Đúng ghi Đ, sai ghi S vào ô trống:
a) \(\dfrac{5}{9} + \dfrac{2}{3} = \dfrac{{5 + 2}}{{9 + 3}} = \dfrac{7}{{12}}\)
b) \(\dfrac{5}{9} + \dfrac{2}{3} = \dfrac{{5 + 2}}{{9 \times 3}} = \dfrac{7}{{27}}\)
c) \(\dfrac{5}{9} + \dfrac{2}{3} = \dfrac{5}{9} + \dfrac{6}{9} = \dfrac{{11}}{9}\)
d) \(\dfrac{5}{9} + \dfrac{2}{3} = \dfrac{{5 \times 2}}{{9 \times 3}} = \dfrac{{10}}{{12}}\)
Ta có : \(\dfrac{5}{9}+\dfrac{2}{3}=\dfrac{5}{9}+\dfrac{6}{9}=\dfrac{11}{9}.\)
Vậy ta có kết quả như sau :
a) \(\dfrac{5}{9} + \dfrac{2}{3} = \dfrac{{5 + 2}}{{9 + 3}} = \dfrac{7}{{12}}\)
b) \(\dfrac{5}{9} + \dfrac{2}{3} = \dfrac{{5 + 2}}{{9 \times 3}} = \dfrac{7}{{27}}\)
c) \(\dfrac{5}{9} + \dfrac{2}{3} = \dfrac{5}{9} + \dfrac{6}{9} = \dfrac{{11}}{9}\)
d) \(\dfrac{5}{9} + \dfrac{2}{3} = \dfrac{{5 \times 2}}{{9 \times 3}} = \dfrac{{10}}{{12}}\)
-- Mod Toán lớp 4