Sao cx gặp bài phân số không -_-
\(a)\frac{19}{21}-\frac{1}{3}+\frac{3}{7}=\frac{19}{21}-(\frac{1}{3}+\frac{3}{7})=\frac{19}{21}-\frac{16}{21}=\frac{3}{21}\)
\(b)1+\frac{2}{3}-\frac{5}{3}=\frac{3}{3}+\frac{2}{3}-\frac{5}{3}=\frac{5}{3}-\frac{5}{3}=0\)
\(c)\frac{7}{12}+\frac{1}{3}-\frac{3}{4}=\frac{7}{12}+\frac{4}{12}-\frac{9}{12}=\frac{7+4-9}{12}=\frac{2}{12}=\frac{1}{6}\)
\(d)\frac{5}{7}-\frac{4}{7}\cdot\frac{1}{4}=\frac{5}{7}-\frac{1}{7}=\frac{4}{7}\)
\(e)\frac{7}{12}\cdot\frac{4}{3}+\frac{2}{9}=\frac{7}{9}+\frac{2}{9}=\frac{9}{9}=1\)
A. \(\frac{19}{21}-\frac{1}{3}+\frac{3}{7}=\frac{4}{7}+\frac{3}{7}=\frac{7}{7}=1\)
B. \(1+\frac{2}{3}-\frac{5}{3}=\frac{5}{3}-\frac{5}{3}=0\)
C. \(\frac{7}{12}+\frac{1}{3}-\frac{3}{4}=\frac{11}{12}-\frac{3}{4}=\frac{1}{6}\)
D. \(\frac{5}{7}-\frac{4}{7}\times\frac{1}{4}=\frac{5}{7}-\frac{1}{7}=\frac{4}{7}\)
E. \(\frac{7}{12}\times\frac{4}{3}+\frac{2}{9}=\frac{7}{9}+\frac{2}{9}=\frac{9}{9}=1\)
\(Linh\)\(Yukiko\)
Sửa lại câu a : \(\frac{19}{21}-\frac{1}{3}+\frac{3}{7}=\frac{19}{21}-\frac{7}{21}+\frac{9}{21}=\frac{19-7+9}{21}=\frac{21}{21}=1\)
