Question

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

Answer 1

Vậy x+1;x+2;x+3;x+4=0

=> x=-1;-2;-3;-4

nha bạn        

Answer 2

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

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

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

\(x+100=0\)

\(x=-100\)