已知关于X的一元二次方程X^2+(2M-1)X+M^2=0有两个实数根X1 和X2,

1.∵一元二次方程X^2+(2M-1)X+M^2=0有两个实数根X1 和X2,
∴△≥0
即：b^2 - 4ac=(2M-1)^2 - 4M^2 =-4M+1≥0
∴M≤1/4
2.x1+x2=1-2m ,x1x2=m^2
(x1-x2)^2 =(x1+x2)^2 - 4x1x2 =(1-2m)^2 - 4m^2 =1-4m
x1-x2=±√1-4m
x1^2 - x2^2 =(x1+x2)(x1-x2)=(1-2m)(√1-4m)=0或(1-2m)(-√1-4m)=0
m=1/2 或 m=1/4
∵方程有实根
∴m=1/2舍
m=1/4

解: (1) 方程有两个实根，
那么△=(2m-1)^2-4m^2>0
=-4m+1 >0
∴m<1/4
(2) x1^2-x2^2=(x1+x2)(x1-x2) =0
∴x1+x2=0 或 x1-x2=0
x1=x2
△ =(2m-1)^2-4m^2=0 , ∴ m=1/4
x1= -x2 ,
x1 *x2=m^2<=0 ∴m=0

1.
方程有两个实数根
△=(2m-1)^2-4m^2≥0
4m^2-4m+1-4m^2≥0
4m≤1
m≤1/4
2.
根据韦达定理，
x1+x2=1-2m
x1*x2=m^2
(x1-x2)^2
=(x1+x2)^2-4x1*x2
=(1-2m)^2-4m^2
=4m^2-4m+1-4m^2
=1-4m=0
m=1/4