若关于x的方程x^-x+m=0 和x^-x+n=0(m,n∈R,且m≠n)

设方程x²-x+m=0两根分别为x1,x4；方程x²-x+n=0两根分别为x2,x3.
由韦达定理得
x1+x4=1 x2+x3=1
x1+x4=x2+x3
x1、x2、x3、x4成等差数列,设公差为d.
又首项为1/4,x1=1/4
所以x4=1-x1=1-1/4=3/4
x4-x1=3d=3/4-1/4=1/2
d=1/6
x2=x1+d=1/4+1/6=5/12,x3=x1+2d=1/4+1/3=7/12
m=x1x4=(1/4)*(3/4)=3/16,n=x2x3=(5/12)*(7/12)=35/144
m+n=3/16+35/144=31/72

x^-x+m=0由韦达定理得x1+x2=1 x1*x2=m
同理x^-x+n=0由韦达定理得x3+x4=1 x3*x4=n
则结合由题意可假设组成的数列为 x1 x3 x4 x2
则x1=1/4 故x2=1-1/4=3/4 公差d=(x2-x1)/3=1/6
则x3=1/4+1/6=5/12 x4=5/12+1/6=7/12
故m=1/4*3/4=3/16 n=5/12*7/12=35/144
故m+n=62/144=31/72