化简求值:(x+1)(x2+1)(x4+1)(x8+1)其中x=2

来源:百度知道 编辑:UC知道 时间:2024/05/31 06:54:19

:(x+1)(x^2+1)(x^4+1)(x^8+1)其中x=2 (x-1)=1
:(x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)
=(x^2-1))(x^2+1)(x^4+1)(x^8+1)
=(x^4-1))(x^4+1)(x^8+1)
=(x^8-1)(x^8+1)
=x^16-1=2^16-1

设y=(x+1)(x2+1)(x4+1)(x8+1)

当x=2时

y=(2-1)(2+1)(2^2+1)(2^3+1)(2^4+1)(2^8+1)

两边同时乘以(2-1)
(2-1)y=(2-1)(2+1)(2^2+1)(2^3+1)(2^4+1)(2^8+1)
利用平方差公式
y=2^16-1

所以(2+1)(2^2+1)(2^3+1)(2^4+1)(2^8+1)=2^16-1

前面乘以一个(x-1)
那么变成 (x^2-1)(x^2+1)(x^4+1)(x^8+1)
=(x^4-1)(x^4+1)(x^8+1)
=(x^8-1)(x^8+1)
=x^16-1

=(2+1)(2^2+1)(2^4+1)(2^8+1)
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)
=2^16-1

2^16-1