\(4\dfrac{1}{5}+\dfrac{4}{5}+2\dfrac{2}{7}+3\dfrac{5}{7}=\left(4\dfrac{1}{5}+\dfrac{4}{5}\right)+\left(2\dfrac{2}{7}+3\dfrac{5}{7}\right)\)
\(=4+\dfrac{1}{5}+\dfrac{4}{5}+2+\dfrac{2}{7}+3+\dfrac{5}{7}\)
\(=4+1+\left(2+3\right)+\left(\dfrac{2}{7}+\dfrac{5}{7}\right)\)
\(=5+5+1=11\)
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Giải:
\(4\dfrac{1}{5}+2\dfrac{2}{7}+\dfrac{4}{5}+3\dfrac{5}{7}\)
\(=\dfrac{21}{5}+\dfrac{16}{7}+\dfrac{4}{5}+\dfrac{26}{7}\)
\(=\left(\dfrac{21}{5}+\dfrac{4}{5}\right)+\left(\dfrac{16}{7}+\dfrac{26}{7}\right)\)
\(=5+6\)
\(=11\)
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