设6x^2-3x-2=0的两个根是x1,x2倒数就是1/x1,1/x2由韦达定理得x1+x2=1/2,x1x2=-1/3设所求一元二次方程为ax^2+bx+c=0韦达定理得1/x1+1/x2=(x1+x2)/x1x2=-3/2=-b/a1/x1*1/x2=1/x1x2=-3=c/a令a=2,则b=3,c=-6∴所求方程为...