Question

[Kfkfj]

tính \(\dfrac{0,5+0,\left(3\right)-0,1\left(6\right)}{2,5+1,\left(6\right)-0,8\left(3\right)}\)

Answer 1

[Nguyễn Trung Hiếu]

\(\frac{0,5+\frac{1}{3}-\frac{1}{6}}{2,5+\frac{5}{3}-\frac{5}{6}}\)

= \(\frac{\frac{3}{6}+\frac{2}{6}-\frac{1}{6}}{\frac{15}{6}+\frac{10}{6}-\frac{5}{6}}\)

= \(\frac{\frac{4}{6}}{\frac{20}{6}}\)

= \(\frac{4}{6}:\frac{20}{6}\)

= \(\frac{2}{3}.\frac{3}{10}\)

= \(\frac{1}{5}\)