Задача 30. Решите неравенство \(x\left( {2x + 1} \right) > -{x^2} + 4\)
\(x\left( {2x + 1} \right) > -{x^2} + 4\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2{x^2} + x + {x^2}-4 > 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,3{x^2} + x-4 > 0.\) Решим неравенство методом интервалов: \(3{x^2} + x-4 = 0;\,\,\,\,\,\,\,\,\,\,D = 1 + 48 = 49;\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{{-1-7}}{6} = -\dfrac{4}{3},}\\{x = \dfrac{{-1 + 7}}{6} = 1.\,\,\,\,\,}\end{array}} \right.\) Следовательно, решение исходного неравенства: \(x\, \in \,\left( {-\infty ;-\dfrac{4}{3}} \right) \cup \left( {1; + \infty } \right).\) Ответ: \(\left( {-\infty ;\;-\dfrac{4}{3}} \right) \cup \left( {1;\; + \infty } \right).\)Решение
