Ta có pt :
$\dfrac{x+4}{96}+\dfrac{x+3}{97} =\dfrac{x+1}{99}+\dfrac{x+2}{98} $
$⇔ (\dfrac{x+4}{96}+1)+(\dfrac{x+3}{97}+1) =(\dfrac{x+1}{99}+1)+(\dfrac{x+2}{98}+1) $
$⇔ \dfrac{x+100}{96}+\dfrac{x+100}{97} =\dfrac{x+100}{99}+\dfrac{x+100}{98} $
$⇔x+100=0$
$⇔x=-100$