y=ax^2+bx+c,
x0=-b/(2a),
y0=c-b^2/(4a) или y0=f(x0)
1.1) <span>y=x^2-4x+3</span>,
x0=-(-4)/(2*1)=2,
y0=3-(-4)^2/(4*1)=-1, {или y0=2^2-4*2+3=-1}
a=1>0 - ветви параболы направлены вверх;
1.4)<span>y= -x^2+6x-8</span>,
x0=-6/(2*(-1))=3,
y0=-8-6^2/(4*(-1))=1;
a=-1<0 - ветви параболы направлены вниз;
2.1) <span>|x^2+5|=6x,</span>
<span>x^2+5=6x,</span>
<span>x^2-6x+5=0,</span>
<span>по теореме обратной к теореме Виета:</span>
<span>x1=1, x2=5;</span>
<span>или</span>
<span>x^2+5=-6x,</span>
<span>x^2+6x+5=0,</span>
<span>по теореме обратной к теореме Виета:</span>
<span>x1=-5, x2=-1;</span>
<span>2.2)<span>|x^2+x|+3x=5,</span></span>
<span><span><span>|x^2+x|=5-3x,</span></span></span>
<span><span><span>x^2+x=5-3x,</span></span></span>
<span><span><span>x^2+4x-5=0,</span></span></span>
<span>по теореме обратной к теореме Виета:</span>
<span>x1=-5, x2=1;</span>
<span>или</span>
<span><span><span>x^2+x=-(5-3x),</span></span></span>
<span><span><span><span><span><span>x^2+x=-5+3x,</span></span></span></span></span></span>
<span><span><span>x^2+2x+5=0,</span></span></span>
D=b^2-4ac=2^2-4*1*5=4-20=-16<0,
нет решений;
2.3) <span>(x+3)^4-13(x+3)^2+36=0,</span>
(x+3)^2=t,
t^2-13t+36=0,
<span>по теореме обратной к теореме Виета:</span>
<span>t1=4,t2=9;</span>
<span>(x+3)^2=4,</span>
<span>x^2+6x+9=4,</span>
<span><span>x^2+6x+5=0,</span></span>
<span>по теореме обратной к теореме Виета:</span>
<span>x1=-5, x2=-1;</span>
<span>или</span>
<span>(x+3)^2=9,</span>
<span>x^2+6x=0,</span>
x(x+6)=0,
x3=0, или x+6=0, x4=-6;
3) <span>3x^2-7x+2<0</span>,
<span>3x^2-7x+2=0,</span>
<span>D=25,</span>
<span>x1=1/3, x2=2,</span>
<span>(x-1/3)(x-2)<0,</span>
<span>1/3<x<2,</span>
<span>xЄ(1/3;2)
</span>
