Question

已知|ab-2|+|a-1|=0。求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+......+1/(a+2011)(b+2011)的值

Answer 1

绝对值都是非负数
两个非负数的和为0,那么这两个数都是0
ab-2=0
a-1=0
解得:a=1,b=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+...+1/2010-1/2013
=1-1/2013
=2012/2013

Answer 2

解：|ab-2|+|a-1|=0
则：|ab-2|=0，|a-1|=0
解得：a=1，b=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)
=1-1/2+1/2-1/3+.......+1/2012-1/2013
=1-1/2013
=2012/2013

类似：1/[a*(a+n)]=(1/n)*[1/a-1/(a+n)]

例子：
1/56=1/(7*8)=1/7-1/8
1/12=1/(2*6)=(1/4)*[1/2-1/6]
....................
等等

以上的式子是很重要的，能记住对解这类的题是有帮助的~
希望能帮到你~~
如果满意，请采纳一下拉~~谢谢啊~~~

Answer 3

由题意有
ab-2=0,a=1
即
a=1,b=2
所以
原式=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

希望对您有所帮助
如有问题，可以追问。
谢谢您的采纳

Answer 4

|ab-2|+|a-1|=0
因：|ab-2|≥0且|a-1|≥0
所以：ab-2=0,a-1=0, 得:a=1,b=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+......+1/(a+2011)(b+2011)
=1/1x2+1/2x3+1/3x4+......+1/2012x2013
=1-1/2+1/2-1/3+1/3-1/4+........+1/2012-1/2013
=1-1/2013
=2012/2013