设x1,x2是二次方程mx2+(3m-2)x+2m-2=0的两根由求根公式得:x1={-(3m-2)+根号[(3m-2)^2-4m(2m-2)]}/2m=[2-3m+根号(m-2)^2]/2m=(2-3m+|m-2|)/2mx2={-(3m-2)-根号[(3m-2)^2-4m(2m-2)]}/2m=[2-3m-根号(m-2)^2]/2m=(2-3m-|...