Question

如何解(x-1)(x的平方-x +1)>0,麻烦讲的详细点

Answer 1

(x-1)(x^2-x+1)>0

由于x^2-x+1=x^2-x+1/4+3/4=(x-1/2)^2+3/4,可知(x-1/2)^2≥0,则x^2-x+1=(x-1/2)^2+3/4>0,所以

(x-1)>0

解得：x>1.

由于x^2-x+1=x^2-x+1/4+3/4=(x-1/2)^2+3/4>0,

不论x取任意数,x^2-x+1永远都大于0,所以不必解这个不等式.