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

Toán Lớp 8: Giải phương trình sau: 6×4 – 5×3 – 38×2 – 5x + 6 = 0

Toán Lớp 8: Giải phương trình sau: 6×4 – 5×3 – 38×2 – 5x + 6 = 0

Comments ( 2 )

  1. Giải đáp:
     
    Lời giải và giải thích chi tiết:
    6x^4 – 5x^3 – 38x^2 – 5x + 6 = 0
    ⇔6x^4+3x³ – 8×3 – 4x²  – 34x² – 17x + 12x + 6 = 0
    ⇔3x³(2x+1)-4x²(2x+1)-17x(2x+1)+6(2x+1)=0
    ⇔(2x+1)(3x³-4x²-17x+6)=0
    ⇔(2x+1)(3x³-x²-3x²+x-18x+6)=0
    ⇔(2x+1)(3x-1)(x²-x-6)=0
    ⇔(2x+1)(3x-1)(x+2)(x-3)=0
    ⇔ 2x+1=0 , 3x-1=0 , x+2=0 , x-3=0
    ⇔2x=-1,3x=1,x=2,x=3
    ⇔x=-1/2,x=1/3,x=2,x=3
    ⇒S = {-1/2;1/3;2;3}

  2. 6x^4 – 5x^3 – 38x^2 – 5x + 6 = 0
    ⇔ 6x^4 – 18x^3 + 13x^3 – 39x^2 + x^2 – 3x – 2x + 6 = 0
    ⇔ 6x^3 (x – 3) + 13x^2 (x – 3) + x (x – 3) – 2 (x – 3) = 0
    ⇔ (6x^3 + 13x^2 + x – 2) (x – 3) = 0
    ⇔ (6x^3 + 12x^2 + x^2 + 2x – x – 2) (x – 3) = 0
    ⇔ [6x^2 (x + 2) + x (x + 2) – (x + 2)] (x – 3) = 0
    ⇔ (6x^2 + x – 1) (x + 2) (x – 3) = 0
    ⇔ (6x^2 – 2x + 3x – 1) (x + 2) (x – 3) = 0
    ⇔ [2x (3x – 1) + (3x – 1)] (x  + 2) (x – 3) = 0
    ⇔ (2x + 1) (3x – 1) (x + 2) (x – 3) = 0
    ⇔ \(\left[ \begin{array}{l}2x + 1=0\\3x – 1 = 0\\ x + 2 = 0\\x – 3=0\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}2x = – 1\\3x = 1\\ x + 2 = 0\\x – 3=0\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x = \dfrac{-1}{2} \\x = \dfrac{1}{3}\\ x  =  – 2\\x =3\end{array} \right.\) 
                                     Vậy x ∈ {$\dfrac{-1}{2}$ ; $\dfrac{1}{3}$ , – 2; 3}

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