1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2012)(b+2012) axb得2,b为

阿拉阿拉阿拉2022-10-04 11:39:540条回答

1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2012)(b+2012) axb得2,b为1

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最后de开始 1年前 已收到1个回答 举报

忧伤也是美丽 花朵

共回答了19个问题采纳率:89.5% 举报

解题思路:根据已知条件计算出a,b的值,代入代数式求值.

∵(a-1)2+|b-2|=0,
∴a=1,b=2,
∴[1/ab+
1
(a+1)(b+1)+
1
(a+2)(b+2)+…+
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(a+2006)(b+2006)],
=[1/2]+[1/2×3]+[1/3×4]+…+[1/2007×2008],
=[1/2]+[1/2]−
1
3+[1/3]-[1/4]+…+[1/2007]-[1/2008],
=1-[1/2008],
=[2007/2008].

点评:
本题考点: 有理数的混合运算;非负数的性质:绝对值;非负数的性质:偶次方.

考点点评: 本题考查的知识点是非负数的性质,明白[1n×(n+1)=1/n]-[1/n+1]是解题的关键.

1年前

1
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最后de开始1年前1
忧伤也是美丽 共回答了19个问题 | 采纳率89.5%
解题思路:根据已知条件计算出a,b的值,代入代数式求值.

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∴a=1,b=2,
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=1-[1/2008],
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点评:
本题考点: 有理数的混合运算;非负数的性质:绝对值;非负数的性质:偶次方.

考点点评: 本题考查的知识点是非负数的性质,明白[1n×(n+1)=1/n]-[1/n+1]是解题的关键.

