解:
要求能用分解因式法求解则m≠0,且△≥0
△=[2(m+1)]²-4m(m+1)(1-m)
=4(m+1)[m+1-m(1-m)]
=4(m+1)(m+1-m+m²)
=4(m+1)(m²+1)≥0
∵m²+1>0
∴m+1≥0
m≥-1
所以m的取值范围是m≥-1且m≠0,
关于x的二次三项式mx²-2(m+1)x+(m+1)(1-m)能用分解因式求解
则mx²-2(m+1)x+(m+1)(1-m)=[mx-(m+1)][x-(1-m)]
即-m(1-m)-(m+1)=-2(m+1)
m-m^2=m+1
m^2=1
所以m=1或m=-1