\({\log _{\dfrac{1}{2}}}\left( {6-x} \right) \ge {\log _{\dfrac{1}{2}}}{x^2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{6-x \le {x^2},}\\{6-x > 0\,\,}\end{array}\,\,\,\,\,\,\,\,\, \Leftrightarrow } \right.\)
\( \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{{x^2} + x-6 \ge 0,}\\{x < 6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x\, \in \,\left( {-\infty ;-3} \right] \cup \left[ {2;\infty } \right),}\\{x < 6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x\, \in \,\left( {-\infty ;-3} \right] \cup \left[ {2;6} \right).} \right.} \right.\)
Натуральные решения: 2, 3, 4, 5. Их сумма равна: \(2 + 3 + 4 + 5 = 14.\)
Ответ: 14.