2小时内满分题)若√a-1 +√ab-2 =0,求1/ab+1/(a+1)(b-1)+1/(a+2)(b-2)+…+1/(a+2004)(b+2004)的值.我可以由√a-1 +√ab-2 =0算出a=1 ,ab=2即b=2,但是代入去算就不会!
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2小时内满分题)若√a-1 +√ab-2 =0,求1/ab+1/(a+1)(b-1)+1/(a+2)(b-2)+…+1/(a+2004)(b+2004)的值.我可以由√a-1 +√ab-2 =0算出a=1 ,ab=2即b=2,但是代入去算就不会!
2小时内满分题)
若√a-1 +√ab-2 =0,求1/ab+1/(a+1)(b-1)+1/(a+2)(b-2)+…+1/(a+2004)(b+2004)的值.
我可以由√a-1 +√ab-2 =0算出a=1 ,ab=2即b=2,但是代入去算就不会!
2小时内满分题)若√a-1 +√ab-2 =0,求1/ab+1/(a+1)(b-1)+1/(a+2)(b-2)+…+1/(a+2004)(b+2004)的值.我可以由√a-1 +√ab-2 =0算出a=1 ,ab=2即b=2,但是代入去算就不会!
√a-1 +√ab-2 =0
a=1 ,ab=2即b=2
1/ab+1/(a+1)(b-1)+1/(a+2)(b-2)+…+1/(a+2004)(b+2004)
=1/1*2+1/2*3+1/3*4+...+1/2005*2006
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/2005-1/2006)
=1-1/2006
=2005/2006
1/ab+1/(a+1)(b-1)+1/(a+2)(b-2)+…+1/(a+2004)(b+2004)=1/3+1/3+1/3+1/3。。。
最后=668
由√(a-1)+(ab-2)?=0知
√(a-1)=0和(ab-2)?=0;?
得a=1,b=2.?
1/ab+1/(a+1)(b+1)+1/(a+2)(a+2)+……+1/(a+2004)(b+2004)
=1/1?2+1/2?3+1/3?4+……+1/(2004?2005)+1/(2005?2006)
=1-1/2006=2005/2006
∵√a-1 +√ab-2 =0,被开方数非负
∴a=1 ,ab=2即b=2
∴1/ab+1/(a+1)(b-1)+1/(a+2)(b-2)+…+1/(a+2004)(b+2004)
=1/1*2+1/2*3+1/3*4+...+1/2005*2006
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/2005-1/2006)
...
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∵√a-1 +√ab-2 =0,被开方数非负
∴a=1 ,ab=2即b=2
∴1/ab+1/(a+1)(b-1)+1/(a+2)(b-2)+…+1/(a+2004)(b+2004)
=1/1*2+1/2*3+1/3*4+...+1/2005*2006
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/2005-1/2006)
=1-1/2+1/2-1/3+....+1/2004+1/2005+1/2005-1/2006(中间各项相互抵消)
=1-1/2006
=2005/2006
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很简单啊,典型的分解因式的题目,1/1*2+1/2*3+1/3*4+...+1/2005*2006
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/2005-1/2006)
=1-1/2006
=2005/2006
初中这样的题目会很多的,试着分解,多做做就会了
√a-1 +√ab-2 =0 被开方数位非负
a=1 ,ab=2即b=2
我觉得你要求的代数式写错了
1/ab+1/(a+1)(b-1)+1/(a+2)(b-2)+…+1/(a+2004)(b+2004)
前面都是“b-”,后面变成“b+”了 应改成
1/ab+1/(a+2)(b+2)+…+1/(a+2004)(b+2004)
再利用拆因式相消法1...
全部展开
√a-1 +√ab-2 =0 被开方数位非负
a=1 ,ab=2即b=2
我觉得你要求的代数式写错了
1/ab+1/(a+1)(b-1)+1/(a+2)(b-2)+…+1/(a+2004)(b+2004)
前面都是“b-”,后面变成“b+”了 应改成
1/ab+1/(a+2)(b+2)+…+1/(a+2004)(b+2004)
再利用拆因式相消法1/(1+n)(2+n)=1/n+1/(n+2)
可得最后结果:2005/2006
收起