{x>0
{x^2+x+1<1⇒x²+x<0⇒x(x+1)<0 x=0 x=-1 -1<x<0<br>Ответ нет решения
{x^2+4x<1⇒x²+4x-1<0 (1)<br>{x^2+4x>-1⇒x²+4x+1>0 (2)
1)D=16+4=20
x1=(-4-2√5)/2=-2-√5 U x2=-2+√5
(-2-√5)2)D=16-4=12
x1=(-4-2√3)/2=-2-√3 U x2=-2+√3
x<-2-√3 U x>-2+√3
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--(-2-√5)--------(-2-√3)-----------------------(-2+√3)-----------(-2+√5)-----
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x∈(-2-√5;-2-√3) U (-2+√3;-2+√5)
{x^2-x>0⇒x(x-1)>0 x=1 x=0 x<0 U x>1
{x^2-x<2⇒x²-x-2<0 x1+x2=1 U x1*x2=-2⇒x1=-1 U x2=2 -1<x<2<br>x∈(-1;0) U (1;2)
{x^2-x<0⇒x(x-1)<0 x=0 x=1 0<x<1<br>{-(x^2-x)<2⇒x²+x+2>0 D=1-8=-7<0⇒x-любое<br>x∈(0;1)