В) (х-1)^2+(х-2)^23х+10-(х+2)^2

Question

В) (х-1)^2+(х-2)^2<1 г) (х+3)(х-2)>3х+10-(х+2)^2

Answer 1

В) (x-1)^2+(x-2)^2<1<br>x^2-2x+1+x^2-4x+4<1<br>x^2-2x+x^2-4x+4<1<br>2x^2-6+4<1 (разделим обе части уравнения на 2)<br>x^2-3x+2<0<br>x^2-x-2x+2<0<br>x*(x-1)-2*(x-1)<0<br>(x-1)*(x-2)<0<br>{x-1<0 x-2>0
{x<1 x>2
{x>1 x<2 <br>x∈ (1;2)

г) (x+3)*(x-2)>3x+10-(x+2)^2
x^2-2x+3x-6>3x+10-(x^2+4x+4)
x^2-2x-6>10-x^2-4x-4
x^2-2x-6>6-x^2-4x
x^2-2x-6-6+x^2+4x>0
2x^2+2x-12>0
x^2+x-6>0
x^2+3x-2x-6>0
x*(x+3)-2*(x+3)>0
(x+3)*(x+2)>0
{x+3>0 x-2>0
{x+3<0 x+2<0<br>{x>-3 x>2
{ x<-3 x<2 <br>x∈(-∞;-3)u(2;+∞)