1. 3x(x - 1) + x - 1 = 0
⇔3x (x - 1) +( x - 1) = 0
⇔(x - 1) (3x + 1) = 0
⇔\(\left[ \begin{array}{l}x - 1=0\\3x + 1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\x=-1/3\end{array} \right.\)
2. 2( x + 3) - x² - 3x = 0
⇔2 ( x + 3) -x (x + 3) = 0
⇔(x + 3) (2 - x ) = 0
⇔\(\left[ \begin{array}{l}x + 3=0\\2 - x= 0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-3\\x=2\end{array} \right.\)
3. 4x² - 25 - (2x + 5)(2x + 7) = 0
⇔(2x + 5) ( 2x - 5) - (2x - 5) ( 2x + 7) = 0
⇔ (2x - 5) (2x + 5 - 2x -7)=0
⇔(2x - 5) -2 = 0
⇔2x - 5 = 0
⇔2x = 5
⇔x = $\frac{5}{2}$
4. x³ + 27 +(x + 3)(x - 9) =0
⇔(x + 3) ( x² -3x + 9) + (x + 3) (x - 9 )=0
⇔(x + 3) (x² -3x +9 +x -9 ) =0
⇔(x + 3)( x² - 2x )=0
⇔( x + 3) x(x - 2)=0
⇔ \(\left[ \begin{array}{l}x + 3=0\\x=0\\x - 2 = 0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-3\\x=0\\x = 2\end{array} \right.\)
5. (x + 2)² - (x - 2) (x + 2) = 0
⇔(x + 2) (x + 2 - x + 2)=0
⇔(x + 2)4 = 0
⇔x + 2 = 0
⇔ x = -2
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