1/x(x+3)+1/(x+3)(x+6)+---+1/(x+27)(x+30）

化简1/x(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)+…+1/(x+96)(x+99)

化简1/x(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)+…+1/(x+96)(x+99)
要过程

newscoms1年前1

BJ123546 共回答了21个问题 | 采纳率81%

原式＝(1/3)[1/X-1/(X+3)]+(1/3)[1/(X+3)-1/(X+6)]+……+(1/3)[1/(X+96)-1/(X+99)]
=(1/3)[1/X-1/(X+3)+1/(X+3)-1/(X+6)+……+1/(X+96)-1/(X+99)]
=(1/3)[1/X-1/(X+99)]
=(1/3)[99/X(X+99)]
=33/X(X+99)

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1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+……+1/(x+2007

1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+……+1/(x+2007)(x+2008)求当X=1时,该式的值
请给出准确步骤

小白兔亚么白又白1年前7

ljbdoit 共回答了12个问题 | 采纳率100%

1/x(x+1)=1/x-1/(x+1)
1/(x+1)(x+2)=1/(x+1)-1/(x+2)
.
.
1/(x+2007)(x+2008)=1/(x+2007)-1/(x+2008)
所有式子相加得,右边两两相消
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+……+1/(x+2007)(x+2008)
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+...+1/(x+2007)-1/(x+2008)
=1/x-1/(x+2008)
=1/1-1/2009
=2008/2009

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一道今天要答案的数学题1/（x+1）-1/(x+2)-1(x+3)+1/(x+4)

dawood1年前2

吸水海棉 共回答了14个问题 | 采纳率92.9%

把整个式子同时乘于(x+1)(x+2)(x+3)(x+4)
同时除于(x+1)(x+2)(x+3)(x+4)
仔细观察分子上的算式,其实很简单的,大概写一下:分母为(x+1)(x+2)(x+3)(x+4)
分子为(x+2)(x+3)(x+4)-(x+1)(x+3)(x+4)-(x+1)(x+2)(x+4)+(x+1)(x+2)(x+3)
两两合并:比如前面两式把同样的(x+3)(x+4)可以提出来剩下(x+2)和(x+1)相减正好等于1 所以前面两式结果即为:(x+3)(x+4)=x^2+7x+12
同理由得后面两式结果为:-(x^2+3x+2) 合并一下 分子为4x+10
综:原式=4x+10/[(x+1)(x+2)(x+3)(x+4)]

1年前查看全部

计算1/x×(x+3)+1/(x+3)×(x+6)+1/(x+6)×(x+9)+1/(x+9)×(x+12)

苦口若干1年前1

老刀123 共回答了21个问题 | 采纳率90.5%

1/x×(x+3)=（1/x-1/(x+3)）/3
1/(x+3)×(x+6)=1/（1+3）x-1/(x+6)）/3
……
以此类推,原式=（1-1/(x+12)）/3=(x+11)/(3x+36)

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求解 1/X(X+3)+1/(X+5)(X+6)+……＋1/(X+99)(X+100)

狼王sailor1年前3

沉睡与穿行 共回答了15个问题 | 采纳率93.3%

1/X(X+3)+1/(X+3)(X+6)+……＋1/(X+99)(X+102)
=1/3[1/x-1/(x+3)]+1/3[1/(x+3)-1/(x+6)]+...+1/3[1/(x+99)-1/(x+102)]
=1/3[1/x-1/(x+102)]
=1/3*101/x(x+102)
=101/3x(x+102)
如果本题有什么不明白可以追问,

1年前查看全部

1/x(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)=3/2x+18

浪漫的梦魂1年前1

liii1976 共回答了18个问题 | 采纳率88.9%

1/x(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)
=[1/x-1/(x+3)+1/(x+3)-1/(x+6)+1(x+6)-1/(x+9)]/3
=[1/x-1/(x+9)]/3
=3/x(x+9)
=3/2x+18
2x+18=x(x+9)
x²+7x-18=0
(x-2)(x+9)=0
x1=2x2=-9(经检验知,不合舍去）
所以,x=2是原分式方程的解

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计算：1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+.+1/(x+2003)(x+200

计算：1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+.+1/(x+2003)(x+2004)
一道分式加法.在线等~~~~~

Google测试员7971年前1

青山君 共回答了19个问题 | 采纳率100%

[1/(x+1)-1/(x+2)]+[1/(x+2)-1/(x+3)]+[1/(x+3)-1/(x+4)]+...[1/(x+2003)-1/(x+2004)]
把相邻的抵消
剩下1/(x+1)-1/(x+2004)=2003/(x+1)(x+2004)

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1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...1/(x+2005)(x+2006)

1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...1/(x+2005)(x+2006)=1/2x+4024

玮姗1年前1

空空如也Mart 共回答了15个问题 | 采纳率86.7%

有等式1/(x+1)(x+2)=1/(x+1)-1/(x+2)
所以左边=1/(x+1)-1/（x+2）+1/(x+2)-1/(x+3).
=1/(x+1)-1/(x+2006)
后面就不用我说了吧 左右移一下统分就可以了

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计算:1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+2009)(x+2

