Bài giải:

Áp dụng công thức $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$ với b>0, d>0 ta có:

$\frac{-1}{5}<\frac{1}{5}\Rightarrow \frac{-1}{5}<\frac{0}{5}<\frac{1}{5}$

$\frac{-1}{5}<\frac{0}{5}\Rightarrow \frac{-1}{5}<\frac{-1}{10}<\frac{0}{5}$

$\frac{-1}{5}<\frac{-1}{10}\Rightarrow \frac{-1}{5}<\frac{-2}{15}<\frac{-1}{10}$

$\frac{-1}{5}<\frac{-2}{15}\Rightarrow \frac{-1}{5}<\frac{-3}{20}<\frac{-2}{15}$

$\frac{-1}{5}<\frac{-3}{20}\Rightarrow \frac{-1}{5}<\frac{-4}{25}<\frac{-3}{20}$

Vậy $\frac{-1}{5}<\frac{-4}{20}<\frac{-3}{20}<\frac{-2}{15}<\frac{-1}{10}<0<\frac{1}{5}$

Đáp án đúng là A