2 Useful Methods for a Tricky Radical Equation!

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  • เผยแพร่เมื่อ 22 ม.ค. 2025

ความคิดเห็น • 5

  • @gregevgeni1864
    @gregevgeni1864 18 วันที่ผ่านมา +1

    Obviously
    x²-x-2=(x-2)(x+1)≠0 , i.e. x≠2 and x≠-1.(*)
    The given equation is equivalent to
    ³√(((x²-x+1)(x+1))/(x-2)(x+1)) -
    ³√(((x²+2x+2²)(x-2))/(x+1)(x-2))=
    =³√(27/(x+1)(x-2))
    ³√((x³+1)/(x-2)(x+1))-
    -³√((x³-8)/(x+1)(x-2))-
    -³√((27/(x+1)(x-2))= 0 (1).
    Let A=³√((x³+1)/(x-2)(x+1)²),
    B=³√((x³-8)/(x+1)(x-2)),
    C=³√((27/(x+1)(x-2)) (2)
    Because (1) => A+(-B)+(-C)=0 =>
    A³-B³-C³=3•A•B•C (**) known identity.
    Due to (**) and (2) we have
    [³√((x³+1)/(x-2)(x+1)²)]³-
    -[³√((x³-8)/(x-2)²(x+1))]³-
    -[³√((27/(x-2)(x+1)]³=
    = 3•³√((27•(x³+1)(x³-8))/(x-2)³(x+1)³)) => (x³+1)/(x²-x-2) -(x³-8)/(x²-x-2) -
    -27/(x²-x-2) =
    =3•³√(3³•(x³+1)(x³-8)/(x²-x-2)³) =>
    -18/(x²-x-2)=3•3(³√(x³+1)(x³-8))/(x²-x-2) => -2 =³√(x³+1)(x³-8), because (*),
    (-2)³=(x³+1)(x³-8)=>-8=x⁶-7x³-8 =>
    x⁶-7x³=0 => x³(x³-7)=0 =>
    x=0 or x=³√7 .

  • @Fjfurufjdfjd
    @Fjfurufjdfjd 18 วันที่ผ่านมา +1

    Πρεπει χ=/2 χ=/-1
    (χ^2-χ+1)^(1/3)/(χ-2)^(1/3)-(χ^2+2χ+4)^(1/3)/(χ+1)^(1/3)=3/[(χ-2)(χ+1)]^(1/3)
    Πολλαπλασιαζω με [(χ-2)(χ+1)]^(1/3)
    ...[χ^3+1^3]^(1/3)-[χ^3-2^3]^(1/3)=3
    Θετω χ^3+1=α^3 και χ^3-8=β^3
    α-β=3 α^3-β^3=9
    ....α-β=3 αβ=-2
    α=1 , β=-2 ή α=2 , β=-1
    χ=0 (τριπλη) ή χ=(7)^(1/3) (τριπλη) δεκτες.

  • @Quest3669
    @Quest3669 18 วันที่ผ่านมา

    X= 7^1/3; 0

  • @walterwen2975
    @walterwen2975 13 วันที่ผ่านมา

    A Tricky Radical Equation:
    ³√[(x² - x + 1)/(x - 2)] - ³√[(x² + 2x + 4)/(x + 1)] = ³√[27/(x² - x - 2)], x ϵ R; x =?
    x ϵ R, x - 2 ≠ 0, x + 1 ≠ 0
    ³√[(x² - x + 1)/(x - 2)] - ³√[(x² + 2x + 4)/(x + 1)]
    = {³√[(x + 1)(x² - x + 1)] - ³√[(x - 2)(x² + 2x + 4)]}/{³√[(x - 2)(x + 1)]}
    ³√[27/(x² - x - 2)] = 3/{³√[(x - 2)(x + 1)]}
    ³√[(x + 1)(x² - x + 1)] - ³√[(x - 2)(x² + 2x + 4)] = ³√(x³ + 1) - ³√(x³ - 2³) = 3
    Let: u = ³√(x³ + 1), v = ³√(x³ - 2³); u - v = 3, u³ - v³ = (x³ + 1) - (x³ - 2³) = 9
    u³ - v³ = (u - v)(u² + uv + v²) = (u - v)[(u - v)² + 3uv] = 3(9 + 3uv) = 9(3 + uv) = 9
    3 + uv = 1, uv = - 2; u - v = 3, v = u - 3, uv = u(u - 3) = - 2, u² - 3u + 2 = 0
    (u - 1)(u - 2) = 0, u - 1 = 0, u = 1; uv = - 2, v = - 2 or u - 2 = 0, u = 2, v = - 1
    u = ³√(x³ + 1) = 1, x³ + 1 = 1, x = 0 or u = ³√(x³ + 1) = 2, x³ = 7, x = ³√7
    v = ³√(x³ - 2³) = - 2, x³ - 2³ = - 2³, x = 0 or v = ³√(x³ - 2³) = - 1, x³ = 7, x = ³√7
    Answer check:
    ³√[(x + 1)(x² - x + 1)] - ³√[(x - 2)(x² + 2x + 4)] = ³√(x³ + 1) - ³√(x³ - 2³) = 3
    x = 0: ³√(x³ + 1) - ³√(x³ - 2³) = 1 - ³√(- 2³) = 1 - (- 2) = 3; Confirmed
    x = ³√7: ³√(7 + 1) - ³√(7 - 8) = 2 - (- 1) = 3; Confirmed
    Final answer:
    x = 0 or x = ³√7: