Tập nghiệm của bất phương trình\({\log _{\frac{1}{3}}}\left( {x + 1} \right) > {\log _3}\left( {2 - x} \right)\) là \(S = \left( {

* Hướng dẫn giải

\(\begin{array}{l}
\left\{ {\begin{array}{*{20}{c}}
{x + 1 > 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
{2 - x > 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
{lo{h_{\frac{1}{3}}}\left( {x + 1} \right) > {{\log }_3}\left( {2 - x} \right)}
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{x >  - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
{x < 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
{ - {{\log }_3}\left( {x + 1} \right) > {{\log }_3}\left( {2 - x} \right)}
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{ - 1 < x < 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
{{{\log }_3}\left( {2 - x} \right) + {{\log }_3}\left( {x + 1} \right) < 0}
\end{array}} \right.\\
 \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{ - 1 < x < 2}\\
{ - {x^2} + x + 1}
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{ - 1 < x < 2}\\
{{x^2} - x - 1 > 0}
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{ - 1 < x < 2}\\
{\left[ {\begin{array}{*{20}{c}}
{x > \frac{{1 + \sqrt 5 }}{2}}\\
{x < \frac{{1 - \sqrt 5 }}{2}}
\end{array}} \right.}
\end{array}} \right.\\
 \Rightarrow S = \left( { - 1;\frac{{1 - \sqrt 5 }}{2}} \right) \cup \left( {\frac{{1 + \sqrt 5 }}{2};2} \right)\\
 \Rightarrow a + b + c + d =  - 1 + \frac{{1 - \sqrt 5 }}{2} + \frac{{1 + \sqrt 5 }}{2} + 2 = 2
\end{array}\)