一道初一整式的题已知a的2次方减a减1等于0,求a的18次方加上323a的-6次方的值.

Question

一道初一整式的题
已知a的2次方减a减1等于0,求a的18次方加上323a的-6次方的值.

Answer 1

a^2-a-1=0
a^2=a+1
a^4=(a^2)^2=a^2+2a+1=(a+1)+2a+1=3a+2
a^8=(a^4)^2=(3a+2)^2=9a^2+12a+4=9(a+1)+12a+4=21a+13
a^16=(a^8)^2=(21a+13)^2=441a^2+546a+169=441(a+1)+546a+169=987a+610
a^24=a^8*a^16=(21a+13)(987a+610)=20727a^2+25641a+7930=20727(a+1)+25641a+7930=46368a+28657
a^6=a^2*a^4=(a+1)(3a+2)=3a^2+5a+2=3(a+1)+5a+2=8a+5
所以a^18+323a^-6
=a^18+323/a^6
=(a^24+323)/a^6
=(46368a+28657+323)/(8a+5)
=5796(8a+5)/(8a+5)
=5796