设f(x)=x²+2mx+2m+1
x²+2mx+2m+1=0的两个根均在区间（0,1）内
∴△=4m²-4(2m+1)≥0
f(0)=2m+1＞0
f(1)=4m+2＞0
0＜-2m/2=-m＜1
∴m∈（-1/2,1-√2]