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

Toán Lớp 10: 4x^4 – 2x^3 – 6x^2 + 2x + 2 = 0

Toán Lớp 10: 4x^4 – 2x^3 – 6x^2 + 2x + 2 = 0

Comments ( 2 )

  1. Giải đáp:
     
    Lời giải và giải thích chi tiết:
    4x^4 – 2x^3 – 6x^2 + 2x + 2 = 0
    ⇔ 2(2x^4 – x^3 – 3x^2 + x + 1) = 0
    ⇔ 2(x^2 – 2x + 1)(2x^2 + 3x + 1) = 0
    ⇔ 2(x^2 – 2x + 1)(x+1)(2x + 1) = 0
    ⇔ 2(x – 1)^2(x + 1)(2x + 1) = 0
    ⇔ (2x + 1)(x-1)^2(x + 1) = 0
    Ta có có 3 TH
    1) 2x + 1 = 0
    ⇔ 2x = -1
    ⇔ x = -1/2
    2) (x – 1)^2 = 0
    ⇔ x – 1 = 0
    ⇔ x = 1
    3) x + 1 = 0
    ⇔ x = -1
    Vậy x ∈ {-1/2 ; 1 ; -1}

  2. 4x^4-2x^3-6x^2+2x+2=0
    <=>2(2x^4-x^3+x+1)=0
    <=>2x^4-x^3-3x^2+x+1=0
    <=>(2x^4+2x^3)-(3x^3+3x^2)+(x+1)=0
    <=>2x^3(x+1)-3x^2(x+1)+(x+1)=0
    <=>(2x^3-3x^2+1)(x+1)=0
    <=>(2x^3-2x^2-x^2+1)(x+1)=0
    <=>[(2x^3-2x^2)-(x^2-1)](x+1)=0
    <=>[2x^2(x-1)-(x+1)(x-1)](x+1)=0
    <=>[2x^2-(x+1)](x-1)(x+1)=0
    <=>(2x^2-x-1)(x-1)(x+1)=0
    <=>(2x+1)(x-1)(x-1)(x+1)=0
    <=>(2x+1)(x-1)^2(x+1)=0
    <=>[(2x+1=0),((x-1)^2=0),(x+1=0):}
    <=>[(2x=-1),(x-1=0),(x=-1):}
    <=>$\left[\begin{matrix} x=-\dfrac{1}{2}\\ x=1\\ x=-1\end{matrix}\right.$
    Vậy S={-1/2; 1; -1}

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