Question

已知a满足等式a^2-a-1=0,求代数式a^8+7a^-4的值

Answer 1

这种题一般不要算出X，
因为它给你A^2=A+1,你就不停代，
如A^4=(A+1)^2=3A+2
A^8=.....=21A+13
A^12=A^8*A^4=....144A+89
代数式a^8+7a^-4=144A+96/3A+2
=48

Answer 2

a^2-a-1=0a^2=a+1a^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