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要过程
pagedown33991年前2
hahahaer 共回答了20个问题 | 采纳率80%
=1、a=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2011)(b+2011)=1/1×2+1/2×3+.+1/2012×2013
=1-1/2+1/2-1/3+.+1/2012-1/2013=1-1/2013=2012/2013
已知:字母a,b满足√(a-1)+√(b-2)=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……
已知:字母a,b满足√(a-1)+√(b-2)=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2012)(b+2013)
冰冷的泪雨1年前3
梦灵儿sheng 共回答了18个问题 | 采纳率88.9%
根号求解是非负数~
两个非负数相加要等于0,只有两个都为0
所以根号a-1=0
根号b-2=0
即a=1,b=2
然后代进表达式,就可以求解了!
表达式分解下
=[1/(b-a)]*[(1/a)-(1/b)]+[1/(b-a)]*[(1/(a+1))-(1/(b+1))]+...
=(1-1/2) + 1/2 *(1/2-1/3).1/2 * (1/2012 - 1/2013)
=( 1-1/2+1/2-1/3.+1/2012-1/2013)
=(1-1/2013)
= 2012/2013
请点击右面的“采纳答案”按钮!
已知平方根a-1+b的平方+1的绝对值=1 求1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2012
已知平方根a-1+b的平方+1的绝对值=1 求1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2012)(b+2012)的值
萝卜缨zi1年前1
jfrg 共回答了21个问题 | 采纳率85.7%
平方根a-1+b的平方+1的绝对值=1
∴√(a-1)+b²+1=1
∴√(a-1)+b²=0
∴a-1=0
b=0
∴a=1
b=0
1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2012)(b+2012)
=1/1×2+1/2×3+1/3×4+……+1/2012×2013
=1-1/2+1/2-1/3+1/3-1/4+……+1/2012-1/2013
=1-1/2013
=2012/2013
已知:字母a,b满足根号a-1+根号b-2=0,求1/ab+1/(a+1)(b+1)+1/(a+2
已知:字母a,b满足根号a-1+根号b-2=0,求1/ab+1/(a+1)(b+1)+1/(a+2
)(b+2)+*******+1/(a+2013)(b+2013)的值
aifeidexuan7431年前0
共回答了个问题 | 采纳率
已知a与-1的差为2,b是-2的相反数,试求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+
已知a与-1的差为2,b是-2的相反数,试求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2010)(b+2010)值
wuwrxp1年前3
华虹NEC 共回答了16个问题 | 采纳率93.8%
a与-1的差为2
a-(-1)=2
a=1
b是-2的相反数
b+(-2)=0
b=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2010)(b+2010)
=1/1*2+1/2*3+1/3*4+1/4*5+.+1/2011*2012
=1-1/2+1/2-1/3+1/3-1/4+.+1/2011-1/2012
=1-1/2012
=2011/2012
当(b+1)/a=3时,代数式{(1+b)/a}的平方-2a/(b+1)+1的值为
why7679872091年前1
hexinoo 共回答了23个问题 | 采纳率91.3%
∵(b+1)/a=3
∴a/(b+1)=1/3
∴{(1+b)/a}的平方-2a/(b+1)+1
=3²-2/3+1
=9又3分之1
如果有理数a,b满足丨ab-2丨加丨1-b丨=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...
如果有理数a,b满足丨ab-2丨加丨1-b丨=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2010)(b+2010)的
独孤恨天恨1年前1
zhgxueyon 共回答了15个问题 | 采纳率73.3%
丨ab-2丨+丨1-b丨=0
所以ab-2=1-b=0
ab=2,b=1
则a=2
所以原式=1/1×2+1/2×3+……+1/2011×2012
=1-1/2+1/2-1/3+……+1/2011-1/2012
=1-1/2012
=2011/2012
如果/ab-2/+/b-1/=0,试求下式的值:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)…+1/(a
如果/ab-2/+/b-1/=0,试求下式的值:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)…+1/(a+2000)(b+2000)
行者5211年前2
冰冷的呜咽 共回答了21个问题 | 采纳率85.7%
/ab-2/+/b-1/=0
得到ab-2=0,b-1=0
解得a=2,b=1
所以要求的式子变成了
1/2+1/(2)(3)+1/(3)(4)…+1/(2001)(2002)
=1-1/2+1/2-1/3+……+1/2001-1/2002
=2001/2002
这个是列项相消法
已知丨ab+2丨+丨a+1丨=0 ,则代数式1/(a-1)(b+1)+1/(a-2)(b+2)+…+1/(a-2009)
已知丨ab+2丨+丨a+1丨=0 ,则代数式1/(a-1)(b+1)+1/(a-2)(b+2)+…+1/(a-2009)x(b+2009)
j急!GAOFEN!
4932728541年前1
科科科 共回答了13个问题 | 采纳率100%
a=-1,b=2
1/(a-1)(b+1)+1/(a-2)(b+2)+…+1/(a-2009)x(b+2009) =-(1/2-1/3+1/3-1/4+…-1/2011)=-2009/4022
若丨ab-2丨+丨a-2丨=0,试求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2012
若丨ab-2丨+丨a-2丨=0,试求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2012)(b+2012)的值
