f(x)=(x²-3x+1)/(x+1)=[(x+1)²-5(x+1)+5]/(x+1)=(x+1)-5+5/(x+1)=(x+1)+5/(x+1)-5≥2根号[(x+1)×5/(x+1)]-5=2(根号5)-5即当x+1=5/(x+1),x=(根号5)-1时,函数取得最小值2(根号5)-5所以函数的值域是[2(根号5...