设f(x)=(m＋1)x²－mx＋m－1 (1)若方程f(x)=0有实根,求实数m的取值范围 (2)若不等式f

x^2-mx+m-1
=(x-m/2)^2+m-1-(m/2)^2
=(x-m/2)^2-(m/2-1)^2
=(x-1)(x-m+1)
a≤f(x)≤b
a≤x≤b存在
f(x)=0
x=1 x=m+1
m>=0
a=1 b=m+1
m