Giải bài 1 trang 167 - SGK Toán lớp 4
Tính:
a) \(\dfrac{2}{7}+ \dfrac{4}{7};\) \(\dfrac{6}{7}-\dfrac{2}{7};\)
\(\dfrac{6}{7}-\dfrac{4}{7};\) \(\dfrac{4}{7}+ \dfrac{2}{7};\)
b) \(\dfrac{1}{3}+ \dfrac{5}{12};\) \(\dfrac{9}{12}-\dfrac{1}{3};\)
\(\dfrac{9}{12}-\dfrac{5}{12};\) \(\dfrac{5}{12}+ \dfrac{1}{3}.\)
a) \(\dfrac{2}{7}+ \dfrac{4}{7} = \dfrac{2\,+\,4}{7}=\dfrac{6}{7}\) \(\dfrac{6}{7}-\dfrac{2}{7}= \dfrac{6\,-\,2}{7}=\dfrac{4}{7}\)
\(\dfrac{6}{7}-\dfrac{4}{7}= \dfrac{6\,-\,4}{7}=\dfrac{2}{7}\) \(\dfrac{4}{7}+ \dfrac{2}{7}= \dfrac{4\,+\,2}{7}=\dfrac{6}{7}\)
b) \(\dfrac{1}{3}+ \dfrac{5}{12}=\dfrac{4}{12}+ \dfrac{5}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)
\(\dfrac{9}{12}-\dfrac{1}{3}=\dfrac{9}{12}-\dfrac{4}{12}=\dfrac{3}{12}=\dfrac{1}{4}\)
\(\dfrac{9}{12}-\dfrac{5}{12}=\dfrac{9\,-\,5}{12}=\dfrac{2}{12}=\dfrac{1}{6}\)
\(\dfrac{5}{12}+ \dfrac{1}{3}=\dfrac{5}{12}+ \dfrac{4}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)