Question

bài 1 tính hopự lý nếu có thể

a) (\(6\dfrac{4}{9}\)+\(3\dfrac{7}{11}\))-\(4\dfrac{4}{9}\)

b) (\(11\dfrac{2}{7}\)+\(3\dfrac{2}{5}\))-\(7\dfrac{2}{7}\)

Answer 1

\(a)\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)

\(=\left(\dfrac{58}{9}+\dfrac{40}{11}\right)-\dfrac{40}{9}\)

\(=\dfrac{58}{9}+\dfrac{40}{11}-\dfrac{40}{9}\)

\(=\left(\dfrac{58}{9}+\dfrac{40}{9}\right)+\dfrac{40}{11}\)

\(=2+\dfrac{40}{11}\)

\(=\dfrac{62}{11}\)

\(b)\left(11\dfrac{2}{7}+3\dfrac{2}{5}\right)-7\dfrac{2}{7}\)

\(=\left(\dfrac{79}{7}+\dfrac{17}{5}\right)-\dfrac{51}{7}\)

\(=\left(\dfrac{79}{7}-\dfrac{51}{7}\right)+\dfrac{17}{5}\)

\(=4+\dfrac{17}{5}\)

\(=\dfrac{37}{5}\)