已知a满足等式a^2-a-1=0,求代数式a^8+7a^-4的值

a^2+2a+1=(a^2+a)+a+1
又a^2=a+1
∴a^2+2a+1=3a+2
同理
=9a^2+12a+4=9a+12a+4+9=21a+13

由a^2-a-1=0得，a^2=a+1,
于是a^4=(a+1)^2=a^2+2a+1=3a+2
a^8=(3a+2)^2=9a^2+12a+4=21a+13
a^8+7a^-4=21a+13+7/(3a+2)
=(63a^2+81a+26+7)/(3a+2)
=(63a+63+81a+33)/(3a+2)
=(144a+96)/(3a+2)
=48

a^4=(a+1)^2=a^2+2a+1=3a+2

a^8=(3a+2)^2=9a^2+12a+4=21a+13

别人计算在这里的，过程写的这么清楚

a^4=(a+1)^2=a^2+2a+1=a+1+2a+1=3a+2
a^8=(3a+2)^2=9a^2+12a+4=9a+9+12a+4=21a+13