求1/xy+1/(x+1)(y+1)+1/(x+2)(y+2)+…+1/(x+2009)(y+2009)+1/(x+20

学习 时间:2026-04-07 20:33:05 阅读:4012
求1/xy+1/(x+1)(y+1)+1/(x+2)(y+2)+…+1/(x+2009)(y+2009)+1/(x+2010)(y+2010)的值搞错了.若|x-1|+(xy-2)2=0,求求1/xy+1/(x+1)(y+1)+1/(x+2)(y+2)+…+1/(x+2009)(y+2009)+1/(x+2010)(y+2010)的值

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俊秀的水杯

现实的蜜蜂

2026-04-07 20:33:05

如果 |x - 1| + (xy - 2)^2 = 0那么|x - 1| = 0,xy - 2 = 0从中解得 x = 1,y = 2那么1/xy + 1/(x+1)(y+1) + ··· + 1/(x+2010)(y+2010)= 1 / (1·2) + 1 / (2·3) + ··· + 1/ (2011·2012)= 1 - 1/2 + 1/2 - 1/3 +1/3 - ··· +1/2011 - 1/2012= 1- 1/2012= 2011 / 2012做题思路:首先通过题目所给的式子可以求得x和y的值后面利用1/[n(n+1)] = 1/n - 1/(n+1)这个裂项公式可以求出最后答案

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  • 忧虑的钢笔
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    2026-04-07 20:33:05

    如果 |x - 1| + (xy - 2)^2 = 0那么|x - 1| = 0,xy - 2 = 0从中解得 x = 1,y = 2那么1/xy + 1/(x+1)(y+1) + ··· + 1/(x+2010)(y+2010)= 1 / (1·2) + 1 / (2·3) + ··· + 1/ (2011·2012)= 1 - 1/2 + 1/2 - 1/3 +1/3 - ··· +1/2011 - 1/2012= 1- 1/2012= 2011 / 2012做题思路:首先通过题目所给的式子可以求得x和y的值后面利用1/[n(n+1)] = 1/n - 1/(n+1)这个裂项公式可以求出最后答案

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