Tính bằng hai cách : a) b) c)

a) C1: C2: b) C1: C2: c) C1: C2:

Bài 109 (Sách giáo khoa trang 49)

Gửi bởi: Sách Giáo Khoa Vào 26 tháng 12 2018 lúc 11:14:37

Tính bằng hai cách :

a) \(2\dfrac{1}{4}+1\dfrac{1}{6}\)

b) \(7\dfrac{1}{8}-5\dfrac{3}{4}\)

c) \(4-2\dfrac{6}{7}\)

C1: \(2\dfrac{1}{4}+1\dfrac{1}{6}\)

\(=\dfrac{9}{4}+\dfrac{7}{6}\)

\(=\dfrac{27}{12}+\dfrac{14}{12}\)

\(=\dfrac{41}{12}=3\dfrac{5}{12}\)

C2: \(2\dfrac{1}{4}+1\dfrac{1}{6}\)

\(=2\dfrac{3}{12}+1\dfrac{2}{12}\)

\(=3\dfrac{5}{12}\)

C1: \(7\dfrac{1}{8}-5\dfrac{3}{4}\)

\(=\dfrac{57}{8}-\dfrac{23}{4}\)

\(=\dfrac{57}{8}-\dfrac{46}{8}\)

\(=\dfrac{11}{8}=1\dfrac{3}{8}\)

C2: \(7\dfrac{1}{8}-5\dfrac{3}{4}\)

\(=7\dfrac{1}{8}-5\dfrac{6}{8}\)

\(=6\dfrac{9}{8}-5\dfrac{6}{8}\)

\(=1\dfrac{3}{8}\)

c) C1: \(4-2\dfrac{6}{7}\)

\(=\dfrac{28}{7}-\dfrac{20}{7}\)

\(=\dfrac{8}{7}=1\dfrac{1}{7}\)

C2: \(4-2\dfrac{6}{7}\)

\(=3\dfrac{7}{7}-2\dfrac{6}{7}\)

\(=1\dfrac{1}{7}\)