$a.$ $(9x^{2} - 4)( x + 1) = (3x + 2)(x^{2} – 1)$
$⇔ (3x - 2)(3x + 2)(x + 1) = (3x + 2)(x - 1)(x + 1)$
$⇔ (3x + 1)(x + 1)(2x - 1) = 0$
$⇔$ \(\left[ \begin{array}{l} 3x \ + \ 2 \ = \ 0 \\ x \ + \ 1 \ = \ 0 \\ 2x \ - 1 \ = \ 0 \end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x \ = \ \frac {-2}{3} \\ x \ = \ -1 \\ x \ = \ \frac {1}{2} \end{array} \right.\)
$b.$ $(x -1 )^{2} - 1 + x^{2} = (1 - x)(x + 3)$
$⇔ (x – 1)(x – 1) + (x – 1)(x + 1) + (x – 1)(x + 3) = 0$
$⇔ (x - 1)(x - 1 + x + 1 + x + 3) = 0$
$⇔ (x - 1)(3x + 3) = 0$
$⇔$ \(\left[ \begin{array}{l} x \ - \ 1 \ = \ 0 \\ 3x \ + \ 3 \ = \ 0 \end{array} \right.\) $⇔$ \(\left[ \begin{array}{l} x \ = \ 1 \\ x \ = \ -1 \end{array} \right.\)
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