[x] la tri tuyet doi ha phuc
khong biet phuc do hay la hoi nua.
TH1: x>=0
x^2-2006x-x+2006=0
x(x-1)-2006(x-1)=0
(x-1)(x-2006)=0
x=1 (nhan)
x=2006 (nhan)
TH2: x<0
x^2+2006x-x+2006=0
x^2+2005x+2006=0
delta=2005^2-4.2006=2005^2-4.(2005+1)=2005^2-4.2005-4=(2005-2)^2-8=2003^2-8 le qua ke no thoi (hay sai o dau do)
can (delta)~~2002<2005 nhan het
x=(-2005+-can(delta)/2