$c)$
Ta có: $x^2-3x+2=(x^2-x)-(2x-2)=(x-1)(x-2)$
$x^2-4x+3=(x^2-x)-(3x-3)=(x-1)(x-3)$
$⇒Đkxđ:x\neq1,x\neq2,x\neq3$
$\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}$
$⇔\frac{(x+4)(x-3)}{(x-1)(x-2)(x-3)}+\frac{(x+1)(x-2)}{(x-1)(x-2)(x-3)}=\frac{(2x+5)(x-2)}{(x-1)(x-2)(x-3)}$
$⇒x^2+4x-3x-12+x^2-2x+x-2=2x^2-4x+5x-10$
$⇔x=-4$ $(tm$ $đkxđ)$
Vậy $S=${$-4$}
$d)\frac{2}{x^2-4}-\frac{1}{x(x-2)}+\frac{x-4}{x(x+2)}=0$ $Đkxđ:x\neq±2,x\neq0$
$⇒2x-x-2+x^2-4x-2x+8=0$
$⇔x^2-5x+6=0$
$⇔(x^2-2x)-(3x-6)=0$
$⇔(x-2)(x-3)=0$
$⇔\left[ \begin{array}{l}x-2=0\\x-3=0\end{array} \right.⇔\left[ \begin{array}{l}x=2(loại)\\x=3(tm)\end{array} \right.$
Vậy $S=${$3$}.
