$a)\frac{4}{x-1}-\frac{5}{x-2}=-3$ $Đkxđ:x\neq1,x\neq2$
$⇒4x-8-5x+5=3x-6-3x^2+6x$
$⇔3x^2-10x+3=0$
$⇔(3x^2-9x)-(x-3)=0$
$⇔(x-3)(3x-1)=0$
$⇔\left[ \begin{array}{l}x-3=0\\3x-1=0\end{array} \right.⇔\left[ \begin{array}{l}x=3(tm)\\x=\frac{1}{3}(tm)\end{array} \right.$
Vậy $S=${$3;\frac{1}{3}$}
$b)3x-\frac{1}{x-2}=\frac{x-1}{2-x}$ $Đkxđ:x\neq2$
$⇒3x^2-6x-1=1-x$
$⇔3x^2-5x-2=0$
$⇔(3x^2-6x)+(x-2)=0$
$⇔(x-2)(3x+1)=0$
$⇔\left[ \begin{array}{l}x-2=0\\3x+1=0\end{array} \right.⇔\left[ \begin{array}{l}x=2(loại)\\x=\frac{-1}{3}(tm)\end{array} \right.$
Vậy $S=${$\frac{-1}{3}$}.
