\(f’\left( x \right) = {\left( {\dfrac{1}{{1-x}}} \right)^\prime } = \dfrac{{{{\left( 1 \right)}^\prime } \cdot \left( {1-x} \right)-1 \cdot {{\left( {1-x} \right)}^\prime }}}{{{{\left( {1-x} \right)}^2}}} = \dfrac{{0 \cdot \left( {1-x} \right)-1 \cdot \left( {-1} \right)}}{{{{\left( {1-x} \right)}^2}}} = \dfrac{1}{{{{\left( {1-x} \right)}^2}}}.\)
Тогда:
\(1 + 5f\left( x \right) + 6f’\left( x \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,1 + \dfrac{5}{{1-x}} + \dfrac{6}{{{{\left( {1-x} \right)}^2}}} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{1-2x + {x^2} + 5-5x + 6}}{{{{\left( {1-x} \right)}^2}}} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\)
\( \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{{x^2}-7x + 12}}{{{{\left( {1-x} \right)}^2}}} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{{x^2}-7x + 12 = 0,}\\{1-x \ne 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\left[ {\begin{array}{*{20}{c}}{x = 3,}\\{x = 4,}\end{array}} \right.}\\{x \ne 1\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 3,}\\{x = 4.}\end{array}} \right.\)
Сумма корней: \(3 + 4 = 7.\)
Ответ: 7.