4月3号前告我
问题描述:
4月3号前告我
已知a满足等式a^2-a-1=0,求代数式a^8+7a^-4的值.
答
a^2=a+1a^4=(a^2)^2=(a+1)^2=a^2+2a+1=(a+1)+2a+1=3a+2a^8=(a^4)^2=(3a+2)^2=9a^2+12a+4=9(a+1)+12a+4=21a+13a^12=a^8*a^4=(3a+2)(21a+13)=63a^2+81a+26=63(a+1)+81a+26=144a+89a^8+7a^-4=a^8+7/a^4=(a^12+7)/a^4=(1...