\({\log _{\dfrac{7}{9}}}\dfrac{{x + 1}}{{x-1}} > 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\log _{\dfrac{7}{9}}}\dfrac{{x + 1}}{{x-1}} > {\log _{\dfrac{7}{9}}}\dfrac{7}{9}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\dfrac{{x + 1}}{{x-1}} < \dfrac{7}{9},}\\{\dfrac{{x + 1}}{{x-1}} > 0}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\dfrac{{2x + 16}}{{x-1}} < 0,}\\{\dfrac{{x + 1}}{{x-1}} > 0\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x\, \in \,\left( {-8;1} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x\, \in \,\left( {-\infty ;-1} \right) \cup \left( {1;\infty } \right)}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x\, \in \,\left( {-8;-1} \right).} \right.} \right.\)
Целые решения: \(-7,\,\,-6,\,\,-5,…,-2.\) Их сумма равна: \(\dfrac{{-7-2}}{2} \cdot 6 = -27.\)
Ответ: \(-27\).