a) $2-\dfrac{x+1}{x-2} = \dfrac{x+19}{(x-3).(x-2)} + \dfrac{2}{x-3}$ $(1)$
$ĐKXĐ : x \neq 2,x \neq 3$
Phương trình $(1)$ tương đương :
$\dfrac{2.(x-3).(x-2)}{(x-3).(x-2)}-\dfrac{(x+1).(x-3)}{(x-2).(x-3)} = \dfrac{x+19}{(x-3).(x-2)} + \dfrac{2.(x-2)}{(x-2).(x-3)}$
$⇒2.(x-3).(x-2) -(x+1).(x-3) = x+19+2.(x-2)$
$⇔(x-3).(x-5) = 3x +15$
$⇔x^2+15-8x = 3x+15$
$⇔x^2-11x = 0 $
$⇔\left[ \begin{array}{l}x=0\\x=11\end{array} \right.$ ( Thỏa mãn )
Vậy $S =\big\{0,11\big\}$
b) $(2x+3).(x-5) = 4x^2+6x$
$⇔(2x+3).(x-5) = 2x.(2x+3)$
$⇔(2x+3).(x-5-2x)=0$
$⇔(2x+3).(-x-5)=0$
$⇔(2x+3).(x+5)=0$
$⇔\left[ \begin{array}{l}2x+3=0\\x+5=0\end{array} \right.$
$⇔\left[ \begin{array}{l}x=\dfrac{-3}{2}\\x=-5\end{array} \right.$
Vậy $S= \bigg\{-\dfrac{3}{2}, -5\bigg\}$
