\(x+2x+3x+....+100x=550\)
\(\Leftrightarrow x.\left(1+2+3+....+100\right)=550\)
Ta có : \(1+2+3+...+100=\frac{\left(100+1\right).100}{2}=5050\)
\(\Rightarrow x.5050=550\)
\(\Leftrightarrow x=\frac{550}{5050}=\frac{11}{101}\)
Vậy : \(x=\frac{11}{101}\)