Tìm `y` biết `1/3+1/6+1/10+1/15+…+2/(y.(y+1))=2009/2011`.

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1. `1/3 + 1/6 + 1/10 + 1/15 + … + 2/(y . (y + 1)) = 2009/2011`

   `=> 2/6 + 2/12 + 2/20 + 2/30 + … + 2/(y . (y + 1)) = 2009/2011`

   `=> 2 . (1/(2 . 3) + 1/(3.4) + 1/(4.5) + 1/(5.6) + … + 1/(y . (y + 1))  2009/2011`

   `=> 2 . (1/2 – 1/3 + 1/3 – 1/4 + 1/4 – 1/5 + 1/5 – 1/6 + … + 1/y – 1/(y + 1)) = 2009/2011`

   `=> 2 . (1/2 – 1/(y + 1)) = 2009/2011`

   `=> 1/2 – 1/(y + 1) = 2009/2011 : 2`

   `=> 1/2 – 1/(y + 1) = 2009/2011 . 1/2`

   `=> 1/2 – 1/(y + 1) = 2009/4022`

   `=> 1/(y + 1) = 1/2 – 2009/4022`

   `=>1/(y + 1) = 2011/4022 – 2009/4022`

   `=> 1/(y + 1) = 1/2011`

   `=> y + 1 = 2011`

   `=> y = 2011 – 1`

   `=> y = 2010`

   Vậy `y = 2010`

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2. `1/3 + 1/6  + 1/10+ 1/15 +…+ 2/(y(y+1)) = 2009/2011`

    `1/2 ( 1/3 + 1/6 + 1/10 + 1/15 +…+ 2/(y(y+1)) )= 1/2 . 2009/2011`

   `1/6 + 1/12 + 1/20 + 1/30+…+ 1/(y(y+1))= 2009/4022`

   `1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 +…+ 1/(y(y+1))  = 2009/4022`

   `1/2 – 1/3 + 1/3 – 1/4 + 1/4 -1/5+….+ 1/y – 1/(y+1) = 2009/4022`

   `1/2 – 1/(y+1) = 2009/4022`

   `1/(y+1) = 1/2 – 2009/4022`

   `1/(y+1)= 901/2011`

   `=>2011 = 901(y+1)`

   `=> 2011 = 901 y + 901`

   `=> 901y= 2011 -901`

   `=> 901y=1110`

   `=> y= 1110: 901`

   `=> y= 1110/901`

   Vậy `y = 1110/901`

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