根号[2010×2011×2012×2013+1]/4=
展开全部
设t=2012
(t-2)(t-1)t(t+1)+1=(t-2)t(t^2-1)+1
=(t^2-2t)(t^2-1)+1
=t^4-2t^3-t^2+2t+1
=(t^4-2t^3+t^2)-(2t^2-2t)+1
=[t^2(t-1)^2]-2t(t-1)+1
=[t(t-1)-1]^2
代入即可:
根号(2012*2011-1)^2/4=(2012*2011-1)/2
(t-2)(t-1)t(t+1)+1=(t-2)t(t^2-1)+1
=(t^2-2t)(t^2-1)+1
=t^4-2t^3-t^2+2t+1
=(t^4-2t^3+t^2)-(2t^2-2t)+1
=[t^2(t-1)^2]-2t(t-1)+1
=[t(t-1)-1]^2
代入即可:
根号(2012*2011-1)^2/4=(2012*2011-1)/2
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