Giải thích các bước giải:

$a)\ (x+1)(x^2-3x+2)=(x+1)x^2$

$\Rightarrow (x+1)[(x^2-3x+2)-x^2]=0$

$\Rightarrow (x+1)(-3x+2)=0$

$\Rightarrow x+1= 0$ hoặc $-3x+2 = 0$

$\Rightarrow x=-1$ hoặc $x = \dfrac23$

Vậy $S =\left\{-1;\dfrac23\right\}$

$b) \dfrac{x-5}{3}=\dfrac{2x+1}{2}-1$

$\Rightarrow 2(x-5)=3(2x+1)-6$

$\Rightarrow 2x-10 = 6x-3$

$\Rightarrow 4x=-7$

$\Rightarrow x=-\dfrac74$

Vậy $S=\left\{-\dfrac74\right\}$

$c)\ 3(x-2)+x^2=x(x-4)$

$\Rightarrow 3x-6+x^2=x^2-4x$

$\Rightarrow 7x=6$

$\Rightarrow x=\dfrac67$

Vậy $S=\left\{\dfrac67\right\}$

$d)\ \dfrac{3x-1}{2}+\dfrac x3=\dfrac{2x+1}{6}$

$\Rightarrow 3(3x-1)+2x=2x+1$

$\Rightarrow 9x-3+2x=2x+1$

$\Rightarrow 9x=4$

$\Rightarrow x=\dfrac49$

Vậy $S=\left\{\dfrac49\right\}$