1/(x+1)(x+2)+1/(x+2)(x+3).1/(x+2006)(x+2007)=?

feidie0022022-10-04 11:39:542条回答

已提交,审核后显示!提交回复

共2条回复
yaomin929 共回答了15个问题 | 采纳率86.7%
1/(x+1)(x+2)=1/(x+1)-1/(x+2)
以后各项也这样拆开,会了吧
1年前
rao_james 共回答了3个问题 | 采纳率
1/(x+1)-1/(x+2007)
1年前

相关推荐

解方程:1/(x+2)+1/(x-2)=1 3x/(x²-1)+(x²-1)/3x=5/2
adrqwes1年前3
wf8421391 共回答了24个问题 | 采纳率95.8%
1)两边同时乘以(x+2)(x-2)
x-2+x+2=x^2-4
x^2-2x-4=0
x=1±√5
经检验,它们都是原方程的根
2) 令t=3x/(x^2-1)
方程化为t+1/t=5/2
t^2+5t/2+1=0
(t+2)(t+1/2)=0
t=-2,-1/2
当t=-2时,3x/(x^2-1)=-2,得2x^2+3x-2=0,(2x-1)(x+2)=0,得x=1/2,-2
当t=-1/2时,3x/(x^2-1)=-1/2,得:x^2+6x-1=0,得x=-3±√10
经检验,以上4个都为原方程的根.
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
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+.+1/(x+2004)(x+2005)
wcpoaiui1年前1
极光精灵 共回答了21个问题 | 采纳率85.7%
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+.+1/(x+2004)(x+2005)
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+-----+(1/(x+2004)-1/(x+2005)
=1/x-1/(x+2005)
=2005/(x^2+2005x)
用裂项相消法
1/(x+n)(x+n+1)=1/(x+n)-1/(x+n+1)
已知两个分式A=4/(x^2-4),B=1/(x+2)+1/(2-x),其中x≠正负2
已知两个分式A=4/(x^2-4),B=1/(x+2)+1/(2-x),其中x≠正负2
:①A=B②A、B互为倒数③A、B互为相反数.请问哪个正确?为什么?
下面有三个结论
dzy51年前3
shinchou 共回答了17个问题 | 采纳率82.4%
A=4/(x^2-4)
=4/(x+2)(x-2)
B=1/(x+2)+1/(2-x)
= [(x-2)-(x+2)]/(x+2)(x-2)
=-4/(x+2)(x-2)
因此A和B 互为相反数
第十七天 最后一题1/x(x+2)+1/(x+1)(x+3)+1(x+2)(x+4).+1/(x+2007)+1/(x+
第十七天 最后一题
1/x(x+2)+1/(x+1)(x+3)+1(x+2)(x+4).+1/(x+2007)+1/(x+2007)(x+2009)
第三个打错了1/(x+2)(x+4)
dww3811年前1
19740307 共回答了12个问题 | 采纳率91.7%
1/X(X+2)=1/2(1/X-1/X+2);都这样化简(分母相差都为2),最后得1/2(1/X-1/(X+2009))=2008/2X(X+2009)=1004/X(X+2009)
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+.+1/(x+999)(x+1000)
yingguo07681年前3
dql_dql 共回答了17个问题 | 采纳率94.1%
1/x(x+1)=1/x-1/(x+1)
1/(x+1)(x+2)=1/(x+1)-1/(x+2)
...
