1/1*2*3*4+1/2*3*4*5+1/3*4*5*6+1/4*5*6*7+1/5*6*7*8+1/6*7*8*9+1/7*8*9*10=

1\/1*2+1\/2*3+1\/3*4+1\/4*5+1\/5*6+1\/6*7+1\/7*8+1\/8*9+1\/9*10用简便方法...
1\/(x(x+1))=1\/x-1\/(x+1)所以，原等式就可以简化为1\/1-1\/10=0.9

1\/1*2*3*4+1\/2*3*4*5+1\/3*4*5*6+...+1\/7*8*9*10
题目有误，应为：5\/1*2*3*4+7\/2*3*4*5+9\/3*4*5*6+...+17\/7*8*9*10 =1\/1*3-1\/2*4+1\/2*4-1\/3*5+1\/3*5-1\/4*6+...+1\/7*9-1\/8*10 = 1\/1*3-1\/8*10 =1\/3-1\/80 =77\/240

1\/1×2×3×4×5 + 1\/2×3×4×5×6 + 1\/3×4×5×6×7 + 1\/4×5...
解：1\/[n(n+1)(n+2)(n+3)(n+4)]=¼×[(n+4)-n]\/[n(n+1)(n+2)(n+3)(n+4)]=¼×{ 1\/[n(n+1)(n+2)(n+3)] -1\/[(n+1)(n+2)(n+3)(n+4)]} 1\/(1×2×3×4×5)+ 1\/(2×3×4×5×6)+1\/(3×4×5×6×7)+ 1\/(4×5×6×7×8...

1\/1*2+2*3+3*4+4*5+5*6+6*7+7*8+8*9+9*10
1\/1*2=1\/1-1\/2 1\/2*3=1\/2-1\/3 ……1\/9*10=1\/9-1\/10 以上全部左右各相加得：1\/1*2+1\/2*3+1\/3*4+1\/4*5+1\/5*6+1\/6*7+1\/7*8+1\/8*9+1\/9*10 =1\/1-1\/2 + 1\/2-1\/3+……+1\/9-1\/10 =1-1\/10=9\/10 ...

1*2*3+2*3*4+3*4*5+4*5*6+5*6*7+6*7*8+7*8*9+8*9*10=
因为4(n-1)n(n+1)=(n-1)n(n+1)(n+2)-(n-2)(n-1)n(n+1)所以 1*2*3+2*3*4+3*4*5+...+n(n+1)(n+2)=n(n+1)(n+2)(n+3)÷4 1*2*3+2*3*4+3*4*5+4*5*6+5*6*7+6*7*8+7*8*9+8*9*10 =8×9×10×11\/4 =1980 ...

1乘2乘3乘4分之一加2乘3乘4乘5分之一加3乘4乘5乘6分之一加等等加17乘...
1\/1*2*3*4+1\/2*3*4*5+1\/3*4*5*6+...+1\/7*8*9*10 =（1\/1*2*3-1\/2*3*4+1\/2*3*4-1\/3*4*5+1\/3*4*5-1\/4*5*6+...+1\/7*8*9-1\/8*9*10 ）÷3 =（1\/1*2*3-1\/18*19*20）÷3 =（1\/6-1\/6840）÷3 =1139\/6840÷3 =1139\/20520 就是裂项 比如1\/...

一除以一乘二乘三的积加上一除二乘三乘四一路下去直到一除以八乘九...
原式=1÷（1×2×3）+1÷（2×3×4）+1÷（3×4×5）+1÷（4×5×6）+1÷（5×6×7）+1÷（6×7×8）+1÷（7×8×9）+1÷（8×9×10）希望我的答案能够帮到你~数不胜数 团队竭诚为您服务

1\/2×4+1\/3×5+1\/4×6+1\/5×7+……+1\/8×10+1\/9×11 数学题
n+2)即算式中每个数都可写作1\/n(n+2) 其中n=2，3，4，5，6，7，8，9 因2\/n(n+2)=（n+2-n)\/（n(n+2)）=(1\/n)-(1\/（n+2 ）)所以原式=（1\/2-1\/4+1\/3-1\/5+1\/4-1\/6+...+1\/9-1\/11)\/2 =(1\/2+1\/3-1\/10-1\/11)\/2 =53\/165 ...

1\/1*2x+1\/2*3x+1\/3*4x+1\/4*5x+1\/6*7x+1\/7*8x+1\/8*9x+1\/9*10x=1\/10...
按费拉里公式，原方程可拆分为两个方程 2x²+.55405383961193x+9.0521839270015=0 2x²-4.55405383961193x+2.2094115808167=0 求解得：x1= 0.70089820882972 x2= 1.57612871097624 x3= -0.138513459902982-2.12294747578136i x4= -0.138513459902982+2.12294747578136i 希望有帮助，不清楚请追问...

数列计算 1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9=?
1＋1\/2＋1\/3＋1\/4＋ … ＋1\/n 这个级数是发散的。简单的说，结果为∞ －－－ 用高中知识也是可以证明的，如下：1\/2≥1\/2 1\/3＋1\/4＞1\/2 1\/5＋1\/6＋1\/7＋1\/8＞1\/2 ……1\/[2^(k-1)+1]＋1\/[2^(k-1)+2]＋…＋1\/2^k＞[2^(k-1)](1\/2^k)＝1\/2 对于任意一...