×^2-×+3≥0配方:(x-1/2)^2+11/4≥0由于平方数非负,11/4>0因此实际上在实数范围内,(x-1/2)^2+11/4>0恒成立则所求解为x∈R
x²-x+3≥0(x-1/2)²-1/4+3≥0(x-1/2)²+2又4分之3≥0因为:(x-1/2)²≥0恒成立,所以:(x-1/2)²+2又4分之3≥0恒成立x的解集是:(-∞,+∞),x∈R
(X-1)x+3没可能等于0