(2+1)(2^2+1)(2^4+1)(2^8+1)如何用平方差公式解?
来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/22 20:25:09
(2+1)(2^2+1)(2^4+1)(2^8+1)如何用平方差公式解?(2+1)(2^2+1)(2^4+1)(2^8+1)如何用平方差公式解?(2+1)(2^2+1)(2^4+1)(2^8+1)如何
(2+1)(2^2+1)(2^4+1)(2^8+1)如何用平方差公式解?
(2+1)(2^2+1)(2^4+1)(2^8+1)如何用平方差公式解?
(2+1)(2^2+1)(2^4+1)(2^8+1)如何用平方差公式解?
(2+1)(2^2+1)(2^4+1)(2^8+1)……（2^64+1)+1
=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)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)
=(2^4-1)(2^4+1)(2^8+1)
=(2^8-1)(2^8+1)
=2^16-1
1/2,1/4,
(1-1/2^2)(1-1/3^2)(1-1/4^2).(1-1/2009^2),
(2+1)({2}^{2}+1)({2}^{4}+1)({2}^{8}+1).({2}^{64}+1)+1
巧算((2^1+1)(2^2+1)(2^4+1)(2^8+1)+1)/2^15
计算(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1
计算:(2+1)(2^2+1)(2^4+1)(2^8+1)-2^16.计算：(2+1)(2^2+1)(2^4+1)(2^8+1)-2^16.
[(1+2^-(1/32)]*[(1+2^-(1/16)]*[(1+2^-(1/8)]*[(1+2^-(1/4)]*[(1+2^-(1/2)]
(1-1/2^2)*(1-1/3^2)*(1-1/4^2)*.*(1-1/2002^2)*(1-1/2003^2)
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)(1+1/2^16),
1,2,4,4,1,( )
(2^1+1)(2^2+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+1)(2-1)(2^2+1)(2^4+1).(2^8+1)化简
(1+2)*(1+2^2)*(1+2^4)*(1+2^8)*(1+2^16)
化简(1+2)(1+2^2)(1+2^4)(1+2^8)(1+2^16)
化简(2+1)(2^2+1)(2^4+1)(2^8+1)…(2^256+1)
(2-1)(2+1)(2^2+1)(2^4)...(2^64+1)+1=?
(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^1024+1)