Рациональные неравенства. Задача 30

ОТВЕТ: \(\left( {-\infty ;\;-\dfrac{4}{3}} \right) \cup \left( {1;\;\infty } \right).\)

\(x\left( {2x + 1} \right) > -{x^2} + 4\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2{x^2} + x + {x^2}-4 > 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,3{x^2} + x-4 > 0.\)

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: \(\left( {-\infty ;\;-\dfrac{4}{3}} \right) \cup \left( {1;\; + \infty } \right).\)