tx^2+(1-2t)x-1/2+t=0时x的值为多少
问题描述:
tx^2+(1-2t)x-1/2+t=0时x的值为多少
答
x1=(2t-1+(1-2t)^1/2)/2t
x2=(2t-1-(1-2t)^1/2)/2t