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

Toán Lớp 8: Tìm x biết : a/ ( x +1 )^2 – 3 ( x + 1 ) = 0 b/ 2 ( 3x – 2 )^2 = 9x^2 – 4

Toán Lớp 8: Tìm x biết :
a/ ( x +1 )^2 – 3 ( x + 1 ) = 0
b/ 2 ( 3x – 2 )^2 = 9x^2 – 4

Comments ( 2 )

  1. Giải đáp + Lời giải và giải thích chi tiết:
    a)
    (x+1)^{2}-3(x+1)=0
    <=>(x+1)(x+1-3)=0
    <=>(x+1)(x-2)=0
    <=> \(\left[ \begin{array}{l}x+1=0\\x-2=0\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x=-1\\x=2\end{array} \right.\) 
    Vậy S={-1;2}
    b)
    2(3x-2)^{2}=9x^{2}-4
    <=>2(3x-2)^{2}=(3x-2)(3x+2)
    <=>2(3x-2)^{2}-(3x-2)(3x+2)=0
    <=>(3x-2)[2(3x-2)-(3x+2)]=0
    <=>(3x-2)(6x-4-3x-2)=0
    <=>(3x-2)(3x-6)=0
    <=> \(\left[ \begin{array}{l}3x-2=0\\3x-6=0\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x=\dfrac{2}{3}\\x=2\end{array} \right.\) 
    Vậy S={(2)/(3);2}

  2. Giải đáp:
     
    Lời giải và giải thích chi tiết:
    a) ( x +1 )^2 – 3 ( x + 1 ) = 0
    ⇔ x^2 + 2x + 1 – 3x – 3 = 0
    ⇔ x^2 -x – 2 = 0
    ⇔ x^2 + x – 2x -2 = 0
    ⇔ x(x+1) – 2 (x+1) = 0
    ⇔ (x+1)(x-2)=0
    ⇔\(\left[ \begin{array}{l}x+1=0\\x-2=0\end{array} \right.\) 
    ⇔\(\left[ \begin{array}{l}x=-1\\x=2\end{array} \right.\) 
    Vậy x=-1; x=2
    b) 2  ( 3x – 2 )^2 = 9x^2 – 4
    ⇔2(3x-2)^2=(3x)^2-2^2
    ⇔2(3x-2)^2=(3x-2)(3x+2)
    ⇔2(3x-2)^2-(3x-2)(3x+2)=0
    ⇔(3x-2)[2(3x-2)-(3x+2)]=0
    ⇔(3x-2)(6x-4-3x-2)=0
    ⇔(3x-2)(3x-6)=0
    ⇔\(\left[ \begin{array}{l}3x-2=0\\3x-6=0\end{array} \right.\) 
    ⇔\(\left[ \begin{array}{l}3x=2\\3x=6\end{array} \right.\) 
    ⇔\(\left[ \begin{array}{l}x=\dfrac{2}{3}\\x=2\end{array} \right.\) 
    Vậy x= 2/3; x=2

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

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