若x-1的绝对值+(xy-2)²=0
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若x-1的绝对值+(xy-2)²=0
1/xy+1/(x+1)(y+1)+1/(x+2)(y+2)+……+1/(x+2012)(y+2012)
求那个算式的值
1/xy+1/(x+1)(y+1)+1/(x+2)(y+2)+……+1/(x+2012)(y+2012)
求那个算式的值
![若x-1的绝对值+(xy-2)²=0](/uploads/image/z/18632324-20-4.jpg?t=%E8%8B%A5x-1%E7%9A%84%E7%BB%9D%E5%AF%B9%E5%80%BC%2B%28xy-2%29%26%23178%3B%3D0)
由 |x - 1| +(xy-2)²=0可知 x - 1 = 0 ,xy - 2 = 0 ,解这两个方程可得 x = 1 ,y = 2
所以:1/xy+1/(x+1)(y+1)+1/(x+2)(y+2)+……+1/(x+2012)(y+2012)
=1/(1×2)+ 1/(2×3) + 1/(3×4)+……+ 1/(2013×2014)
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ……+ 1/2013 - 1/2014
= 1 - 1/2014
= 2013/1014
所以:1/xy+1/(x+1)(y+1)+1/(x+2)(y+2)+……+1/(x+2012)(y+2012)
=1/(1×2)+ 1/(2×3) + 1/(3×4)+……+ 1/(2013×2014)
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ……+ 1/2013 - 1/2014
= 1 - 1/2014
= 2013/1014
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