已知|ab-2|与|b-1|互为相反数,试求代数式1⼀ab+1⼀(a+1)(b+1)+1⼀(a+2)(b+2)+…+1⼀(a+2011)(b+2011)的值

2025-01-06 12:45:34
推荐回答(3个)
回答1:

|ab-2|与|b-1|互为相反数
|ab-2|>=0,|b-1|>=0
ab-2=0
b-1=0
b=1,a=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2011)(b+2011)
=1-1/2+1/2-1/3+1/3-1/4+..........+1/2012-1/2013
=1-1/2013
=2012/2013

回答2:

ab-2=0
b-1=0
b=1,a=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2011)(b+2011)
=1-1/2+1/2-1/3+1/3-1/4+..........+1/2012-1/2013
=1-1/2013
=2012/2013

希望能帮你忙,不懂请追问,懂了请采纳,谢谢

回答3:

2012/2013