Giải câu 6 trang 29 sách toán VNEN 6 tập 2

a) $\frac{3}{10}\;-\;\frac{-2}{5}\;-\;\frac{11}{-20} = \frac{3}{10}\;+\;\frac{4}{10}\;-\;\frac{11}{-20} = \frac{3\;+\;4}{10}\;+\;\frac{11}{20} = \frac{7}{10}\;+\;\frac{11}{20} = \frac{14}{20}\;+\;\frac{11}{20} = \frac{25}{20} = \frac{5}{4}$.

b) $\frac{3}{4}\;+\;\frac{-5}{6}\;-\;\frac{7}{18} = \frac{3}{4}\;-\;\frac{5}{6}\;-\;\frac{7}{18} = \frac{9}{12}\;-\;\frac{10}{12}\;-\;\frac{7}{18} = \frac{-1}{12}\;-\;\frac{7}{18} = \frac{-3}{36}\;-\;\frac{14}{36} = \frac{-17}{36}$

c) $\frac{5}{14}\;-\;\frac{7}{-8}\;+\;\frac{-1}{2} = \frac{5}{14}\;+\;\frac{7}{8}\;-\;\frac{1}{2} = \frac{20}{56}\;+\;\frac{49}{56}\;-\;\frac{1}{2} = \frac{69}{56}\;-\;\frac{1}{2} = \frac{69}{56}\;-\;\frac{28}{56} = \frac{41}{56}$.

d) $\frac{1}{2}\;+\;\frac{1}{-4}\;+\;\frac{2}{3}\;-\;\frac{-5}{6} = \frac{1}{2}\;-\;\frac{1}{4}\;+\;\frac{2}{3}\;+\;\frac{5}{6} = \frac{2}{4}\;-\;\frac{1}{4}\;+\;\frac{2}{3}\;+\;\frac{5}{6} = \frac{1}{4}\;+\;\frac{2}{3}\;+\;\frac{5}{6}$

$= \frac{3}{12}\;+\;\frac{8}{12}\;+\;\frac{5}{6} = \frac{11}{12}\;+\;\frac{5}{6} = \frac{11}{12}\;+\;\frac{10}{12} = \frac{21}{12}$