geejon11年前4
媛媛3920627 共回答了13个问题 | 采纳率92.3%
若丨ab-2丨+丨a-2丨=0,试求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2012)(b+2012)的值
由丨ab-2丨+丨a-2丨=0,得ab=2,a=2,b=1;故
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2012)(b+2012)
=1/2+1/(3×2)+1/(4×3)+.+1/(2014×2013)
=1/2+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+.+(1/2013-1/2014)
=1/2+1/2-1/2014=2013/2014
已知;|ab-2|与|b-1|互为相反数.试求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...
已知;|ab-2|与|b-1|互为相反数.试求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2009)(b+2009)
凉心20051年前3
肉肉和球球 共回答了18个问题 | 采纳率88.9%
|ab-2|与|b-1|互为相反数.
所以ab-2=0,b-1=0
a=2,b=1
原式=1/(1x2)+1/(2x3)+1/(3x4)+...+1/(2010x2011)
=1-1/2+1/2-1/3+1/3-1/4+...+1/2010-1/2011
=1-1/2011
=2010/2011
甲天数是乙丙合作的a倍;乙天数等于甲、丙合作的b倍;丙天数等于甲乙合作的c倍.求1/(a+1)+1/(b+1)+1(c+
甲天数是乙丙合作的a倍;乙天数等于甲、丙合作的b倍;丙天数等于甲乙合作的c倍.求1/(a+1)+1/(b+1)+1(c+1)
鸡鸣外欲曙1年前1
sxw19830123 共回答了21个问题 | 采纳率85.7%
为了方便书写,假设一下
假设甲=X,乙=Y,丙=Z
ax=y+z.(a+1)x=x+y+z.1/(a+1)=x/(x+y+z)
by=x+z.(b+1)y=x+y+z.1/(b+1)=y/(x+y+z)
cz=x+y.(c+1)z=x+y+z.1/(c+1)=z/(x+y+z)
所以1/(a+1)+1/(b+1)+1(c+1)=1
如果a-1的绝对值+(b-2)的平方=0,求1/ab+1/(a+1)x(b+1)+1/(a+2)x(b+2)+…+1/(
如果a-1的绝对值+(b-2)的平方=0,求1/ab+1/(a+1)x(b+1)+1/(a+2)x(b+2)+…+1/(a+2009)x(b+2009)的值
phoenixboss1年前1
yy树 共回答了19个问题 | 采纳率89.5%
a=1,b=2
1/ab+1/(a+1)x(b+1)+1/(a+2)x(b+2)+…+1/(a+2009)x(b+2009)
=1/(1*2)+1/(2*3)+1/(3*4)+...+1/(2010*2011)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/2010-1/2011)
=1-1/2011
=2010/2011
已知|ab-2|与|b-1|互为相反数.试求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/
已知|ab-2|与|b-1|互为相反数.试求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2009)(b+2009)
娃哈哈ur1401年前4
哦饿的而饿 共回答了20个问题 | 采纳率85%
由|b-1|+|ab-2|=0可以得到|b-1|=0且|ab-2|=0,故b=1a=2.
我们知道1/n(n+1)=[(n+1)-n]/n(n+1)=1/n-1/(n+1),将a,b数值带入可知道结果为2010/2011.不以后尽管向我提问.理科类保管满意.
甲天数是乙丙合作的a倍;乙天数等于甲、丙合作的b倍;丙天数等于甲乙合作的c倍.求1/(a+1)+1/(b+1)+1(c+
甲天数是乙丙合作的a倍;乙天数等于甲、丙合作的b倍;丙天数等于甲乙合作的c倍.求1/(a+1)+1/(b+1)+1(c+1)
香肩上的蝴蝶1年前2
在济南看海 共回答了17个问题 | 采纳率76.5%
解1:
设整个工程为1.由甲队独做所需天数是乙、丙两队合作所需天数的a倍知,乙丙合做的工效是甲的a倍,若甲乙丙三人合做,则甲做1/(a+1),同样的道理,乙做b/(b+1),丙做c/(c+1).于是 1/(a+1)+1/(b+1)+1/(c+1)就是用甲乙丙合做这项工程的工作总量,这相当于算了两个甲乙丙合做的总量,答案是1.
解2:
设甲、乙、丙单独做用的天数分别为x、y、z天.
那么:a/x = 1/x + 1/y,进一步变形可得:1/(a+1) = yz / (xy + yz + xz) ;
同理可得:1/(b+1) = xz / (xy + yz + xz) ; 1/(c+1) = xy / (xy + yz + xz) ;
1/(a+1) +1/(b+1)+ 1/(c+1) =yz / (xy + yz + xz) + xz / (xy + yz + xz) +xy / (xy + yz + xz) = 1 看看吧 应该没问题
若ab-2的绝对值+(b-2)的平方=0.求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a
若ab-2的绝对值+(b-2)的平方=0.求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2009)(b+2009)
宋明志1年前3
gufengcat520520 共回答了14个问题 | 采纳率100%
易得b=2;a=1
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2009)(b+2009)
=1/2+1/(2×3)+1/(3×4)+……+1/(2010×2011)
=1/2+1/2-1/3+1/3-1/4+······1/2010-1/2011
=1-1/2011
=2010/2011
已知|ab-2|与(b-1)2次方互为相反数,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...1/
已知|ab-2|与(b-1)2次方互为相反数,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...1/(a+2006)(b+2006)的值
网评猿0071年前2
caits 共回答了30个问题 | 采纳率90%
|ab-2|>=0
(b-1)^2>=0
|ab-2|与(b-1)^2互为相反数

