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$\begin{array}{l}
1)\,\,\,\,\,\left\{ \begin{array}{l}
xy \ne 1\\
0 < x + y \ne 1
\end{array} \right.\\
\,\,\,\,\,(1) \Leftrightarrow \,\,\,{\log _{x + y}}\left| {xy} \right|.{\log _{\left| {xy} \right|}}4\left| {xy} \right| = 1\\
\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,{\log _{x + y}}4\left| {xy} \right| = 1\\
\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,4\left| {xy} \right| = x + y
\end{array}$
Từ đó ta có hệ : $\left\{ \begin{array}{l}
\,4\left| {xy} \right| = x + y\\
x - y = 2\sqrt {3\,}
\end{array} \right.$
$\begin{array}{l}
\Leftrightarrow \,\,\left\{ \begin{array}{l}
y = x - 2\sqrt 3 \\
4\left| {x\left( {x - 2\sqrt 3 } \right)} \right| = 2x - 2\sqrt 3
\end{array} \right.\\
\Leftrightarrow \,\,\left\{ \begin{array}{l}
x \ge \sqrt 3 \\
{\left[ {2\left( {x - 2\sqrt 3 } \right)} \right]^2} - {\left( {x - \sqrt 3 } \right)^2} = 0
\end{array} \right.\\
\Leftrightarrow \,\,\left\{ \begin{array}{l}
x \ge \sqrt 3 \\
\left[ \begin{array}{l}
2{x^2} - \left( {4\sqrt 3 - 1} \right)x - \sqrt 3 = 0\\
2{x^2} - \left( {4\sqrt 3 + 1} \right)x + \sqrt 3 = 0
\end{array} \right.
\end{array} \right.\,\,\,\, \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{3 + 2\sqrt 3 }}{2}\\
x = 2 + \sqrt 3
\end{array} \right.
\end{array}$
- Với $x = \frac{{3 + 2\sqrt 3 }}{2}\,\,\,\,\, \Rightarrow \,\,\,y = x - 2\sqrt 3 = 2 - \sqrt 3 $
$xy = \left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right) = 1:$ Không thỏa điều kiện
Vậy hệ có một nghiệm là$\left( {\frac{{3 + 2\sqrt 3 }}{2},\,\,\frac{{3 - 2\sqrt 3 }}{2}} \right)$
$2)$ ĐS : $\left( {\frac{{5 + \sqrt 5 }}{2},\,\frac{{5 - \sqrt 5 }}{2}} \right)$
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