$\text{Bài 11:}$
$a) \dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}$
$\to \dfrac{x-23}{24}+\dfrac{x-23}{25}-\dfrac{x-23}{26}-\dfrac{x-23}{27}=0$
$\to (x-23)(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27})=0$
$\text{Do:} \dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\ne 0$
$\to x-23=0$
$\to x=23$
$\text{Vậy S={23}}$
$c) \dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}$
$\to \dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1-(\dfrac{x+3}{2002}+1)-(\dfrac{x+4}{2001}+1)=0$
$\to \dfrac{x+1+2004}{2004}+\dfrac{x+2+2003}-\dfrac{x+3+2002}{2002}-\dfrac{x+4+2001}{2001}=0$
$\to (x+2005)(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001})=0$
$\text{Do:} \dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\ne 0$
$\to x+2005 =0$
$\to x=-2005$
$\text{Vậy S={-2005}}$
$d)\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0$
$\to (\dfrac{201-x}{99}+1)+(\dfrac{203-x}{97}+1)+(\dfrac{205-x}{95}+1)=0$
$\to \dfrac{201-x+99}{99}+\dfrac{203-x+97}{97}+\dfrac{205-x+95}{95}=0$
$\to \dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0$
$\to (300-x)(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95})=0$
$\text{Do:} \dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\ne 0$
$\to 300-x=0$
$\to x=300$
$\text{Vậy S={300}}$
