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

* Hướng dẫn 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 \(a + b + c + d = - 1 + \frac{{1 - \sqrt 5 }}{2} + \frac{{1 + \sqrt 5 }}{2} + 2 = 2\)