方程x^2+4mx-4m+3=0,x^2+(m-1)x+m^2=0,x^2+2mx-2m=0中至少有一个有实数根,m的取

当△=b^2-4ac≥0时,至少有一个实数根
x^2+4mx-4m+3=0中
b=4m,a=1,c=-4m+3
(4m)^2-4*(-4m+3)≥0
16m^2+16m-12≥0
4m^2+4m-3≥0
(2m+3)(2m-1)≥0
m≤-3/2 或 m≥1/2
x^2+(m-1)x+m^2=0
中b=(m-1),a=1,c=m^2
(m-1)^2-4m^2 ≥0
m^2-2m+1-4m^2≥0
-3m^2-2m+1≥0
3m^2+2m-1≤0
(3m-1)(m+1)≤0
1/3≥m≥-1
x^2+2mx-2m=0
中b=2m,a=1,c=-2m
(2m)^2-4*(-2m)≥0
4m^2+8m≥0
m^2+2m≥0
m(m+2)≥0
m≥0或m≤-2