数学简算1\/2*3+1\/3*4+1\/4*5+...1\/99*100
解:原式=1\/2-1\/3+1\/3-1\/4……-1\/99+1\/99-1\/100=49\/100 注:1\/2×3=1\/2-1\/3 不信你试试 具体的考察一般情形 1\/n(n+1)=1\/n - 1\/(n+1)可证明 但如果是小学的话 只能找上述规律
简算1\/2*3+1\/3*4+1\/4*5+...+1\/18*19+1\/19*20
1\/(n(n+1))=1\/n-1\/(n+1)以此类推 1\/2*3+1\/3*4+1\/4*5+...+1\/18*19+1\/19*20 =1\/2-1\/3+1\/3-1\/4+...+1\/18-1\/19+1\/19-1\/20 =1\/2-1\/20 =9\/20
1×1\/2+2×1\/3+3×1\/4+5×1\/4+,加99×1\/100等于多少?
1\/1*2+1\/2*3+1\/3*4+...+1\/99*100 =(1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+...+(1\/99-1\/100)=1-1\/100 =99\/100 简便计算是一种特殊的计算,它运用了运算定律与数字的基本性质,从而使计算简便,使一个很复杂的式子变得很容易计算出得数。
1*2+1\/2*3+1\/3*4···+1\/99*100
要算式思路 1\/1*2+1\/2*3+1\/3*4···+1\/99*100 =1\/(1×2)+1\/(2×3)+...+1\/(99×100) =(1-1\/2)+(1\/2-1\/3)+...+(1\/99-1\/100) =1-1\/2+1\/2-1\/3+...+1\/99-1\/100 =1-1\/100 =99\/100 类似:1\/[a*(a+n)]=(1\/n)*[1\/a-1\/(a+n)] ...
1*2+1\/2*3+1\/3*4+1\/4*5...+1\/99*100怎样简算?
=1-1\/2+1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+...-1\/99+1\/99-1\/100 =1-1\/100 =99\/100 1又2分之1-6分之5+12分之7-20分之9+30分之11-42分之13+56分之15 =1+1\/2-1\/2-1\/3+1\/3+1\/4-1\/4-1\/5+1\/5+1\/6-1\/6-1\/7+1\/7+1\/8 =1+1\/8 =9\/8 ...
1*1\/2+1\/2*1\/3+1\/3*1\/4+...1\/99*1\/100简算
运用拆分法做这个题目,即把每一项都拆成两个分数相减的形式.望你掌握,以后会用到,早高三 1\/1*2+1\/2*3+1\/3*4+...1\/99*100==(1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+...+(1\\99-1\\100)=骸礌汾啡莴独风扫袱激1-1\/100=99\/100 ...
简便计算:1\/2*3+1\/3*4+1\/4*5……+1\/98*99+1\/99+100急急急,快的加二十...
第一道题:=1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+、、、+1\/98-1\/99+1\/99+100=1\/2+100=100.5 第二道题:(1901+1999)+(1902+1998)+、、、+(1949+1951)+1950 =3900+3900+、、、+3900+1950 =175500+1950=177450
1\/1×2+1\/2×3+1\/3×4+1\/4×5...1\/98×99+1\/99×100
1\/1×2+1\/2×3+1\/3×4+1\/4×5...1\/98×99+1\/99×100 =1-1\/2+1\/2-1\/3+1\/3-1\/4+……+1\/99-1\/100 =1-1\/100 =99\/100
1*2*3分之1+2*3*4分之1+3*4*5分之1+...+8*9*10分之1(简算)
本题可以设辅助未知数来解;设m=1\/2+1\/3+1\/4,n=1\/2+1\/3+1\/4+1\/5,则n-m=1\/5,∴原式=(1+m)×n-(1+n)×m=n+mn-m-mn=n-m=1\/5.
...之一加二乘三分之一加三乘四分之一省略号加九十九乘一百分之一简算...
1\/2+2\/3+3\/4+4\/5……+99\/100 =1-1\/2+1-1\/3+1-1\/4+1-1\/5……+1-1\/100 =99-(1\/2+1\/3+1\/4+1\/5+……+1\/100)=99-(99*100-1)\/(99*100)=98又9899\/9900