Question

Помогите решить, пожалуйста
|x^2+x-5| больше 3х

Answer 1

$|x^2+x-5| textgreater 3xLongleftrightarrow left[ {{x^2+x-5 textgreater 3x} atop {x^2+x-5 textless -3x}} right. ,\\1) x^2+x-5 textgreater 3x,\x^2+x-5-3x textgreater 0,\x^2-2x-5 textgreater 0,\x^2-2x-5=0,\D=(-2)^2-4cdot1cdot(-5)=4+20=24,\\x_{1,2}=frac{2pmsqrt{24}}{2}=1pmfrac{2sqrt6}{2}=1pmsqrt6,\x_1=1+sqrt6,\x_2=1-sqrt6,\a textgreater 0 Longrightarrow x textless 1-sqrt6, x textgreater 1+sqrt6;$

$2) x^2+x-5 textless -3x,\x^2+x+3x-5 textless 0,\x^2+4x-5 textless 0,\x^2+4x-5=0\a+b+c=0 Longrightarrow x_1=1, x_2=-5,\\a textgreater 0 Longrightarrow -5<x<1,\\3) left[ {{x textless 1-sqrt6, x textgreater 1+sqrt6} atop {-5<x<1}} right..$

Ответ: $x> 1+sqrt6$ или $x<1.$

Answer 2

x²+x-5<-3x U x²+x-5>3x
1)x²+x-5<-3x
x²+4x-5<0
x1+x2=-4 U x1*x2=-5
x1=-5 U x2=1
-5<x<1
2)x²+x-5>3x
x²-2x-5>0
D=4+20=24
x1=(2-2√6)/2=1-√6
x2=1+√6
x<1-√6 U x>1+√6
x∈(-∞;1) U (1+√6;∞)