a) $x(2x -3) – 2(3 – 2x)=0$
⇔$x(2x -3) + 2(2x – 3) = 0$
⇔$(2x-3)(x+2)=0$
⇒\(\left[ \begin{array}{l}2x-3=0\\x+2=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=\frac{3}{2}\\x=-2\end{array} \right.\)
b) $(x + \frac{1}{2} )^2 − (x + \frac{1}{2} ) (x + 6) = 8 $
⇔$(x + \frac{1}{2})(x + \frac{1}{2}-x-6)=8$
⇔$(x + \frac{1}{2})(-\frac{11}{2})=8$
⇔$x+\frac{1}{2}=-\frac{16}{11}$
⇒$x=-\frac{43}{22}$
c)$(x^2 + 2x)^2 - 2x^2 – 4x = 3$
⇔$(x^2 + 2x)^2-2(x^2+2x)-3=0$
⇔$(x^2 + 2x)^2+(x^2+2x)-3(x^2+2x)-3=0$
⇔$(x^2 + 2x)[(x^2 + 2x)+1]-3[(x^2 + 2x)+1]=0$
⇔$(x^2 + 2x+1)(x^2 + 2x-3)=0$ ⇔$(x+1)^2(x-1)(x+3)=0$
⇔$x+1=0$ $hoặc$ $x-1=0$ $hoặc$ $x+3=0$
⇒$x=-1$ $hoặc$ $x=1$ $hoặc$ $x=-3$
