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+2007)(b+2007)的值

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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+2007)(b+2007)的值
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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+2007)(b+2007)的值
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+2007)(b+2007)的值

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+2007)(b+2007)的值
因为(a-1)²+|ab-2|=0
所以:a-1=0,a=1
ab-2=0,b=2
1/ab+1/[a+1][b+1]+1/[a+2][b+2]+……+1/[a+2007][b+2007]
=1/(1*2)+1/(2*3)+1/(3*4)+...+1/(2008*2009)
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2008-1/2009
=1-1/2009
=2008/2009

|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 这是参考,一样的题目把后面缺的加上去就行了


由条件式:即两个非负数的和=0,则每一个数都=0,

ab-2=0
1-b=0
∴a=2、b=1
∴原式=1/﹙1×2﹚+1/﹙2×3﹚+1/﹙3×4﹚+……+1/﹙2008×2009﹚
=﹙1-1/2﹚+﹙1/2-1/3﹚+﹙1/3-1/4﹚+……+﹙1/2008-1/2009﹚
=1-1/2009
=2008/2009...

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由条件式:即两个非负数的和=0,则每一个数都=0,

ab-2=0
1-b=0
∴a=2、b=1
∴原式=1/﹙1×2﹚+1/﹙2×3﹚+1/﹙3×4﹚+……+1/﹙2008×2009﹚
=﹙1-1/2﹚+﹙1/2-1/3﹚+﹙1/3-1/4﹚+……+﹙1/2008-1/2009﹚
=1-1/2009
=2008/2009

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百度的 参考下 啊~~
有理数a、b满足|ab-2|+|1-b|=0则必有 AB-2=0 1-B=0
===>A=2,B=1
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+.....+1/2010-1/2011
=1-1/2011
=2010/2011

条件是两个非负数和为0,所以里面两个式子都等于0.b=1,a=2.再代入后面式子中,结果是1/2+1/(2*3)+1/(3*4)+...+1/(2008*2009)=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/2008-1/2009)=2008/2009

b=1
b+2007=2008

2008÷2009

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