1)
(x^2-5x)=t;
t^2-2t-24=0;
Po teoreme obratnoy Vieta
t1=6;
t2=-4;
x^2-5x=6 or x^2-5x=-4;
x^2-5x-6=0 or x^2-5x+4=0;
1 urav) Po teoreme obratnoy Vieta
x1=-1; x2=6;
2 urav)
x=1; x=4;
2) x^3+2x^2-3x<=0;</p>
x(x^2-2x-3)<=0;</p>
po teoreme obratnoy Vieta, uravnenie v skobkah
x1=1; x2=-3;
x((x-1)(x+3))<=0;</p>
Nahodim nuli: x=0, x=1, x=-3;
Metodom intervalov
xE(-beskonechnosti; -3] & [0; 1]