Задача 53. Решите неравенство \(\dfrac{{\left( {x-3} \right)\left( {{x^2}-8x + 16} \right)}}{{{{\left( {2-x} \right)}^2}}} \leqslant 0\)
\(\dfrac{{\left( {x-3} \right)\left( {{x^2}-8x + 16} \right)}}{{{{\left( {2-x} \right)}^2}}} \le 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{\left( {x-3} \right){{\left( {x-4} \right)}^2}}}{{{{\left( {x-2} \right)}^2}}} \le 0.\) Решим неравенство методом интервалов. Найдём нули числителя: \(\left( {x-3} \right){\left( {x-4} \right)^2} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x-3 = 0,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\left( {x-4} \right)}^2} = 0\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 3,\,\,\,}\\{x = 4.\,\,}\end{array}} \right.\) Найдём нули знаменателя: \({\left( {x-2} \right)^2} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = 2.\) Следовательно, решение исходного неравенства: \(x\, \in \,\left( {-\infty ;2} \right) \cup \left( {2;3} \right] \cup \left\{ 4 \right\}.\) Ответ: \(\left( {-\infty ;\;2} \right) \cup \left( {2;\;3} \right] \cup \left\{ 4 \right\}.\)