设m^2-3m+1=0,n^2-3n+1=0,且m≠n,求代数式1/m^2+1/n^2的值

设m^2-3m+1=0,n^2-3n+1=0,且m≠n,求代数式1/m^2+1/n^2的值
设m^2-3m+1=0,n^2-3n+1=0,且m≠n,求代数式1/m^2+1/n^2的值

设m^2-3m+1=0,n^2-3n+1=0,且m≠n,求代数式1/m^2+1/n^2的值
m^2-3m+1=0,n^2-3n+1=0
m,n是方程x^2-3x+1=0的两根
m+n=3,mn=1
1/m^2+1/n^2=(m^2+n^2)/(mn)^2=[(m+n)^2-2mn]/(mn)^2
=(9-2)/1
=7

7，要过程吗？

由题意可知m、n为方程x^2-3x+1=0的两个解，由韦达定理可知m+n=3,m*n=1.把代数式变形得1/m^2+1/n^2=(n^2+m^2)/m^2*n^2=7