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

Toán Lớp 8: Phân tích đa thức thành nhân tử: A = 125x^3 -10x^2 + 2x -1 B = x^4+4 C = x^4 + 4y^4 Tìm x a) x^3 – 7x^2 – 9x +63 = 0

Toán Lớp 8: Phân tích đa thức thành nhân tử:
A = 125x^3 -10x^2 + 2x -1
B = x^4+4
C = x^4 + 4y^4
Tìm x
a) x^3 – 7x^2 – 9x +63 = 0

Comments ( 1 )

  1. Bài 1:
    A=125x^3-10x^2+2x-1
    =125x^3+15x^2-25x^2+5x-3x-1
    =(125x^3+15x^2+5x)-(25x^2+3x+1)
    =5x(25x^2+3x+1)-(25x^2+3x+1)
    =(5x-1)(25x^2+3x+1)
    B=x^4+4
    =x^4+4x^2-4x^2+4
    =(x^4+4x^2+4)-4x^2
    =[(x^2)^2+2.x^2 .2+2^2]-4x^2
    =(x^2+2)^2-(2x)^2
    =(x^2+2x+2)(x^2-2x+2)
    C=x^4+4y^4
    =x^4+4y^4+4x^2y^2-4x^2y^2
    =(x^4+4x^2y^2+4y^4)-4x^2y^2
    =[(x^2)^2+2.x^2 .2y^2+(2y^2)^2]-(2xy)^2
    =(x^2+2y^2)^2-(2xy)^2
    =(x^2+2xy+2y^2)(x^2-2xy+2y^2)
    Bài 2:
    a)x^3-7x^2-9x+63=0
    ⇔(x^3-7x^2)-(9x-63)=0
    ⇔x^2(x-7)-9(x-7)=0
    ⇔(x-7)(x^2-9)=0
    ⇔(x-7)(x^2-3^2)=0
    ⇔(x-7)(x+3)(x-3)=0
    ⇔$\left[\begin{matrix} x-7=0\\ x+3=0\\x-3=0\end{matrix}\right.$
    ⇔$\left[\begin{matrix} x=7\\ x=-3\\x=3\end{matrix}\right.$
    Vậy x∈{7;-3;3}

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

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