Vì \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+\left|x+\frac{3}{101}\right|+...+\left|x+\frac{100}{101}\right|>0\forall x\)
mà \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+\left|x+\frac{3}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)
nên x>0
Với x>0, ta được:
\(x+\frac{1}{101}+x+\frac{2}{101}+x+\frac{3}{101}+...+x+\frac{100}{101}=101x\)
\(\Leftrightarrow100x-101x+\frac{5050}{101}=0\)
\(\Leftrightarrow-x+50=0\)
hay x=50
Vậy: S={50}