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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: Thực hiện phép cộng: P= $\frac{1}{1-x}$+ $\frac{1}{1+x}$+ $\frac{2}{1+x^2}$+$\frac{4}{1+x^4}$+$\frac{8}{1+x^8}$

Tháng Ba 6, 2023 Toán học 0 Comments 0 views

Toán Lớp 8: Thực hiện phép cộng: P= $\frac{1}{1-x}$+ $\frac{1}{1+x}$+ $\frac{2}{1+x^2}$+$\frac{4}{1+x^4}$+$\frac{8}{1+x^8}$

Comments ( 1 )

  1. landinhngoc97
    landinhngoc97
    6 Tháng Ba, 2023 at 2:04 chiều
    Reply
    Giải đáp + Lời giải và giải thích chi tiết:
     $P = \dfrac{1}{1 – x} + \dfrac{1}{1 + x} + \dfrac{2}{1 + x^2} + \dfrac{4}{1 + x^4} + \dfrac{8}{1 + x^8}$
    $P = \dfrac{-1}{x – 1} + \dfrac{1}{x + 1} + \dfrac{2}{x^2 + 1} + \dfrac{4}{x^4 + 1} + \dfrac{8}{x^8 + 1}$
    $P = \dfrac{-(x + 1)}{(x – 1)(x + 1)} + \dfrac{x – 1}{(x + 1)(x – 1)} +\dfrac{2}{x^2 + 1} + \dfrac{4}{x^4 + 1} + \dfrac{8}{x^8 + 1}$
    $P = \dfrac{-x – 1 + x – 1}{(x – 1)(x + 1)} + \dfrac{2}{x^2 + 1} + \dfrac{4}{x^4 + 1} + \dfrac{8}{x^8 + 1}$
    $P = \dfrac{-2}{x^2 – 1} + \dfrac{2}{x^2 + 1} + \dfrac{4}{x^4 + 1} + \dfrac{8}{x^8 + 1}$
    $P = \dfrac{-2(x^2 + 1)}{(x^2 – 1)(x^2 + 1)} + \dfrac{2(x^2 – 1)}{(x^2 – 1)(x^2 + 1)} + \dfrac{4}{x^4 + 1} + \dfrac{8}{x^8 + 1}$
    $P = \dfrac{-2x^2 – 2}{x^4 – 1} + \dfrac{2x^2 – 2}{x^4 – 1} + \dfrac{4}{x^4 + 1} + \dfrac{8}{x^8 + 1}$
    $P = \dfrac{-2x^2 – 2 + 2x^2 – 2}{x^4 – 1} + \dfrac{4}{x^4 + 1} + \dfrac{8}{x^8 + 1}$
    $P = \dfrac{-4}{x^4 – 1} + \dfrac{4}{x^4 + 1} + \dfrac{8}{x^8 + 1}$
    $P = \dfrac{-4(x^4 + 1)}{(x^4 – 1)(x^4 + 1)} + \dfrac{4(x^4 – 1)}{(x^4 – 1)(x^4 + 1)} + \dfrac{8}{x^8 + 1}$
    $P = \dfrac{-4x^4 – 4}{x^8 – 1} + \dfrac{4x^4 – 4}{x^8 – 1} + \dfrac{8}{x^8 + 1}$
    $P = \dfrac{-4x^4 – 4 + 4x^4 – 4}{x^8 – 1} + \dfrac{8}{x^8 + 1}$
    $P = \dfrac{-8}{x^8 – 1} + \dfrac{8}{x^8 + 1}$
    $P = \dfrac{-8(x^8 + 1)}{(x^8 – 1)(x^8 + 1)} + \dfrac{8(x^8 – 1)}{(x^8 – 1)(x^8 + 1)}$
    $P = \dfrac{-8x^8 – 8}{(x^8 – 1)(x^8 + 1)} + \dfrac{8x^8 + 8}{(x^8 – 1)(x^8 + 1)}$
    $P = \dfrac{-8x^8 – 8 + 8x^8 – 8}{x^{16} – 1}$
    $P = \dfrac{-16}{x^{16} – 1}$

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222-9+11+12:2*14+14 = ? ( )

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