会的请速速回答!已知|a-1|+|ab-2|=0,求:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2000)(b+2000).
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![会的请速速回答!已知|a-1|+|ab-2|=0,求:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2000)(b+2000).](/uploads/image/z/2493364-4-4.jpg?t=%E4%BC%9A%E7%9A%84%E8%AF%B7%E9%80%9F%E9%80%9F%E5%9B%9E%E7%AD%94%21%E5%B7%B2%E7%9F%A5%7Ca-1%7C%2B%7Cab-2%7C%3D0%2C%E6%B1%82%3A1%2Fab%2B1%2F%28a%2B1%29%28b%2B1%29%2B1%2F%28a%2B2%29%28b%2B2%29%2B%E2%80%A6%2B1%2F%28a%2B2000%29%28b%2B2000%29.)
会的请速速回答!已知|a-1|+|ab-2|=0,求:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2000)(b+2000).
会的请速速回答!
已知|a-1|+|ab-2|=0,
求:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2000)(b+2000).
会的请速速回答!已知|a-1|+|ab-2|=0,求:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2000)(b+2000).
因为平方和绝对值都是非负的,已知中两者相加为0,只能是两者都为0
于是 a=1,b=2
同时1/ab=1/2=1-1/2
1/(a+1)(b+1)=1/(2*3)=1/2-1/3
...
1/(a+2000)(b+2000)=1/(2001*2002)=1/2001-1/2002
累加可得
等式右边的被减数与下一等式右边的减数抵消
故结果为
=1-1/2002=2001/2002
由题意知:
a -1=0,且 ab -2=0,
解得:a= 1, b= 2.
故:
原式=1/(1*2) +1/(2*3) +1/(3*4) +...+1/(2001*2002)
=(2 -1)/(1*2) +(3 -2)/(2*3) +(4 -3)/(3*4)+...+(2002 -2001)/(2001*2002)
=(1 -1/2) +(1/2 -1/3) +(1/3 -1/4)+...+(1/2001 -1/2002)
=1 -1/2002
=2001/2002
从"已知"看出:|1-1|+|1*2-2|=0,符合题目,所以A=1,B=2
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由绝对值概念易得a=1,b=2。
则1/ab+1/(a+1)(b+1)+……+1/(a+2000)(b+2000)
=1/(1*2)+1/(2*3)+……+1/(2001*2002)
=1-1/2+1/2-1/3+1/3-……-1/2001+1/2001-1/2002
=1-1/2002
=2001/2002