f(x)=根号下xf(x2-2x+1)=√(x-1)^20.5=√0.25f(x2-2x+1)<0.5即(x-1)^2<0.25（x-1-0.5）(x-1+0.5)<0(x-1.5)(x-0.5)<00.5
f(x2-2x+1)=f[(x-1)^2]=│x-1│………(x-1)^2表示(x-1)的平方原不等式即│x-1│<0.5 解得0.5
