Question

解不等式(3-4x)(x的平方+x+2)分之x(x-1)(x-2)的三次方(x平方-1)(x三次方-1)大于0

Answer 1

x(x-1)(x-2)³(x+1)(x-1)(x-1)(x²+x+1)/[(3-4x)(x²+x+2)]>0
x(x-1)³(x-2)³(x+1)(x²+x+1)/[(3-4x)(x²+x+2)]>0

x²+x+1=(x+1/2)²+3/4>0
x²+x+2=(x+1/2)²+7/4>0
所以x(x-1)³(x-2)³(x+1)/(3-4x)>0
若x-1=0或x-2=0，不等式不成立
所以(x-1)²(x-2)²>0
两边除以(x-1)²(x-2)²
x(x-1)(x-2)(x+1)/(3-4x)>0
x(x-1)(x-2)(x+1)(3-4x)>0
所以x(x-1)(x-2)(x+1)(4x-3)<0
零点是-1,0,3/4,1,2
小于0
所以x<-1,0<x<3/4,1<x<2

Answer 2

结果是(-1,0)U(3/4,1)
(x的平方+x+2)恒>0，