问一条韦达定理的数学题.

a²+1=3a,
b²+1=3b,
得出x²+1=3x,2根为a,b
x²-3x+1=0
2根之和a+b=-b/a=-(-3/1)=3
2根之积ab=c/a=(1/1)=1
(1/a²)+(1/b²)
=(a²+b²)/(a²b²),其中a²+b²=a²+2ab+b²-2ab=(a+b)²-2ab
=[(a+b)²-2ab]/(ab)²
=(3²-2)/1²
=9-2
=7