a, $x^2-5x+6=0$
$⇔x^2-2x-3x+6=0$
$⇔x(x-2)-3(x-2)=0$
$⇔(x-2)(x-3)=0$
\(⇔\left[ \begin{array}{l}x-2=0\\x-3=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\)
Vậy $S=\{2;3\}$
b, $2x^3+6x^2=x^2+3x$
$⇔2x^3+5x^2-3x=0$
$⇔x(2x^2+5x-3)=0$
$⇔x(2x^2+6x-x-3)=0$
$⇔x(x+3)(2x-1)=0$
\(⇔\left[ \begin{array}{l}x=0\\x+3=0\\2x-1=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=0\\x=-3\\x=\dfrac{1}{2}\end{array} \right.\)
Vậy $S=\bigg\{-3;0;\dfrac{1}{2}\bigg\}$
c, $(3x-1)(x^2+2)=(3x-1)(7x-10)$
$⇔(3x-1)(x^2+2)-(3x-1)(7x-10)=0$
$⇔(3x-1)(x^2+2-7x+10)=0$
$⇔(3x-1)(x^2-7x+12)=0$
$⇔(3x-1)(x^2-3x-4x+12=0)=0$
$⇔(3x-1)(x-3)(x-4)=0$
\(⇔\left[ \begin{array}{l}3x-1=0\\x-3=0\\x-4=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=3\\x=4\end{array} \right.\)
Vậy $S=\bigg\{\dfrac{1}{3};3;4\bigg\}$
