∵关于x的方程x^2+2(1+m)x+(3m^2+4mn＋4n^2+2)=0有实数根
∴方程x^2+2(1+m)x+(3m^2+4mn＋4n^2+2)=0的△≥0
即[2(1+m)]^2-4*1*(3m^2+4mn＋4n^2+2)≥0
∴2m^2-2m+4mn+4n^2+1≤0
∵2m^2-2m+4mn+4n^2+1=m^2+4mn+2n+m^2-2m+1=（m+2n)^2+(m-1)^2
∴（m+2n)^2+(m-1)^2≤0
∴m+2n=0且m-1=0
∴m=1,n=-1/2