解
∵a+b+c=0
∴方程必有一根为x=1我需要解的过程！解a+b+c=0c=-a-b∴ax^2+bx-a-b=0∴a(x^2-1)+b(x-1)=0a(x+1)(x-1)+b(x-1)=0提取(x-1)(x-1)[a(x+1)+b)]=0∴x=1