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

Toán Lớp 8: Phân tích đa thức x^3 – 7x – 6 thành nhân tử bằng 3 cách

Toán Lớp 8: Phân tích đa thức x^3 – 7x – 6 thành nhân tử bằng 3 cách

Comments ( 1 )

  1. +) Cách $1$:
    \qquad x^3-7x-6=x^3-x-6x-6
    =x(x^2-1)-6(x+1)=x(x-1)(x+1)-6(x+1)
    =(x+1)(x^2-x-6)=(x+1)(x^2-3x+2x-6)
    =(x+1).[x(x-3)+2(x-3)]
    =(x+1)(x+2)(x-3)
    $\\$
    +) Cách $2$:
    \qquad x^3-7x-6=x^3-4x-3x-6
    =x(x^2-4)-3(x+2)=x(x-2)(x+2)-3(x+2)
    =(x+2)(x^2-2x-3)=(x+2)(x^2-3x+x-3)
    =(x+2)[x(x-3)+x-3]=(x+2)(x-3)(x+1)
    $\\$
    +) Cách $3$:
    \qquad x^3-7x-6=x^3-3^3-7x+21
    =(x-3)(x^2+3x+9)-7(x-3)
    =(x-3)(x^2+3x+9-7)=(x-3)(x^2+x+2x+2)
    =(x-3)[x(x+1)+2(x+1)]=(x-3)(x+1)(x+2)
    $\\$
    Vậy x^3-7x+6=(x+1)(x+2)(x-3)

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

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