\(=\dfrac{1}{2}\cdot3-\dfrac{1}{4}+\dfrac{1}{3}=\dfrac{3}{2}-\dfrac{1}{4}+\dfrac{1}{3}=\dfrac{19}{12}\)
\(=0,5.3-\dfrac{1}{4}+\dfrac{1}{3}\\=\dfrac{3}{2}-\dfrac{1}{4}+\dfrac{1}{3}\\ =\dfrac{18}{12}+\dfrac{-3}{12}+\dfrac{4}{12}\\ \dfrac{19}{12} \)
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