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

Toán Lớp 8: tìm x ( x^2 – 4x^2) – (x-4) =0

Toán Lớp 8: tìm x
( x^2 – 4x^2) – (x-4) =0

Comments ( 2 )

  1. Bạn tham khảo
    ( x^2 – 4x^2) – (x-4) =0
    <=>-3x^2-x=4=0
    <=>-(x-1)(3x+4)=0
    <=>[(x-1=0),(3x+4=0):}
    <=>[(x=1),(x=-4/3):}
    Vậy x\in{1;-4/3}
    _______________________________
    ( x^2 – 4x^2) – (x-4) =0
    <=>x^2(x-4)-(x-4)=0
    <=>(x-4)(x^2-1)=0
    <=>(x-4)(x-1)(x+1)=0
    <=>[(x-4=0),(x-1=0),(x+1=0):}
    <=>[(x=4),(x=1),(x=-1):}
    Vậy x\in{4;1;-1}

  2. Giải đáp:
    x∈{1;-4/3}
    Lời giải và giải thích chi tiết:
    (x^2-4x^2)-(x-4)=0
    -3x^2-x+4=0
    -3x^2+3x-4x+4=0
    -(3x^2-3x)-(4x-4)=0
    -3x(x-1)-4(x-1)=0
    (x-1)(3x+4)=0
    => \(\left[ \begin{array}{l}x-1=0\\3x+4=0\end{array} \right.\) 
    => \(\left[ \begin{array}{l}x=1\\x=\dfrac{-4}{3}\end{array} \right.\) 
    Vậy x∈{1;-4/3}
    __________________________________________
    Nếu đề sai:
    (x^3-4x^2)-(x-4)=0
    x^2(x-4)-(x-4)=0
    (x-4)(x^2-1)=0
    (x-4)(x-1)(x+1)=0
    => $\left[\begin{matrix} x-4=0\\x+1=0\\ x-1=0\end{matrix}\right.$
    => $\left[\begin{matrix} x=4\\x=-1\\ x=1\end{matrix}\right.$
    Vậy x∈{4;1;-1}
    ** Đặt nhân tử chung
    ** HĐT 3:A^2-B^2=(A+B)(A-B)

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