计算:1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+2009)(x+2010)=1/2x+4020

wsvip20001年前2

xue_jiong 共回答了16个问题 | 采纳率87.5%

应该是1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+2009)(x+2010)=1/（2x+4020 ）吧
1/(x+1)-1/(x+2)+1/(x+2）-1/(x+3)+1/(x+3)-1/(x+4)+...+1/(x+2009)-1/(x+2010)=1/（2x+4020）
1/(x+1)-1/(x+2010)=1/[2（x+2010）]
2009/(x+1)(x+2010)=1/[2(x+2010)]
2009/(x+1)=1/2
x=4017

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1/(x-1)(x+1)+1/(x+1)(x+3)+1/(x+3)(x+5)+1/(x+5)(x+7)+1/(x+7)(

1/(x-1)(x+1)+1/(x+1)(x+3)+1/(x+3)(x+5)+1/(x+5)(x+7)+1/(x+7)(x+9)+1/(x+9)(x+11)
x=√3-5
化简求值

cfahydt1年前2

ef_s2_5cclx0f9_9 共回答了25个问题 | 采纳率92%

1/(x-1)(x+1)+1/(x+1)(x+3)+1/(x+3)(x+5)+1/(x+5)(x+7)+1/(x+7)(x+9)+1/(x+9)(x+11)
=(1/2)×[1/(x-1)-1/(x+1)+1/(x+1)-1/(x+3)+1/(x+3)-1/(x+5)+1/(x+5)-1/(x+7)+1/(x+7)-1/(x+9)+1/(x+9)-1/(x+11)]
=(1/2)×(1/(x-1)-1/(x+11)]
=(1/2)×(x+11-x+1)/[(x-1)(x+11)]
=6/[(x-1)(x+11)]
=6/[(√3-6)(√3+6)]
=6/(3-36)
=-6/33
=-2/11

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计算：1/x(x+3)+1/(x+3)(x+6)+...+1/(x+2004)(x+2007)

fe453c43ba0b1c221年前4

嗷嗷坏 共回答了15个问题 | 采纳率93.3%

1/x(x+3)+1/(x+3)(x+6)+...+1/(x+2004)(x+2007)
=(1/3)[1/x-1/(x+3)+1/(x+3)-1/(x+6)+---+1/(x+2004)-1/(x+2007)]
=(1/3)[1/x-1/(x+2007)]
=669/(x^2+2007x)

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1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+2008)(x+2009

1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+2008)(x+2009)

eabo1年前3

酒和MM 共回答了16个问题 | 采纳率93.8%

1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+2008)(x+2009)
=[1/(x+1)-1/(x+2)]+[1/(x+2)-1/(x+3)]+[1/(x+3)-1/(x+4)]+...+[1/(x+2008)-1/(x+2009)]
=1/(x+1)-1/(x+2009)

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1/X(X+3)+1/(X+3)(X+6)+1/(X+6)(X+9)=3/2X+18 规律是1/X(X+3)=1/X-1

1/X(X+3)+1/(X+3)(X+6)+1/(X+6)(X+9)=3/2X+18 规律是1/X(X+3)=1/X-1/(X+3)

suny2001年前1

真爱米兰 共回答了18个问题 | 采纳率88.9%

1/X(X+3)=1/X-1/(X+3)错误.
正确：1/X(X+3)=1/3[1/X-1/(X+3)]

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1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+1/(x+4)(x+5)

1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+1/(x+4)(x+5)=5/x^2+11x-708

мe℡恬1年前2

zhy3166 共回答了17个问题 | 采纳率88.2%

答：分式的裂项知识
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+1/(x+4)(x+5)=5/x^2+11x-708
1/x-1/(x+1)+1/(x+1)-1/(x+2)+.+1/(x+4)-1/(x+5)=5/(x^2+11x-708)
1/x-1/(x+5)=5/(x^2+11x-708)
(x+5-x)/(x^2+5x)=5/(x^2+11x-708)
5x^2+55x-3540=5x^2+10x
45x=3540
x=236/3
经检验,x=236/3是原分式方程的根

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计算1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3）(x+4)+……+1/(x+2007)(x+200

计算1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3）(x+4)+……+1/(x+2007)(x+2008）
快一点好吗5555……

wilianguan1年前2

cid01 共回答了14个问题 | 采纳率85.7%

=1/(x+1)-1/(x+2007)

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1/x(x+3)+1/(x+3)(x+6)+...+1/(x+15)(x+18)

zzhonghe1年前2

白白小葱葱 共回答了24个问题 | 采纳率91.7%

1/x(x+3)+1/(x+3)(x+6)+...+1/(x+15)(x+18)
=1/3[1/x-1/(x+3)]+1/3[1/(x+3)-1/(x+5)]+……+1/3[1/(x+15)-1/(x+18)]
=1/3[1/x-1/(x+3)+1/(x+3)-1/(x+5)+……+1/(x+15)-1/(x+18)]
=1/3[1/x-1/(x+18)]
=1/3[(x+18)/x(x+18)-x/x(x+18)]
=1/3[18/x(x+18)]
=6/x(x+18)

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已知两个分式：A=6/x²-9,B=1/(x+3)+1/(3-x),

已知两个分式：A=6/x²-9,B=1/(x+3)+1/(3-x),
已知两个分式：A=6/（x²-9）,B=1/(x+3)+1/(3-x),下面有三个结论：（1）A=B （2）A,B互为倒数 （3）A,B互为相反数.请问哪个正确?说明理由

心魔和魔心1年前1

jah520025 共回答了19个问题 | 采纳率89.5%

B=(3-X+X+3)/(9-X^2)=6/(9-X^2)=-A
(3)

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