Question

решите неравенство 2х^2+х-1>0​

Answer 1

$\displaystyle 2x^2+x-1 > 0\\2x^2+2x-x-1 > 0\\2x(x+1)-(x+1) > 0\\(x+1)(2x-1) > 0\\\\\left \{ {{x+1 > 0} \atop {2x-1 > 0}} \right. \rightarrow \left \{ {{x > -1} \atop {x > \frac{1}{2} }} \right. \rightarrow x \in (\frac{1}{2},+ \infty)\\ \\\left \{ {{x+1 < 0} \atop {2x-1 < 0}} \right. \rightarrow \left \{ {{x < -1} \atop {x < \frac{1}{2} }} \right. \rightarrow x \in(-\infty,-1)\\\\x \in (-\infty,-1)\cup(\frac{1}{2}, + \infty)$