如果以X为未知数的方程MX^2-(1-m)X+m=0有实数根，则M的取值范围是？

1.当m=0时 有实数根x=0成立
2.当m不为0的时候,方程有实数根,(1-m)＾2-4m＾2>=0
=>3m＾2+2m-1<=0
=>(3m-1)(m+1)<=0
=>-1=<m<=1/3 且M不等于零;
综合1 2得-1=<m<=1/3

delta = (m-1)^2 - 4m*m
= -3m^2 - 2m + 1
=(m+1)(-3m+1)
>= 0,

so, m<= -1 或m >=1/3

利用韦达定理
delta>=0
得到3m^2+2m-1<=0
解出-1<=m<=1/3