(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)等于多少RT
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/24 04:41:45
x)036R&BAhcOv=]2醉A!6IEj/!g
s_l_~ϬsV'?2Ϧ|}Ov/}}ix:i[˙Kl5h>}c]B*
-t 04)gun F 1 =K
(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)等于多少RT
(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)等于多少
RT
(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)等于多少RT
思路:在原式乘上(2^2-1),不断的产生平方差,可以巧解.最后再除以3原式=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1) /3 =(2^4-1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)/3 =(2^8-1)(2^8+1)(2^16+1)(2^32+1)/3 =(2^16-1)(2^16+1)(2^32+1)/3 =(2^32-1)(2^32+1)/3 =(2^64-1)/3
(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)(1+1/2^2)(1+1/2^4)(1+1/2^8)(1+1/2^16),
计算1/(1+2)+2/(1+2^2)+4/(1+2^4)+8/(1+2^8)+16/(1+2^16)
(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)(2^2+1)(2^4+1)(2^8+1)……(2^1024+1)
(2+1)(2^2+1)(2^4+1)(2^8+1).(2^2048+1)=?
初中数学题&(1+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+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^16