\(\left( {\sqrt {36{x^2} + 7} -\sqrt {35{x^2} + 16} } \right)\;\sqrt {2-x} = 0.\)
Запишем ОДЗ: \(\left\{ \begin{array}{l}36{x^2} + 7 \ge 0,\\35{x^2} + 16 \ge 0,\\2-x \ge 0.\end{array} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ \begin{array}{l}x \in R,\\x \in R,\\x \le 2\end{array} \right.\;\;\;\; \Leftrightarrow \;\;\;\;x \le 2\;\;\;\; \Leftrightarrow \;\;\;\;x \in \left( {-\infty ;2} \right].\)
\(\left( {\sqrt {36{x^2} + 7} -\sqrt {35{x^2} + 16} } \right)\;\sqrt {2-x} = 0\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\sqrt {36{x^2} + 7} = \sqrt {35{x^2} + 16} ,}\\{\sqrt {2-x} = 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,\,\;\,\,\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{36{x^2} + 7 = 35{x^2} + 16,}\\{2-x = 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,\;\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{x^2} = 9,}\\{x = 2\;\;}\end{array}\;\;\;\; \Leftrightarrow } \right.\;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = -3,\;\,\;\;\;\;\;\;\;\;\;\;\,}\\{x = 3 \notin \left( {-\infty ;2} \right],}\\{x = 2\,\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = -3,}\\{x = 2.\;\,}\end{array}} \right.\)
ОТВЕТ: -3; 2.