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

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

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

Comments ( 1 )

  1. Giải đáp:
     a, (x+z)³-(y+x)³+(y-z)³
    + Đặt x-y=a; y-z=b; z-x=c
    ⇒ a+b+c=(x-y)+(y-z)+(z-x)
    ⇒ a+b+c=0 ⇒a+b=−c⇒(a+b)³=−c³
    ⇒a³+3a²b+3ab²+b²=−c³
    ⇒a³+b³+3ab(a+b)+c³=0
    ⇒a³+b³+c³+3ab(−c)=0
    ⇒a³+b³+c³−3abc=0
    ⇒a³+b³+c³=3abc
    ⇔(x−y)³+(y−z)³+(z−x)³=3(x−y)(y−z)(z−x) 
     
    b, (x-y-z)³-x³+y³+z³
    =x³+y³+z³+3x²y+3xy²+3x²z+3xz²+3y²z+3yz²+6xyz
    =3(xyz+x²y+x²z+xz²+xy²+y²z+xyz+yz²)
    =3[xy(x+z)+xz(x+z)+y²(x+z)+yz(x+z)]
    =3(x+z)[xy+xz+y²+yz]
    =3(x+z)[x(y+z)+y(y+z)]
    =3(x+z)(y+z)(x+y)
    c, (a-b)³+(b-c)³+(c-a)³
    =a³- 3a²b + 3ab² – b³ + b³ – 3b²c + 3bc² – c³ + c³ – 3ac² + 3a²c – a³
    = (a³ – a³) + (b³ – b³) + (c³ – c³) – 3(a²b + ac² – ab² – bc² + b²c – a²c)
    = 3[(a²b – ab²) + (ac² – bc²) – (a²c – b²c)]
    = 3[ab(a – b) + c²(a – b) – c(a² – b²)]
    = 3[ab(a – b) + c²(a – b) – c(a + b)(a – b)]
    = 3(a – b)[ab + c² – c(a + b)]
    = 3(a – b)(ab + c² – ac – bc)
    = 3(a – b)[(ab – ac) – (bc – c²)]
    = 3(a – b)[a(b – c) – c(b – c)]
    = 3(a – b)(a – c)(b – c)
     

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

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