Задача 69. Решите неравенство \(\dfrac{{{x^3}-2{x^2} + 5x + 2}}{{{x^2} + 3x + 2}} \geqslant 1\)
\(\dfrac{{{x^3}-2{x^2} + 5x + 2}}{{{x^2} + 3x + 2}} \ge 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{{x^3}-2{x^2} + 5x + 2}}{{{x^2} + 3x + 2}}-1 \ge 0\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{{x^3}-2{x^2} + 5x + 2-{x^2}-3x-2}}{{{x^2} + 3x + 2}} \ge 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{{x^3}-3{x^2} + 2x}}{{{x^2} + 3x + 2}} \ge 0.\) Решим неравенство методом интервалов. Найдём нули числителя: \({x^3}-3{x^2} + 2x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x\left( {{x^2}-3x + 2} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2}-3x + 2 = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 0,}\\{x = 1,}\\{x = 2.}\end{array}} \right.\) Найдём нули знаменателя: \({x^2} + 3x + 2 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -2,}\\{x = -1.}\end{array}} \right.\) Следовательно, решение исходного неравенства: \(x\, \in \,\left( {-2;-1} \right) \cup \left[ {0;1} \right] \cup \left[ {2; + \infty } \right).\) Ответ: \(\left( {-2;\;-1} \right) \cup \left[ {0;\;1} \right] \cup \left[ {2;\; + \infty } \right).\)