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+.+1/(x+999)(x+1000)
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+...+1/(x+999)-1/(x+1000)
=1/x-1/(x+1000)=1000/x(x+1000)
求1/(x+1)(x+2)+1/(x+3)(x+4)+...+1/(x+1999)(x+2000)的值
54551121年前1
burstflame 共回答了16个问题 | 采纳率87.5%
=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)+1/(x+4)-1/(x+5)...+1/(x+1998)-1/(x+1999)+1/(x+1999)-1/(x+2000)
=1/(x+1)-1/(x+2000)
=1999/(x+1)(x+2000)
=1999/(x^2+2001x+2000)
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)
解方程:1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+9)(x+10)=5/1
解方程:1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+9)(x+10)=5/12 急
78651431年前4
ZHT1201 共回答了22个问题 | 采纳率95.5%
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+9)(x+10)=5/12
1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+...+1/(x+9)-1/(x+10)=5/12
1/x-1/(x+10)=5/12
10/x(x+10)=5/12
所以x²+10x-24=0
(x+12)(x-2)=0
x=-12,x=2
经检验,x=-12和x=2是方程的解
计算1/x(x+2)+1/(x+2)(x+4)+1/(x+4)(x+6)+...1/(x+2004)(x+2006)
我该去哪里1年前2
stock_beginner 共回答了20个问题 | 采纳率85%
1/x(x+2)+1/(x+2)(x+4)+1/(x+4)(x+6)+...1/(x+2004)(x+2006)
=1/x(x+2)+{1/(x+2)-1/(x+4)+1/(x+4)-1/(1+6)+...+1/(x+2004)-1/(x+2006)}/2
=1/x-1/(x+2)+{1/(x+2)-1/(x+2006)}/2
=1/x-1/(2x+4)-1/(2x+4012)
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
化简:1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+8)(x+9)
霍夫蔓1年前2
jdyb 共回答了24个问题 | 采纳率87.5%
化简:
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+8)(x+9)
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+...+1/(x+8)-1/(x+9)
=1/x-1/(x+9)
=9/x(x+9)
化简:1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+.+1/(x+2013)(x+2014)
脑波控制1年前1
lizhangpeng 共回答了21个问题 | 采纳率71.4%
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+.+1/(x+2013)(x+2014)
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+.+1/(x+2013)-1/(x+2014)
=1/x-1/(x+2014)
=2014/x(x+2014)
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
1/X(X+1)+1/(X+1)(X+2)+1/(X+2)(X+3)其中X的平方+3X+1=0
solky1年前1
xb0431 共回答了20个问题 | 采纳率95%
第一题
由X^2+3X+1=0,得X^2=-3X-1代入,
1/(X+1)(X+2)=1,故原式=1/X(X+1)+1+1/(X+2)(X+3)=
1+1/(-1-2X)+1/(2X+5)通分,原式=1+4/-(2X+5)(2X+1)=
1+4/-[4(X^2+3X+1)+1]=1-4=-3
第二题
移项,1/(x-10)-1/(x-7)=1/(x-9)-1/(x-6),通分得
3/(x^2-17x+70)=3/(x^2-15x+54),即(x^2-17x+70)=(x^2-15x+54),
得x=8
化解求值1/x(x+2)+1/(x+2)(x+4)+1/(x+4)(x+6)+```+1/(x+20)(x+22)
秋天打老虎1年前1
挖土机 共回答了12个问题 | 采纳率91.7%
1/x(x+2)=(1/2)(1/x-1/(x+2))
……
累加
=1/2(1/x-1/(x+22))
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)
找规律计算:1/x(x+2)+1/(x+2)(x+4)+1/(x+4)(x+6)+……1/(x+2004)(x+2006
找规律计算:1/x(x+2)+1/(x+2)(x+4)+1/(x+4)(x+6)+……1/(x+2004)(x+2006)
vavol51年前1
寒云烟 共回答了21个问题 | 采纳率90.5%
1/x(x+2)+1/(x+2)(x+4)+1/(x+4)(x+6)+……1/(x+2004)(x+2006)
=1/2[1/x-1/(x+2)]+1/2[1/(x+2)-1/(x+4)]+1/2[1/(x+4)-1/(x+6)]+……+1/2[1/(x+2004)-1/(x+2006)]
=1/2[1/x-1/(x+2)+1/(x+2)-1/(x+4)+1/(x+4)-1/(x+6)+……+1/(x+2004)-1/(x+2006)]
=1/2[1/x-1/(x+2006)]
=1003/x(x+2006)
1/X(X+2)+1/(X+1)(X+3)+1\(X+2)(X+4)+.+1/(X+2007)(X+2009)
1/X(X+2)+1/(X+1)(X+3)+1(X+2)(X+4)+.+1/(X+2007)(X+2009)
要详解
s48610021年前1
sharen302 共回答了17个问题 | 采纳率82.4%
1/X(X+2)+1/(X+1)(X+3)+1(X+2)(X+4)+.+1/(X+2007)(X+2009)
=(1/2)(1/x-1/(x+2)+1/(x+2)-1/(x+4)+1/(x+4).1/(x+2007)-1/(x+2009))
=(1/2)*(1/2-1/(x+2009))
=(x+2007)/4(x+2009)
A=4/X²-4,B=1/(X+2)+1/(2-X) X≠±2
A=4/X²-4,B=1/(X+2)+1/(2-X) X≠±2
则1·A=B 2·AB互为倒数 3·AB互为相反数
哪个对,为啥?
长发金牙1年前2
三连音 共回答了15个问题 | 采纳率100%
帮帮帮帮帮帮帮帮手.解方程:1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+2005)(x+200
帮帮帮帮帮帮帮帮手.
解方程:
1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+2005)(x+2006)=1/2x+4012
圈养uu各种LOLI1年前3
564808244 共回答了18个问题 | 采纳率94.4%
1/(x+1)(x+2)=1/(x+1)-1/(x+2)
1/(x+2)(x+3)=1/(x+2)-1/(x+3)
.
