已知,A=(2+1)(2^2+1)(2^4+1)(

已知,A=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^12

1个回答

(1)已知m=(2+1)*(2^2+1)*(2^4+1)*(2^8+1)*(2^16+1)*(2^32+1)*(2^64
(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),
(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)=?
计算:(2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)
(2+1)(2∧2+1)(2∧4+1)(2∧8+1)(2∧16+1)(2∧32+1)-2∧64求值
(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)+1的末位数字.
1/2+1/4+1/8+1/16+1/32+1/64简便运算＝﹙1／2＋1／4＋1／8＋1／16＋1／32＋1／64＋1
试确定(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)的末位数字.
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
试确定(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)+1的末位数字