b) \(7\dfrac{2}{5}+8\dfrac{4}{9}+2\dfrac{3}{5}-3\dfrac{4}{9}\)
= \(\left(7\dfrac{2}{5}+2\dfrac{3}{5}\right)+\left(8\dfrac{4}{9}-3\dfrac{4}{9}\right)\)
= 10 + 5 = 15
d) \(1\dfrac{4}{7}+\dfrac{3}{7}\times\dfrac{1}{5}+\dfrac{3}{7}\times\dfrac{4}{5}\)
\(=1\dfrac{4}{7}+\dfrac{3}{7}\times\left(\dfrac{1}{5}+\dfrac{4}{5}\right)\)
= \(1\dfrac{4}{7}+\dfrac{3}{7}\)
\(=1+\dfrac{4}{7}+\dfrac{3}{7}\)
= \(1+\dfrac{7}{7}\)
=1 + 1 = 2
b) Ta có: \(7\dfrac{2}{5}+8\dfrac{4}{9}+2\dfrac{3}{5}-3\dfrac{4}{9}\)
\(=\dfrac{37}{5}+\dfrac{76}{9}+\dfrac{13}{5}-\dfrac{31}{9}\)
\(=10+5\)
=15
d) Ta có: \(1\dfrac{4}{7}+\dfrac{3}{7}\cdot\dfrac{1}{5}+\dfrac{3}{7}\cdot\dfrac{4}{5}\)
\(=\dfrac{11}{7}+\dfrac{3}{7}\)
=2
toán này là toán lớp 5 mà sao lại lớp 3
