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

Toán Lớp 8: Phân tích đa thức sau thành nhân tử (a-b+c)^3-(a-b-c)^3-(c-a-b)^3-(a+b+c)^3

Toán Lớp 8: Phân tích đa thức sau thành nhân tử
(a-b+c)^3-(a-b-c)^3-(c-a-b)^3-(a+b+c)^3

Comments ( 1 )

  1. Giải đáp:
     (a – b + c)^3 – (a – b – c)^3 – (c – a – b)^3 – (a + b + c)^3
    Đặt a – b – c= x, c – a – b = y, a + b + c = z
    Khi đó ta có: x + y + z = (a – b – c) + (c – a – b) + (a + b + c) = a – b + c
    Đa thức trở thành:
    (x + y + z)^3 – x^3 – y^3 – z^3
    = [(x + y) + z]^3 – x^3 – y^3 – z^3
    = (x + y)^3 + 3(x + y)^2 z + 3(x + y)z^2 + z^3 – x^3 – y^3 – z^3
    = x^3 + 3x^2 y + 3xy^2 + y^3 + 3(x^2 + 2xy + y^2)z + 3xz^2 + 3yz^2 – x^3 – y^3
    = 3x^2 y + 3xy^2 + 3x^2 z + 6xyz + 3y^2 z + 3xz^2 + 3yz^2
    = 3(x^2 y + xy^2 + x^2 z + 2xyz + y^2 z + xz^2 + yz^2)
    = 3[(x^2 y + 2xyz + yz^2) + (xy^2 + y^2 z) + (x^2 z + xz^2)]
    = 3[y(x^2 + 2xz + z^2) + y^2 (x + z) + xz(x + z)]
    = 3[y(x + z)^2 + y^2 (x + z) + xz(x + z)]
    = 3(x + z)[y(x + z) + y^2 + xz]
    = 3(x + z)(xy + yz + y^2 + xz)
    =3(x + z)[(xy + y^2) + (yz + xz)]
    = 3(x + z)[y(x + y) + z(x + y)]
    = 3(x + z)(x + y)(y + z)
    = 3(a – b – c + a + b + c)(a – b – c + c – a – b)(c – a – b + a + b + c)
    =3.2a.(-2b).2c
    = -24abc

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