为什儿是1/3一1/3+1/5一1/7
追答2/1x3=1-1/3
2/3x5=1/3-1/5
2/5x7=1/5-1/7
2/97x99=1/97-1/99
明白了谢谢!!
追答采纳一下呗
2\/1x3十2\/3x5十2\/5x7十……2\/97x99=?
2\/1x3十2\/3x5十2\/5x7十……2\/97x99 =1-1\/3+1\/3-1\/3+1\/5-1\/7+……+1\/97-1\/99 =1-1\/99 =98\/99
2\/1X3十2\/3X5十2\/5X7……2\/99X101
原式=2×[1-(1\/3)]+2×[(1\/3)-(1\/5)]+2×[(1\/5)-(1\/7)]+……+2×[(1\/99)-(1\/101)]=2×[1-(1\/3)+(1\/3)-(1\/5)+(1\/5)-(1\/7)+……+(1\/99)-(1\/101)]=2×[1-(1\/101)]=2×(100\/101)=200\/101 ...
谁知道2\/1x3+2\/3x5+2\/5x7+2\/7x9+...+2\/97x99的答案?要用简便
您好:2\/1x3+2\/3x5+2\/5x7+2\/7x9+...+2\/97x99 =1-1\/3+1\/3-1\/5+1\/5-1\/7+1\/7-1\/9+。。。+1\/97-1\/99 =1-1\/99 =98\/99 如果本题有什么不明白可以追问,如果满意请点击右下角“采纳为满意回答”如果有其他问题请采纳本题后,另外发并点击我的头像向我求助,答题不易,请谅...
2÷1×3十2÷3Ⅹ5十2÷5×7十……2÷95Ⅹ97=请解答快解决?
=2x [ 1\/(1x3) + 1\/(3x5) +...+ 1\/(95x97) ]=[ ( 1-1\/3) +(1\/3+1\/5)+...+(1\/95-1\/97) ]=1 - 1\/97 =96\/97
用简便计算:2\/1x3+2\/5x7+2\/7x9+2\/9x11+2\/11x13+2\/13x15
2\/3=1-1\/3 2\/3x5 = 1\/3-1\/5 2\/5x7 = 1\/5-1\/7 2\/7x9 = 1\/7-1\/9 2\/9x11=1\/9-1\/11 2\/11x13= 1\/11-1\/13 以上各式左右分别相加 原式=1-1\/13=12\/13
1\/1X3十2\/3X5十2\/5×7十…十2\/97Ⅹ99十2\/9×101怎样简便?
1\/1-1\/3=2\/3=2\/1x3 1\/3-1\/5=2\/15=2\/3x5 1\/5-1\/7=2\/35=2\/5x7 ……所以,2\/1x3+2\/3x5+2\/5x7+……+2\/97x99+2\/99x101 =1\/1-1\/3+1\/3-1\/5+1\/5-1\/7+……+1\/97-1\/99+1\/99-1\/101 =1-1\/101 =100\/101 ...
1\/1X3十2\/3X5十2\/5×7十…十2\/97Ⅹ99十2\/99×101怎样算简便?
1\/(1×3)+2\/(3×5)+2\/(5×7)+…+2\/(97×99)+2\/(99×101)=1\/3+1\/3-1\/5+1\/5-1\/7+1\/7-1\/9+…+1\/97-1\/99+1\/99-1\/101 =2\/3-1\/101 =199\/303
2\/1x3+2\/3x5+2\/5x7+…+2\/2011x2013
2\/1x3+2\/3x5+2\/5x7+…+2\/2011x2013 =1-1\/3+1\/3-1\/5+1\/5-1\/7+1\/7-……-1\/2011+1\/2011-1\/2013 =1-1\/2013 =2012\/2013
1x 2分之3加3x 2分之5加5x2分之7一直规律计算到99
=[1×3+3×5+5×7+……+99×101]\/2 =[(4×1^2-1)+(4×2^2-1)+(4×3^2-1)+……+(4×50^2-1)]\/2 =[4×1^2+4×2^2+4×3^2+……+4×50^2-50]\/2 =[4×(1^2+×2^2+×3^2+……+×50^2)-50]\/2 =2×(1^2+×2^2+×3^2+……+×50^2)-25 =2...
1\/1x3+1\/3x5+1\/5x7+……1\/97x99=
首先把每一个分式拆成两项之差,即原式 1\/1x3+1\/3x5+1\/5x7+……1\/97x99= (1\/2)x(1-1\/3)+(1\/2)x(1\/3-1\/5)+(1\/2)x(1\/5-1\/7)+……+(1\/2)x(1\/97-1\/99)然后将每一项的1\/2提出来,即原式=(1\/2)x(1-1\/3+1\/3-1\/5+1\/5-1\/7+……+1\/97-1\/99)观察这个...