方程mx*x-nx+2=0两根相等
判别：n^2-8m=0,n^2=8m>0,m>0.1)
方程x*x-4mx+3n=0,假设x1,x2,x2=3x1
判别(-4m)^2-12n>0,4m^2-3n>0.2)
4x1=4m,x1=m,3x1^2=3n,x1^2=n>0
n=m^2,m^2-8m=0,m=8
n=m^2=64
x*x-(k+n)x+(k-m)=0
x^2-(k+64)+k-8=0
判别(k+64)^2-4(k-8)=(k+62)^2+284>0
所以：x*x-(k+n)x+(k-m)=0一定有实数根