已知实数a满足a2-a-1=0,求a8+7a-4的值
问题描述：
已知实数a满足a2-a-1=0,求a8+7a-4的值
2 8 -4 为次方
答
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
2.
a^2-a-1=0
a^2=a+1
a^8+7a^-4
=a^6*a^2+7a^2-4
=a^6(a+1)+7a^2-4
=a^7+a^6+7a^-4
=a^5(a+1)+a^6+7a^2-4
=2a^6+a^5+7a^-4
=3a^5+2a^4+7a^-4
=5a^4+3a^3+7a^-4
=8a^3+5a^2+7a^-4
=13a^2+8a+7a^-4
=21a+13+7a^-4
=34+21a^-1+7a^-4
=34+21a^-1+7a^-2-7a^-3
=34+14a^-1+14a^-2
=48