Giải bài 68 trang 35 - Sách giáo khoa Toán 6 tập 2

Đề bài

Tính:

Hướng dẫn giải

Giải:

a) \( \frac{3}{5}- \frac{-7}{10}- \frac{13}{-20}= \frac{3}{5}+ \frac{7}{10}+ \frac{13}{20}= \frac{3.4}{20}+ \frac{7.2}{20}+ \frac{13}{20}= \frac{12+14+13}{20}= \frac{39}{20}\)

b) \( \frac{3}{4}+\frac{-1}{3}-\frac{5}{18}=\frac{3}{4}+\frac{-1}{3}+\frac{-5}{18}=\frac{3.9}{36}+\frac{-1.12}{36}+\frac{-5.2}{36}=\frac{27-12-10}{36}=\frac{5}{36}\)

\(\frac{3}{14}-\frac{5}{-8}+\frac{-1}{2}=\frac{3}{4}+\frac{5}{8}+\frac{-1}{2}=\frac{3.4}{56}+\frac{5.7}{56}+\frac{-1.28}{56}=\frac{12+35-28}{56}=\frac{19}{56} \)

d) \( \frac{1}{2}+\frac{1}{-3}+\frac{1}{4}-\frac{1}{-6}=\frac{1}{2}+\frac{-1}{3}+\frac{1}{4}+\frac{1}{6}=\frac{1.6}{12}+\frac{-1.4}{12}+\frac{1.3}{12}+\frac{1.2}{12}\)

\(=\frac{6-4+3+2}{12}=\frac{7}{12}\)