ta có \(\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1=0\)

<=> \(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)

<=> \(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

<=> \(x+100=0\) vì ( \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}>0\)

<=> \(x=-100\)

vậy x = -100