|ab-2|=(b-1)^2=0
ab-2=0
b-1=0
a=2
b=1
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...1/(a+2006)(b+2006)
=1/2*1+1`/(2+1)(1+1)+1/(2+2)(1+2)+.+1/(2+2006)(1+2006)
=1/1*2+1/2*3+1/3*4+.+1/2007*2008
=1-1/2+1/2-1/3+1/3-1/4+.+1/2007-1/2008
=1-1/2008
=2007/2008
已知4a-4ab+2b-4b+4=0 试求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+
已知4a-4ab+2b-4b+4=0 试求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2009)(b+2009)的值 ( "/"为分数线)
好好gg1年前1
northqd 共回答了14个问题 | 采纳率92.9%
4a-4ab+2b-4b+4=(2a)-2*(2a)*b+b + b-4b+4 =(2a-b)+(b-2)=0 得2a-b=0 ,b-2=0 → a=1,b=2 即求1/(1*2)+1/(2*3)+.+1/(2010*2011)=1-1/2+1/2-1/3+1/3-1/4+.+1/2010-1/2011 =1-1/2011=2010/2011 注: 1/x - 1/(x+1)= 【(x+1)-x】/[x*(x+1)]=1/[x*(x+1)]
求采纳
已知a=1,b=2,试求:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.1/(a+2009)(b+20
已知a=1,b=2,试求:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.1/(a+2009)(b+2009)的值.
xgstb1年前1
爱已随风飘去 共回答了14个问题 | 采纳率85.7%
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.1/(a+2009)(b+2009)
=1/(1×2)+1/(2×3)+.1/(2010×2011)
=1-1/2+1/2-1/3+1/3-1/4+.+1/2010-1/2011
=1-1/2011
=2010/2011
如果有理数a,b 满足a等于2,b等于1,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+
如果有理数a,b 满足a等于2,b等于1,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2004)(b+2004)的值
钻石雨201年前2
兽包传说 共回答了13个问题 | 采纳率84.6%
a等于2,b等于1代人
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2004)(b+2004)
=1/1*2+1/2*3+1/3*4+…+1/2005*2006
=1-1/2+1/2-1/3+1/3-1/4+…+1/2005-1/2006
=1-1/2006=2005/2006
初一绝对值练习题 .急已知|ab-2|与|b-1|互为相反数试求代数式1/ab+1/(a+1)(b+1)+1/(a+2)
初一绝对值练习题 .急
已知|ab-2|与|b-1|互为相反数
试求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2009)(b+2009)
冻下小拇指1年前1
likesomuch 共回答了13个问题 | 采纳率100%
因为:|ab-2|与|b-1|互为相反数
∴ ab-2=0,b-1=0
∴ a=2,b=1
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2008)(b+2008)
=1/2*1+1/3*2+1/4*3+...+1/2010*2009
=1-1/2+1/2-1/3+1/3-1/4+...+1/2009-1/2010
=1-1/2010
=2009/2010
如果有理数a,b满足|ab-2|+(1-b)^2=0,试求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)…
如果有理数a,b满足|ab-2|+(1-b)^2=0,试求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)…+1/(a+2004(b+2004)的值)
ivan101011年前3
mcj251x 共回答了20个问题 | 采纳率90%
|ab-2|+(1-b)^2=0则|ab-2|=(1-b)^2=0则ab=2,b=1则a=2,b=1则1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)…+1/(a+2004(b+2004)=1/(1*2)+1/(2*3)+...+1/(2005*2006)=1/1-1/2+1/2-1/3+...+1/2005-1/2006=1-1/2006=2005/2006
已知|a-1|+|b-2|=0,试求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2011
已知|a-1|+|b-2|=0,试求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2011)(b+2011)的值.
此名无重复1年前1
oliver-rabbit 共回答了16个问题 | 采纳率87.5%
|a-1|+|b-2|=0可知a=1,b=2
从而1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2011)(b+2011)=1-1/2+1/2-1/3+1/3-1/4+.+1/2012-1/2013=1-1/2013=2012/2013
已知2a^2+4ab+5b^2+4a-2b+5=0,求1/ab+1/(a-1)(b+1)+1/(a-2)(b+2)+..
已知2a^2+4ab+5b^2+4a-2b+5=0,求1/ab+1/(a-1)(b+1)+1/(a-2)(b+2)+...+1/(a-2007)(b+2007)
斜飞细雨1年前1
好老虎 共回答了22个问题 | 采纳率90.9%
2a^2+4ab+5b^2+4a-2b+5=0
==> (a^2+4ab+4b^2)+(a^2+4a+4)+(b^2-2b+1)=0
==> (a+2b)^2+(a+2)^2+(b-1)^2=0
==> a=-2, b=1
代进1/ab+1/(a-1)(b+1)+1/(a-2)(b+2)+...+1/(a-2007)(b+2007)
=-(1/2+1/(2*3)+1/(3*4)+.+1/(2008*2009)
=-(1/2+1/2-1/3+1/3-1/4……+1/2008-1/2009)
(中间的部分全部减没了)
=-(1/2+1/2-1/2009)
=-2010/2009
已知|ab-2|与|a-1|互为相反数,试求下式的值:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……
已知|ab-2|与|a-1|互为相反数,试求下式的值:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……1/(a+2004)(b+2004)
lanbula1年前4
迷惑的选择 共回答了25个问题 | 采纳率88%
因为|ab-2|和|a-1|互为相反数,所以二者都为0,推出a=1,b=2
所以原式=1/(1*2)+1/(2*3)+.+1/(2005*2006)
=1-1/2+1/2-1/3+.+1/2005-1/2006 (约去中间项)
=1-1/2006
=2005/2006
这一类题目都是这么用拆项法解.
ab-6的绝对值与b-2的绝对值互为相反数 求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2).1/(a+2
ab-6的绝对值与b-2的绝对值互为相反数 求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2).1/(a+2005)(b+2005)的值
wsndlg1年前1
复印生活 共回答了9个问题 | 采纳率100%
绝对值大于等于0
相加等于0,若有一个大于0,则另一个小于0,不成立.
所以两个都等于0
所以ab-6=0,b-2=0b=2,a=6/b=3
所以原式=1/2*3+1/3*4+1/4*5+……+1/2007*2008
=1/2-1/3+1/3-1/4+1/4-1/5+……+1/2007-1/2008
=1/2-1/2008
=1003/2008

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