a)
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: \(2\dfrac{1}{4}+1\dfrac{1}{6}\)
\(=2\dfrac{3}{12}+1\dfrac{2}{12}\)
\(=3\dfrac{5}{12}\)
b) 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: \(4-2\dfrac{6}{7}\) \(=3\dfrac{7}{7}-2\dfrac{6}{7}\) \(=1\dfrac{1}{7}\)
Đáp án và hướng dẫn giải bài 109:
a) Cách 1
cách 2
b) Cách 1
cách 2
c, cách 1
cách 2