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Điều kiện : $\left\{ \begin{array}{l}
1 + x > 0\\
1 - y > 0\\
1 + 2x > 0\\
1 + 2y > 0
\end{array} \right.$
Hệ có thể biến dổi thành :
$\left\{ \begin{array}{l}
2{\log _{1 + x}}\left( {1 - y} \right) + 2{\log _{1 - y}}\left( {1 + x} \right) =
4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\\
{\log _{1 + x}}\left( {1 + 2y} \right) + {\log _{1 - y}}\left( {1 + 2x} \right) =
2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)
\end{array} \right.$
$(1) \Leftrightarrow \,{\log _{1 + x}}\left( {1 - y} \right) + \frac{1}{{{{\log }_{1 + x}}\left( {1
- y} \right)}} = 2$
$X = {\log _{1 + x}}\left( {1 - y} \right)$
Ta có: ${X^2} - 2X + 1 = 0\,\,\,\,\,\, \Leftrightarrow X = 1$
$ \Leftrightarrow \,\,y = - x$. Từ đó :
$\begin{array}{l}
(2) \Leftrightarrow \,{\log _{1 + x}}\left( {1 - 2x} \right) + {\log _{1 + x}}\left( {1 + 2x} \right) = 2\\
\,\,\,\,\,\,\,\, \Leftrightarrow 1 - 4{x^2} = {\left( {1 + x} \right)^2}\,\,\,\,\,\,\,\, \Leftrightarrow
\,\,\,\,5{x^2} + 2x = 0
\end{array}$
$ \Leftrightarrow \left[ \begin{array}{l}
x = 0\,\,\,\,\,\left( l \right)\\
x = \frac{{ - 2}}{5}
\end{array} \right.$ Vậy $\left\{ \begin{array}{l}
x = - \frac{2}{5}\\
y = \frac{2}{5}
\end{array} \right.$
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