Lời giải:Ta có:
\(\left\{ \begin{array}{l}
x + 1 > 0\\
2 - x > 0\\
{\log _{\frac{1}{3}}}\left( {x + 1} \right) > {\log _3}\left( {2 - x} \right)
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > - 1\\
x < 2\\
- {\log _3}\left( {x + 1} \right) > {\log _3}\left( {2 - x} \right)
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
- 1 < x < 2\\
{\log _3}\left( {2 - x} \right) + {\log _3}\left( {x + 1} \right) < 0
\end{array} \right.\)
\( \Leftrightarrow \left\{ \begin{array}{l}
- 1 < x < 2\\
{x^2} + x + 1 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
- 1 < x < 2\\
\left[ \begin{array}{l}
x > \frac{{1 + \sqrt 5 }}{2}\\
x < \frac{{1 - \sqrt 5 }}{2}
\end{array} \right.
\end{array} \right.\)
\( \Rightarrow S = \left( { - 1;\frac{{1 - \sqrt 5 }}{2}} \right) \cup \left( {\frac{{1 + \sqrt 5 }}{2};2} \right)\)
Suy ra \(a + b + c + d = - 1 + \frac{{1 - \sqrt 5 }}{2} + \frac{{1 + \sqrt 5 }}{2} + 2 = 2\)
