a: \(=\left(\dfrac{19}{29}+\dfrac{3}{29}+\dfrac{7}{29}\right)+\left(\dfrac{9}{7}+\dfrac{12}{7}\right)=1+3=4\)
b: \(=\dfrac{45}{30}\cdot\dfrac{1414}{2525}\cdot\dfrac{20}{28}=\dfrac{3}{2}\cdot\dfrac{5}{7}\cdot\dfrac{14}{25}=\dfrac{210}{350}=\dfrac{3}{5}\)
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