$ĐK : x \neq 49,x \neq 50$
Ta có :
$\dfrac{x-49}{50}+\dfrac{x-50}{49} =\dfrac{49}{x-50}+\dfrac{50}{x-49} $
$ ⇔ \dfrac{x-49}{50}-1+\dfrac{x-50}{49}-1 =\dfrac{49}{x-50}-1+\dfrac{50}{x-49}-1 $
$⇔\dfrac{x-99}{50}+\dfrac{x-99}{49} =\dfrac{99-x}{x-50}+\dfrac{99-x}{x-49} $
$⇔(x-99).( dfrac{1}{50}+\dfrac{1}{49} +\dfrac{1}{x-50}+\dfrac{1}{x-49}) =0$
$⇔x-99=0$
$⇔x=99$
