Tính nhanh: `1/[x(x+1)] + 1/[(x+1)(x+2)] +…+ 1/[(x+2018)(x+2019)]`

1 bình luận về “Tính nhanh: `1/[x(x+1)] + 1/[(x+1)(x+2)] +…+ 1/[(x+2018)(x+2019)]`”

1. Giải đáp: 2019/(x^2 + 2019x)

    

   Lời giải và giải thích chi tiết:

   1/(x(x+1)) + 1/((x+1)(x+2)) + ….. + 1/((x+2018)(x+2019))

   = 1/x – 1/(x+1) + 1/(x+1) – 1/(x+2) + ….. + 1/(x+2017) – 1/(x+2018) + 1/(x+2018) – 1/(x+2019)

   = 1/x + (1/(x+1) – 1/(x+1)) +(1/(x+2) – 1/(x+2)) + …. + (1/(x+2018) – 1/(x+2018)) – 1/(x+2019)

   = 1/x + 0 – 1/(x+2019)

   = 1/x – 1/(x+2019)

   = (x+2019)/(x^2 + 2019x) – x/(x^2 + 2019x)

   = 2019/(x^2 + 2019x)

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