Sửa lại đề:
$\dfrac{x+1}{99}+\dfrac{x+3}{97}+\dfrac{x+2}{98}=\dfrac{x+7}{93}+\dfrac{x+9}{91}+\dfrac{x+11}{89}$
$\to \dfrac{x+1}{99}+1)+(\dfrac{x+3}{97}+1)+(\dfrac{x+2}{98}+1)=(\dfrac{x+7}{93}+1)+(\dfrac{x+9}{91}+1)+(\dfrac{x+11}{89}+1)$
$\to \dfrac{x+1+99}{99}+\dfrac{x+3+97}{97}+\dfrac{x+2+98}{98}=\dfrac{x+7+93}{93}+\dfrac{x+9+91}{91}+\dfrac{x+11+89}{89}$
$\to \dfrac{x+100}{99}+\dfrac{x+100}{97}+\dfrac{x+100}{98}=\dfrac{x+100}{93}+\dfrac{x+100}{91}+\dfrac{x+100}{89}$
$\to \dfrac{x+100}{99}+\dfrac{x+100}{97}+\dfrac{x+100}{98}-\dfrac{x+100}{93}-\dfrac{x+100}{91}-\dfrac{x+100}{89}=0$
$\to (x+100)(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{93}-\dfrac{1}{91}-\dfrac{1}{89})=0$
Do:
$\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{98}-\dfrac{1}{93}-\dfrac{1}{89}-\dfrac{1}{91}\ne0$
$\to x+100=0$
$\to x=-100$
Vậy $S={-100}$
