查人人中国名人网连接器模具零配件:数学题目
来源：百度文库 编辑：查人人中国名人网 时间：2024/04/29 09:57:47
计算
(2+1)(2^2+1)(2^4+1)(2^8+1)+5
------------------------------(分数线)
2^14+1
(2+1)(2^2+1)(2^4+1)(2^8+1)+5
------------------------------(分数线)
2^14+1
原式=[(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)+5 ]/(2^14+1)
=[(2^2-1)(2^2+1)(2^4+1)(2^8+1)+5 ]/(2^14+1)
=……
=(2^16-1+5)/(2^14+1)
=[2^(14+2)+4]/(2^14+1)
=4(2^14+1)/(2^14+1)
=4
上下同时乘以2-1 则上面=2^16-1+5=4*(2^14+1)
则原试=4
分子乘上(2-1)得:
(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)+5
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)+5
=(2^4-1)(2^4+1)(2^8+1)+5
=(2^8-1)(2^8+1)+5
=2^16-1+5
=4(2^14+1)
所以原式=4
记A=(2+1)(2^2+1)(2^4+1)(2^8+1)+5
B=2^14+1
则 A=(2-1)A
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)+5
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)+5
=(2^4-1)(2^4+1)(2^8+1)+5
=(2^8-1)(2^8+1)+5
=2^16-1+5
=2^16+4
=4B
故A/B=4