已知实数a满足a²-5a+1=0,求a²/a4+a²+1的值

已知实数a满足a²-5a+1=0,求a²/a4+a²+1的值
已知实数a满足a²-5a+1=0,求a²/a4+a²+1的值

已知实数a满足a²-5a+1=0,求a²/a4+a²+1的值
a^2=5a-1,则a^4=25a^2-10a+1
a4+a²+1=25a^2-10a+1+a^2+1=24a^2+2(a^2-5a+1)=24a^2
原式=a^2/24a^2=1/24

a²-5a+1=0
即a²+1=5a
∴ a+1/a=5
即(a+1/a)²=25
∴ a²+1/a²+2*a*(1/a)=25
∴ a²+1/a²=23
∴ a²/(a^4+a²+1)
=1/(a²+1/a²+1)
=1/24

a^2+1=5a,
a^2=(5a-1)=>a^4=25a^2-10a+1=25(5a-1)-10a+1=115a-24
a4+a²+1=120a-24=24(5a-1)=24a^2所以a²/a4+a²+1=a^2/24a^2=1/24