∵方程x²+mx+1=0有两个不相的负实数根,∴x1+x2=-m0,△=m^2-4>0,m>2或m0,∴m>2
:4x²+4(m-2)x+1=0,无实数根∴△=16(m-2)^2-16x1+x2=-m<0,m怎么>0