...
1/(x+2005)(x+2006)=1/(x+2005)-1/(x+2006)
不难得出 1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+2005)(x+2006)=1/(x+1)-1/(x+2006)=1/2x+4012
再解方程...
解方程:1/(x+1)(x+2)+1/(x+2)(x+3)...+1/(x+99)(x+100)+1/(x+100)=1
解方程:1/(x+1)(x+2)+1/(x+2)(x+3)...+1/(x+99)(x+100)+1/(x+100)=1999/2000
c52320591年前4
Yahgoro 共回答了13个问题 | 采纳率92.3%
1/(x+a)(x+a+1)=1/(x+a)-1/(x+a+1)
1/(x+1)(x+2)+1/(x+2)(x+3)...+1/(x+99)(x+100)+1/(x+100)
=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+……+1/(x+99)-1/(x+100)+1/(x+100)
=1/(x+1)=1999/2000
x=1/1999
解分式方程:1/x(x+2)+1/(x+2)(x+4)-1/2x=1
hazeoob1年前1
我长得不帅 共回答了19个问题 | 采纳率94.7%
解分式方程:1/x(x+2)+1/(x+2)(x+4)-1/2x=1
两边同乘以2x(x+2)(x+4)得:2(x+4)+2x-(x+2)(x+4)=2x(x+2)(x+4)
展开得2x+8+2x-(x²+6x+8)=2x(x²+6x+8)
-x²-2x=2x(x²+6x+8)
2x(x²+6x+8)+x(x+2)=0
x(2x²+13x+18)=0
x(2x+9)(x+2)=0
故x₁=0(增根,舍去);x₂=-9/2; x₃=-2(增根,舍去).
1/(x^2+5x+6)=1/(x+2)+1/(x+3)解方程
byhaini1年前1
yikchin88 共回答了23个问题 | 采纳率87%
x+3+x+2=1 2x=-4 x=-2 又因为x=-2时,分式无意义,所以该方程无解.
这题我不会,化简 1/(x+1)(x+2)+1/(x+2)(x+3)+……+1/(x+9)(x+10).
任性的生活1年前2
每只每条 共回答了17个问题 | 采纳率88.2%
原式=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+……+1/(x+9)=1/(x+10)=1/(x+1)-1/(x+10)=9/(x+1)(x+10).
裂项相消,中间项全消了,只剩第一项减去最后一项,完成化简
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
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)
如果解方程2x/(x+1)=a/(x+2)+1/(x^2+3x+2)时,会产生增根,求A的值?
要ggID1年前1
zhuangliang1984 共回答了17个问题 | 采纳率100%
2x/(x+1)=a/(x+2)+1/[(x+2)(x+1)]
方程两边同时乘以(x+2)(x+1)得:2x(x+2)=a(x+1)+1
增根只可能为-1或-2
x=-1代入上式得:-2(-1+2)=1, 不符
x=-2代入上式得:0=-a+1, 得:a=1
因此只能为a=1
解方程1/(x+)(x+2)+1/(x+2)(x+3)+...1/(x+99)(x+100)+1/x+100=2009/
解方程1/(x+)(x+2)+1/(x+2)(x+3)+...1/(x+99)(x+100)+1/x+100=2009/2010
帮帮忙,明天就交了,谢谢了~
ttffrr1年前3
茸哥 共回答了25个问题 | 采纳率88%
1/(x+1)(x+2) 这个等于1/(x+1)-1/(x+2)
依次类推.
1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+.+1/(x+99)-1/(x+100)+1/(x+100)=2009/2010;
所以1/(x+1)=2009/2010;
解得x=1/2009
计算1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3).+1/(x+2003)(x+2004)
计算1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3).+1/(x+2003)(x+2004)
求全过程 谁能告诉我1/x-1/(x+1)是什么意思?怎么表达,是不是两个分数线?
zhengke441年前1
虞美人m 共回答了18个问题 | 采纳率94.4%
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3).+1/(x+2003)(x+2004)
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3).+1/(x+2003)-1/(x+2004)
=1/x-1/(x+2004)
=2004/(x²+2004x)
1/x-1/(x+1) 是分式1/x与1/(x+1)的差.
=(x+1-x)/x(x+1)
=1/x(x+1)
求方程1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+1994)(x+1995)的解
vnionville1年前1
相思鸟555 共回答了14个问题 | 采纳率92.9%
等号在哪啊?这是方程么?
