1)∵f(x)在x=t/2+2,取得最小值y=-t^2/4 ∴f(x)=[x-(t/2+2)]^2-t^2/4∵f(1)=0 ∴将x=1代入,得：t=-1 ∴f(x)=[x-(t/2+2)]^2-t^2/4=(x-3/2)^2-1/42)①当x=t/2+2∈[-1,1/2]时解得：t∈[-6,-3],最小值y=-t^2/4=-5,解得t=-...