作业帮如果√（a-1)+(ab-2)²=0,那么1÷ab + 1÷[(a+1)(b+1)] + 1÷[(a+2)(b+2)]+．．．+ 1÷[(a+2007)(b+2007)]
来源：学生作业帮助网 编辑：作业帮 时间：2024/12/02 09:58:12
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如果√（a-1)+(ab-2)²=0,那么1÷ab + 1÷[(a+1)(b+1)] + 1÷[(a+2)(b+2)]+．．．+ 1÷[(a+2007)(b+2007)]
如果√（a-1)+(ab-2)²=0,那么1÷ab + 1÷[(a+1)(b+1)] + 1÷[(a+2)(b+2)]+．．．+ 1÷[(a+2007)(b+2007)]
如果√（a-1)+(ab-2)²=0,那么1÷ab + 1÷[(a+1)(b+1)] + 1÷[(a+2)(b+2)]+．．．+ 1÷[(a+2007)(b+2007)]
由√（a-1)+(ab-2)²=0知a=1,ab=2(因为两个式子都是大于等于0的数,而相加为0,那么它们只能为0)
所以a=1,b=2
代入得:1/2+1/(2*3)+1/(3*4)+.+1/(2008*2009)
注意到:1/(2*3)=1/2-1/3
1/(3*4)=1/3-1/4
.
所以原式=1/2+1/2-1/3+1/3-1/4+.+1/2008-1/2009
=1-1/2009=2008/2009