初一数学完全平方和平方差公式 2题
(2x+3y-5)(2x-3y+1)(2^2+1)(2^4+1)(2^8+1)..(2^32+1)请教数学达人帮我解答哦,谢谢...
(2x+3y-5)(2x-3y+1)
(2^2+1)(2^4+1)(2^8+1)..(2^32+1)
请教数学达人帮我解答哦,谢谢 展开
(2^2+1)(2^4+1)(2^8+1)..(2^32+1)
请教数学达人帮我解答哦,谢谢 展开
展开全部
(2x+3y-5)(2x-3y+1)=[2(x-1)+3(y-1)][2(x-1)-3(y-1)]
=[2(x-1)]^2-[3(y-1)]^2=4(x-1)^2-9(y-1)^2
(2^2+1)(2^4+1)(2^8+1)..(2^32+1)=1/(2^2-1)*(2^2-1)(2^2+1)(2^4+1)(2^8+1)..(2^32+1)=1/3*(2^4-1)(2^4+1)(2^8+1)..(2^32+1)=.....
=1/3*(2^64-1)
=[2(x-1)]^2-[3(y-1)]^2=4(x-1)^2-9(y-1)^2
(2^2+1)(2^4+1)(2^8+1)..(2^32+1)=1/(2^2-1)*(2^2-1)(2^2+1)(2^4+1)(2^8+1)..(2^32+1)=1/3*(2^4-1)(2^4+1)(2^8+1)..(2^32+1)=.....
=1/3*(2^64-1)
本回答被提问者采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询