已知a满足等式a²-a-1=0,求代数式a的八次方加7a的负四次方的值
问题描述：
已知a满足等式a²-a-1=0,求代数式a的八次方加7a的负四次方的值
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
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