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

Toán Lớp 9: cho `f(x)=(x^3)/(1-3x+3x^2)` tính `A=f(1/(2022))+f(2/(2022))+f(3/(2022))+…+f((2020)/(2022))+f((2021)/(2022))`

Home/ Toán Lớp 9: cho `f(x)=(x^3)/(1-3x+3x^2)` tính `A=f(1/(2022))+f(2/(2022))+f(3/(2022))+…+f((2020)/(2022))+f((2021)/(2022))`

Toán Lớp 9: cho `f(x)=(x^3)/(1-3x+3x^2)` tính `A=f(1/(2022))+f(2/(2022))+f(3/(2022))+…+f((2020)/(2022))+f((2021)/(2022))`

Tháng Một 15, 2023 Toán học 1 Comment 0 views

Toán Lớp 9: cho f(x)=(x^3)/(1-3x+3x^2)
tính A=f(1/(2022))+f(2/(2022))+f(3/(2022))+…+f((2020)/(2022))+f((2021)/(2022))

Comments ( 1 )

  1. thudan
    thudan
    15 Tháng Một, 2023 at 11:23 sáng
    Reply
    Giải đáp:
    \(A = \dfrac{2021}{2}\)
    Lời giải và giải thích chi tiết:
    Nhận thấy:
    $\dfrac{1}{2022} + \dfrac{2021}{2022} = 1$
    $\dfrac{2}{2022} + \dfrac{2020}{2022} = 1$
    $\cdots$
    Xét: $f(x) + f(1 – x)$
    $= \dfrac{x^3}{1-3x + 3x^2} + \dfrac{(1-x)^3}{1 – 3(1-x) + 3(1-x)^2}$
    $= \dfrac{x^3}{1-3x + 3x^2} + \dfrac{(1-x)^3}{1 – 3x + 3x^2}$
    $= \dfrac{x^3 + (1 – x)^3}{1 – 3x + 3x^2}$
    $= \dfrac{(x + 1 – x)[x^2 – x(1-x) + (1-x)^2]}{1 – 3x + 3x^2}$
    $= \dfrac{3x^2 -3x + 1}{1 – 3x + 3x^2}$
    $= 1$
    Ta được:
    \(\begin{array}{l}
    A = f\left(\dfrac{1}{2022}\right) +  f\left(\dfrac{2}{2022}\right) +  f\left(\dfrac{3}{2022}\right)+\cdots+ f\left(\dfrac{2020}{2022}\right)+ f\left(\dfrac{2021}{2022}\right)\\
    \quad = \left[ f\left(\dfrac{1}{2022}\right) +  f\left(\dfrac{2021}{2022}\right)\right] + \left[ f\left(\dfrac{2}{2022}\right) +  f\left(\dfrac{2020}{2022}\right)\right] + \cdots +\left[ f\left(\dfrac{1010}{2022}\right)+f\left(\dfrac{1012}{2022}\right)\right] +f\left(\dfrac{1011}{2022}\right)\\
    \quad =\underbrace{1+1+\cdots+1}_{\text{1010 số hạng}} + f\left(\dfrac12\right)\\
    \quad = 1010+\dfrac12\\\quad = \dfrac{2021}{2}
    \end{array}\)

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

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