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

Toán Lớp 8: a) x²y² + xy + 1/4 b) x³ – 6x² + 12x – 8 c) (x-y) ³ + 6(x-y) ² + 12(x-y) +8

Toán Lớp 8: a) x²y² + xy + 1/4
b) x³ – 6x² + 12x – 8
c) (x-y) ³ + 6(x-y) ² + 12(x-y) +8

Comments ( 2 )

  1. a/ $x^2y^2+xy+\dfrac{1}{4}\\=(xy)^2+2.xy.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\\=\left(xy+\dfrac{1}{2}\right)^2$
    Vậy $x^2y^2+xy+\dfrac{1}{4}=\left(xy+\dfrac{1}{2}\right)^2$
    b/ $x^3-6x^2+12x-8\\=x^3-3.x^2.2+3.x.4-2^3\\=x^3-3.x^2.2+3.x.2^2-2^3\\=(x-2)^3$
    Vậy $x^3-6x^2+12x-8=(x-2)^3$
    c/ $(x-y)^3+6(x-y)^2+12(x-y)+8\\=(x-y)^3+3.(x-y)^2.2+3.(x-y).4+2^3\\=(x-y)^3+3.(x-y)^2.2+3.(x-y).2^2+2^3\\=[(x-y)+2]^3\\=(x-y+2)^3$
    Vậy $(x-y)^3+6(x-y)^2+12(x-y)+8=(x-y+2)^3$
     

  2. Giải đáp:Đề là phân tích đa thức thành nhân tử?
     
    Lời giải và giải thích chi tiết:
    a)x^2y^2+xy+1/4
    =(xy)^2+2*xy*1/2+(1/2)^2
    =(xy+1/2)^2
    b)x^3-6x^2+12x-8
    =x^3-3.x^{2}.2+3.x.2^2-2^3
    =(x-2)^3
    c)(x-y)^3+6(x-y)^2+12(x-y)+8
    =(x-y)^3+3.(x-y)^{2}.2+3.(x-y).2^2+2^3
    =(x-y+2)^3

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

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