Hệ phương trình \(\left\{ \begin{array}{l}{\log _2}\left( {1 + x} \right) + {\log _{\dfrac{1}{2}}}\left( {1 - y} \right) = 0\\{\log _{1 + x}}\left( {1 + 2y} \right) + {\log _{1 - y}}\left( {1 + 2x} \right) = 2\end{array} \right.\)  có bao nhiêu nghiệm?

ĐK : \(x >  - \dfrac{1}{2};\,0 \ne y < 1\)

Ta có \(\left\{ \begin{array}{l}{\log _2}\left( {1 + x} \right) + {\log _{\dfrac{1}{2}}}\left( {1 - y} \right) = 0\\{\log _{1 + x}}\left( {1 + 2y} \right) + {\log _{1 - y}}\left( {1 + 2x} \right) = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\log _2}\left( {1 + x} \right) - {\log _2}\left( {1 - y} \right) = 0\\{\log _{1 + x}}\left( {1 + 2y} \right) + {\log _{1 - y}}\left( {1 + 2x} \right) = 2\end{array} \right.\)

\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}{\log _2}\left( {1 + x} \right) = {\log _2}\left( {1 - y} \right)\\{\log _{1 + x}}\left( {1 + 2y} \right) + {\log _{1 - y}}\left( {1 + 2x} \right) = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}1 + x = 1 - y\\{\log _{1 + x}}\left( {1 + 2y} \right) + {\log _{1 - y}}\left( {1 + 2x} \right) = 2\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}y =  - x\\{\log _{1 + x}}\left( {1 + 2y} \right) + {\log _{1 - y}}\left( {1 + 2x} \right) = 2\end{array} \right. \Rightarrow {\log _{1 + x}}\left( {1 - 2x} \right) + {\log _{1 + x}}\left( {1 + 2x} \right) = 2\\ \Leftrightarrow {\log _{1 + x}}\left( {1 - 4{x^2}} \right) = 2 \Leftrightarrow \left\{ \begin{array}{l}1 + x > 0\\1 + x \ne 1\\1 - 4{x^2} > 0\\1 - 4{x^2} = {\left( {x + 1} \right)^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x >  - 1\\x \ne 0\\ - \dfrac{1}{2} < x < \dfrac{1}{2}\\5{x^2} + 2x = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \ne 0\\ - \dfrac{1}{2} < x < \dfrac{1}{2}\\\left[ \begin{array}{l}x = 0\\x =  - \dfrac{2}{5}\end{array} \right.\end{array} \right. \Rightarrow x =  - \dfrac{2}{5} \Rightarrow y = \dfrac{2}{5}\left( {tm} \right)\end{array}\)

Vậy hệ có nghiệm duy nhất \(\left( {x;y} \right) = \left( { - \dfrac{2}{5};\dfrac{2}{5}} \right)\)