\(\dfrac{{\sqrt {12-x-{x^2}} }}{{2x-7}} \le \dfrac{{\sqrt {12-x-{x^2}} }}{{x-5}}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{{\left( {x-5} \right)\sqrt {12-x-{x^2}} -\left( {2x-7} \right)\sqrt {12-x-{x^2}} }}{{\left( {2x-7} \right)\left( {x-5} \right)}} \le 0\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\dfrac{{\sqrt {12-x-{x^2}} \left( {x-5-2x + 7} \right)}}{{\left( {2x-7} \right)\left( {x-5} \right)}} \le 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{{\left( {2-x} \right)\sqrt {12-x-{x^2}} }}{{\left( {2x-7} \right)\left( {x-5} \right)}} \le 0.\)
Решим данное неравенство методом интервалов при условии, что:
\(12-x-{x^2} \ge 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x\, \in \,\left[ {-4;\,3} \right].\)
Найдём нули числителя:
\(\left( {2-x} \right)\sqrt {12-x-{x^2}} = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{2-x = 0,}\\{\sqrt {12-x-{x^2}} = 0}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = 2,\\x = -4,\\x = 3.\end{array} \right.\)
Найдём нули знаменателя: \(\left( {2x-7} \right)\left( {x-5} \right) = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{7}{2},\\x = 5.\end{array} \right.\)
Ответ: \(\left\{ {-4} \right\} \cup \left[ {2;3} \right].\)