\(\left( {25-{x^2}} \right)\sqrt {3-x} = 0.\)
Область допустимых значений: \(3-x \ge 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,-x \ge -3\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x \le 3.\)
\(\left( {25-{x^2}} \right)\sqrt {3-x} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \le 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left[ {\begin{array}{*{20}{c}}{25-{x^2} = 0,}\\{\sqrt {3-x} = 0\,\,\,}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \le 3\,\,\,\,\,\,}\\{\left[ {\begin{array}{*{20}{c}}{x = 5,\,\,\,}\\{x = -5,}\\{x = 3\,\,\,}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -5,}\\{x = 3.\,\,\,}\end{array}} \right.\)
Ответ: \(-5;\,\,3.\)