（2+1)(2²+1)(2^4+1)(2^8+1）

（2+1)(2²+1)(2^4+1)(2^8+1）（2^16+1）（2^32+1）的解

2个回答

（2+1)(2²+1)(2^4+1)(2^8+1）（2^16+1）（2^32+1）
=（2-1）（2+1)(2²+1)(2^4+1)(2^8+1）（2^16+1）（2^32+1）
=2^64-1

(2^2+1)*(2^4+1)*(2^8+1)*(2^16+1)*(2^32+1）
(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)
(2+1)(2^2+1)(2^4+1)(2^8+1)（2^16+1）－2^32
(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1=
[(1+2^-(1/32)]*[(1+2^-(1/16)]*[(1+2^-(1/8)]*[(1+2^-(1/4)]*[(
(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),
计算:(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)/（2^32）-1=?
（1+2^(1/32))*(1+2^(1/16))*(1+2^(1/8))*(1+2^(1/4))*(1+2^(1/2)
(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)=?
求值:(2+1)*(2^2+1)*(2^4+1)*(2^8+1)*(2^16+1)-2^32