求值:(2+1)*(2^2+1)*(2^4+1)*

1个回答

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

(2+1)(2∧2+1)(2∧4+1)(2∧8+1)(2∧16+1)(2∧32+1)-2∧64求值
(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=
(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)]*[(
(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)=?
求解以下两式的值：1、（2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64