(1 -1 /x+1)*(1 -1 /x+2)*(1 -1 /x+3)*…*(1 -1 /x+2019)

Question

(1 -1 /x+1)*(1 -1 /x+2)*(1 -1 /x+3)*…*(1 -1 /x+2019)

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Sarah 1 tuần 2021-12-03T22:43:49+00:00 2 Answers 2 views 0

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    0
    2021-12-03T22:45:12+00:00

    `(1 -1 /(x+1))*(1 -1 /(x+2))*(1 -1 /(x+3))*…*(1 -1 /(x+2019))`

    `=(x+1 -1) /(x+1)*(x+2 -1) /(x+2)*(x+3 -1) /(x+3)*…*(x+2019-1) /(x+2019)`

    `=(x) /(x+1)*(x+1) /(x+2)*(x+2) /(x+3)*…*(x+2018) /(x+2019)`

    `=x/(x+2019)`

    0
    2021-12-03T22:45:36+00:00

    Đáp án :

    `A=x/(x+2019)`

    Giải thích các bước giải :

    `A=(1-1/(x+1))(1-1/(x+2))(1-1/(x+3))…(1-1/(x+2019))`

    `<=>A=((x+1)/(x+1)-1/(x+1))((x+2)/(x+2)-1/(x+2))…((x+2019)/(x+2019)-1/(x+2019))`

    `<=>A=(x+1-1)/(x+1).(x+2-1)/(x+2)…(x+2019-1)/(x+2019)`

    `<=>A=x/(x+1).(x+1)/(x+2)…(x+2018)/(x+2019)`

    `<=>A=[x(x+1)(x+2)…(x+2018)]/[(x+1)(x+2)(x+3)…(x+2019)]`

    `<=>A=x/(x+2019)`

    Vậy `A=x/(x+2019)`

    ~Chúc bạn học tốt !!!~

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