a, $3x(x-1)+2(1-x)=0$
$⇔3x(x-1)-2(x-1)=0$
$⇔(x-1)(3x-2)=0$
$⇔$\(\left[ \begin{array}{l}x-1=0\\3x-2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\x=2/3\end{array} \right.\)
b, $x(2x-3)-4x+6=0$
$⇔x(2x-3)-2(2x-3)=0$
$⇔(2x-3)(x-2)=0$
$⇔$\(\left[ \begin{array}{l}2x-3=0\\x-2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=3/2\\x=2\end{array} \right.\)
c, $x^2-4-(x+5)(2-x)=0$
$⇔-(4-x^2)-(x+5)(2-x)=0$
$⇔-(2-x)(2+x)-(x+5)(2-x)=0$
$⇔(2-x)(-2x-x-x-5)=0$
$⇔(2-x)(-4x-5)=0$
⇔\(\left[ \begin{array}{l}2-x=0\\-4x-5=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=2\\x=-5/4\end{array} \right.\)
d, $2x^3+4x^2=x^2+2x$
$⇔2x^3+3x^2-2x=0$
$⇔x(2x^2+3x-2)=0$
$⇔x(x-2)(2x-1)=0$
$⇔$\(\left[ \begin{array}{l}x=0\\x-2=0\\2x-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=2\\x=1/2\end{array} \right.\)
