题目内容：

已知实数a满足a²-a-1=0,则a八次方+7a负四次的值为?

优质解答

a^2-a-1=0
a^2=a+1
a^4=(a+1)^2=a^+2a+1=3a+2
a^2-a=1
a-1=1/a
所以,原式=(3a+2)^2+7(a-1)^4
=9a^2+12a+4+7(a^2-2a+1)^2
=9a^2+12a+4+7(a+1-2a+1)^2
=9a^2+12a+4+7(a^2-4a+4)
=16(a^2-a+2)
=16(1+2)
=48