Giải các phương trình sau : a ) ( 9x^2 – 4 ) ( x+1 ) = ( 3x + 2 ) ( x^2 -1 ) b ) ( x-1 )^2 -1 +x^2=(1-x) (x+3)

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1. $a.$ $(9x^{2} – 4)( x + 1) = (3x + 2)(x^{2} – 1)$

   $⇔ (3x – 2)(3x + 2)(x + 1) = (3x + 2)(x – 1)(x + 1)$

   $⇔ (3x + 1)(x + 1)(2x – 1) = 0$

   $⇔$ \(\left[ \begin{array}{l} 3x \ + \ 2 \ = \ 0 \\ x \ + \ 1 \ = \ 0 \\ 2x \ – 1 \ = \ 0 \end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x \ = \ \frac {-2}{3} \\ x \ = \ -1 \\ x \ = \ \frac {1}{2} \end{array} \right.\)

   $b.$ $(x -1 )^{2} – 1 + x^{2} = (1 – x)(x + 3)$

   $⇔ (x – 1)(x – 1) + (x – 1)(x + 1) + (x – 1)(x + 3) = 0$

   $⇔ (x – 1)(x – 1 + x + 1 + x + 3) = 0$

   $⇔ (x – 1)(3x + 3) = 0$

   $⇔$ \(\left[ \begin{array}{l} x \ – \ 1 \ = \ 0 \\ 3x \ + \ 3 \ = \ 0 \end{array} \right.\) $⇔$ \(\left[ \begin{array}{l} x \ = \ 1 \\ x \ = \ -1 \end{array} \right.\)

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2. a ) ( 9x^2 – 4 ) ( x+1 ) = ( 3x + 2 ) ( x^2 -1 )

        (3x-2)(3x+2)(x+1)-(3x+2)(x-1)(x+1)=0

        (3x+2)(x+1)(3x-2-x+1)=0

         (3x+2)(x+1)(2x-1)=0

   =>3x+2=0 hoặc x+1=0 hoặc 2x-1=0

        3x+2=0 =>x=-2/3

        x+1=0 => x=-1

        2x-1=0 => x=1/2

   vậy S={-2/3 , -1 , 1/2}

   b ) ( x-1 )^2 -1 +x^2=(1-x) (x+3)

        x^2-2x+1-1+x^2-(1-x)(x+3)=0

        2x^2-2x-x-3+x^2+3x=0

        3x^2                           =0

      =>x= 0

   vậy S={0)

    

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