$\frac{x-1}{99} + \frac{x-2}{98} = \frac{x-3}{97} + \frac{x}{100}$
⇔$\frac{x-1}{99} -1 + \frac{x-2}{98} -1 = \frac{x-3}{97} -1 + \frac{x}{100} -1 $
⇔$\frac{x-100}{99} + \frac{x-100}{98} = \frac{x-100}{97} + \frac{x-100}{100}$
⇔ $ (x-100) ( \frac{1}{99} + \frac{1}{98} - \frac{1}{97} - \frac{1}{100} ) = 0$
Do $ \frac{1}{99} + \frac{1}{98} - \frac{1}{97} - \frac{1}{100} $ khác 0
⇒ $ x-100 = 0$
⇒ $ x=100 $
Vậy pt có nghiệm duy nhất $ x= 100 $
