f(x)=x²+2mx+m²-1/2m-3/2=(x+m)²-1/2m-3/2开口向上，对称轴为x=-m若m>=0, 则f(x)在x>0单调增，f(0)=m²-1/2m-3/2>=0即可，得：2m²-m-3>=0, (2m-3)(m+1)>=0, 得：m>=3/2若m<0, 则f(-m)=-1/2m-3/2为最小值，则须有-1/2m-3/2>0, 得：m<-3综合得m的取值范围是：m>=3/2, 或m<-3