如果要是这一长串式子化简的话,解法如下:
首先:给你个公式1/a*(a+1)=1/a-1/(a+1)
因此1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+1994)(x+1995)=
1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+...+1/(x+1994)-1/(x+1995)=
1/(x+1)-1/(x+1995)(中间的一减一加都约掉了)=1994/(x+1)*(x+1995)(通分了)
如果要有方程的话,看看你后面是等于什么,应该可以自己解决了吧
解方程:1/x+1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+……+1/(x+9)(x+10)=
解方程:1/x+1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+……+1/(x+9)(x+10)=0
lzh75531年前1
77730 共回答了14个问题 | 采纳率92.9%
1/x+1/x-1/(x+1)+1/(x+1)-1/(x+2)+.+1/(x+9)-1/(x+10)=0
2/x-1/(x+10)=0
2(x+10)-x=0
x=-20
计算:1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+……+1/(x+9)(x+10)
2008foreverlove1年前2
无身无色 共回答了11个问题 | 采纳率72.7%
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+……+1/(x+9)(x+10)
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+……+1/(x+9)-1/(x+10)
=1/x-1/(x+10)
=(x+10-x)/{x(x+10)}
= 10/{x(x+10)}
计算:1/X(X+2)+1/(X+2)(X+4)+...+1/(X+28)(X+30)
Believe_loveOps1年前2
cacee 共回答了15个问题 | 采纳率80%
1/2*[1/x-1/(x+2)+1/(x+2)-1/(x+4)......+1/(x+28)-1/(x+30)]
=1/2*[1/x-1/(x+30)]
1/x(x+2)+1/x(x+2)(x+4)+1/(x+4)(x+6)+……1(x+2004)(x+2006)
浪迹ff07321年前1
平平1209 共回答了16个问题 | 采纳率75%
第二个多了个X吧?
原式=[1/x-1/(x+2)]/2+[1/(x+2)-1/(x+4)]/2+...+[1/(x+2004)-1/(x+2006)]/2
=[1/x-1/(x+2006)]/2
=2005/2x(x+2006)
如果解方程2x/(x+1)=a/(x+2)+1/(x^2+3x+2)时,会产生增根,求a的值?
爱上一棵柴1年前3
liujiapiaoxu 共回答了14个问题 | 采纳率92.9%
2x/(x+1)=a/(x+2)+1/[(x+2)(x+1)]
方程两边同时乘以(x+2)(x+1)得:2x(x+2)=a(x+1)+1
增根只可能为-1或-2
x=-1代入上式得:-2(-1+2)=1, 不符
x=-2代入上式得:0=-a+1, 得:a=1
因此只能为a=1
非常欣赏你的勤学好问精神,祝你成功!
如果本题有什么不明白可以追问,如果满意记得采纳
如果有其他问题请采纳本题后另发点击向我求助,答题不易,请谅解,谢谢.
祝学习进步!
化简1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3),其中x^2+3x+1=0
安臣1年前1
sumace 共回答了17个问题 | 采纳率82.4%
原式=[1/x-1/(x+1)]+[1/(x+1)-1/(x+2)]+[1/(x+2)-1/(x+3)]
=1/x-1/(x+3)
=2/(x^2+3x)
=2/(-1)
=-2
解方程:1/x(x+2)+1/(x+2)(x+4)+……+1/(x+8)(x+10)=5/24
在这边1年前4
九月鹰扬 共回答了17个问题 | 采纳率100%
1/x(x+2)+1/(x+2)(x+4)+……+1/(x+8)(x+10)=5/24
1/(2x)-1/{2(x+2)) + 1/{2(x+2)}-1/{2(x+4)} +……+1/{2(x+8)} - 1/{2(x+10)} = 5/24
1/(2x)-1/{2(x+10)} = 5/24
1/x-1/(x+10) = 5/12
10/{x(x+10)} = 5/12
2/{x(x+10)} =1 /12
x(x+10)=24
x^2+10x-24=0
(x+12)(x-2)=0
分母中不含(x+12)和(x-2)
x1=-12,x2=2
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+99)(x+100)=_____
jefferyyuan1年前1
清筱荷 共回答了16个问题 | 采纳率81.3%
1/x(x+1)=1/x-1/(x+1)
1/(x+1)(x+2)=1/(x+1)-1/(x+2)
1/(x+2)(x+3)=1/(x+2)-1/(x+3)
.
1/(x+99)(x+100)=1/(X+99)-1/(X+100)
很明显,这些数相加的话,中间部分正好是加减抵消.
最后
原式=1/x-1/(X+100)
=(x+100-x)/x(x+100)
=100/x(x+100)
解方程1/x+1/(x+1)+1/(x+2)+1/(x+3)=19/20.
八度创维1年前4
bilive 共回答了20个问题 | 采纳率85%
X=3;
应该没有简单的方法,只能通分求解了,
这个题应该是考你计算耐心的吧.
希望我的回答对你有所帮助.

大家在问