\(\left( {\sqrt {x-4} -1} \right)\left( {\sqrt x -3} \right) \ge 0.\)
Запишем ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{x-4 \ge 0,}\\{x \ge 0\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \ge 4,\,\,}\\{x \ge 0\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \in \,\left[ {4;\infty } \right).\)
Решим данное неравенство методом интервалов.
\(\left( {\sqrt {x-4} -1} \right)\left( {\sqrt x -3} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sqrt {x-4} -1 = 0,}\\{\sqrt x -3 = 0\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sqrt {x-4} = 1,}\\{\sqrt x = 3\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\, \Leftrightarrow } \right.\)
\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x-4 = 1,}\\{x = 9\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 5,}\\{x = 9.}\end{array}} \right.\)
Ответ: \(\left[ {4;5} \right] \cup \left[ {9;\infty } \right).\)