x3+x2−4x−4x3+6x2+5x−12>0.
1) x3+x2−4x−4=0:
2) x3+6x2+5x−12=0:
3) (x+2)(x+1)(x−2)(x+4)(x+3)(x−1)>0
x<−4; −3<x<−2;\ \ \
−1<x<1; \ \ \ x>2.
ОтветОтвет: