已知关于x的函数y=x+（2t+1）x+t-1,当t取何值时,y的最小值是0?如题.急.

y = x^2 + (2t+1)x + t^2 -1 = [x + (2t + 1)/2]^2 + t^2 - 1 - (2t+1)^2/4 = [x + (2t + 1)/2]^2 +[ 4t^2 - 4 - (2t+1)^2] / 4 = [x + (2t + 1)/2]^2 +[ 4t^2 - 4 - 4t^2 - 4t -1] / 4 = [x + (2t + 1)/2]^2 +[- 4t -5] / 4 = [x + (2t + 1)/2]^2 - (4t + 5) / 4 y min = - (4t + 5) / 4 = 0 ==> 4t + 5 = 0 ==> t = -5/4