Question

Решить неравенство с модулем Пожалуйста нужна помощь

Answer 1

$|x^{2}+3x|<x+4\\\\\left \{ {{x^{2}+3x<x+4 } \atop {x^{2}+3x>-(x+4) }} \right. \\\\\\\left \{ {{x^{2}+2x-4<0 } \atop {x^{2}+4x+4>0 }} \right.\\\\\\1)x^{2}+2x-4<0\\\\x^{2}+2x-4=0\\\\D=2^{2}-4*(-4)=4+16=20=(2\sqrt{5})^{2}\\\\x_{1}=\frac{-2-2\sqrt{5}}{2}=-1-\sqrt{5}\\\\x_{2}=\frac{-2+2\sqrt{5} }{2}=\sqrt{5}-1\\\\(x-(-1-\sqrt{5}))(x-(\sqrt{5}-1))<0\\\\x\in(-1-\sqrt{5};\sqrt{5}-1)$

$2)x^{2} +4x+4>0\\\\(x+2)^{2} >0\\\\x\in(-\infty;-2)\cup(-2;+\infty)\\\\Otvet:\boxed{x\in(-1-\sqrt{5};-2)\cup(-2;\sqrt{5}-1)}$