-x>0 \end{array} \right.
\Leftrightarrow \left\{ \begin{array}{l} 1< x\neq 2 \\ 1\ne y<2\\x<4 \end{array} \right.$
Ta có: $y<2\Rightarrow 5-y>3>1.$ Suy ra $\log_{x-1}{(5-y)}<0\Leftrightarrow x-1<1\Leftrightarrow x<2$
Với $x<2$ thì $4-x>1$ suy ra: $\log_{2-y}{(4-x)}<0\Leftrightarrow 2-y<1 \Leftrightarrow y>1$
ĐS: $\left\{ \begin{array}{l}
1 < x < 2\\
1 < y < 2
\end{array} \right.$
$3)$ Điều kiện: $\left\{ \begin{array}{l} 1\ne x-2>0\\ 1\ne 3-y>0\\ 2y-4>0\\x-4>0 \end{array} \right.
\Leftrightarrow \left\{ \begin{array}{l} x>4\\ 2<y<3 \end{array} \right.$
Ta có: $x>4\Rightarrow x-2>1$ Suy ra $ \log_{x-2}{(2y-4)}>0 \Rightarrow 2y-4>1\Leftrightarrow y>\frac{5}{2}$
Với $y>\frac{5}{2}$ thì $3-y<1$ Suy ra: $\log_{3-y}{(x-4)}>0 \Leftrightarrow x-4<1\Leftrightarrow x<5$
ĐS: $\left\{ \begin{array}{l}
4 < x < 5\\
\frac{5}{2} < y < 3
\end{array} \right.$
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