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

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

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

Comments ( 2 )

  1. Giải đáp:
    (a+b+c)^3-a^3-b^3-c^3=3.(a+b)(a+c)(b+c)
    Lời giải và giải thích chi tiết:
    (a+b+c)^3-a^3-b^3-c^3
    =[(a+b+c)^3-a^3]-(b^3+c^3)
    =(a+b+c-a).[(a+b+c)^2+(a+b+c).a+a^2]-(b+c)(b^2-bc+c^2)
    =(b+c).[(a+b)^2+2c.(a+b)+c^2+a^2+ab+ac+a^2]-(b+c)(b^2-bc+c^2)
    =(b+c).(a^2+2ab+b^2+2ac+2bc+c^2+a^2+ab+ac+a^2]-(b+c).(b^2-bc+c^2)
    =(b+c).(3a^2+b^2+c^2+3ab+3ac+2bc)-(b+c)(b^2-bc+c^2)
    =(b+c)(3a^2+b^2+c^2+3ab+3ac+2bc-b^2+bc-c^2)
    =(b+c)(3a^2+3ab+3ac+3bc)
    =3.(b+c)(a^2+ab+ac+bc)
    =3.(b+c).[a.(a+b)+c.(a+b)]
    =3.(a+b)(a+c)(b+c)

  2. #tnvt
    (a+b+c)^3-a^3-b^3-c^3
    =[(a+b+c)^3-a^3]-(b^3+c^3)
    =(a+b+c-a)[(a+b+c)^2+a(a+b+c)+a^2]-(b+c)(b^2-cb+c^2)
    =(b+c)(a^2+b^2+c^2+2ac+2ba+2cb+a^2+ab+ac+a^2)-(b+c)(b^2-cb+c^2)
    =(b+c)[(3a^2+b^2+c^2+3ac+3ba+2cb)-(b^2-bc+c^2)]
    =(b+c)(3a^2+b^2+c^2+3ac+3ba+2cb-b^2+bc-c^2)
    =(b+c)(3a^2+3ac+3ba+3bc)
    =(b+c)[3a(a+c)+3b(a+c)]
    =(b+c)(a+c)(3a+3b)
    =3(a+c)(a+b)(b+c)
     

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