\(9\times\left(\dfrac{-1}{3}\right)^3+\dfrac{1}{3}:0,25-5\dfrac{3}{4}=\dfrac{9}{1}\times\left(\dfrac{-1}{27}\right)+\dfrac{1}{3}:\dfrac{1}{4}-\dfrac{23}{4}\)
\(=\left(\dfrac{-9}{27}\right)+\dfrac{4}{3}-\dfrac{23}{4}=\left(\dfrac{-1}{3}\right)+\dfrac{4}{3}-\dfrac{23}{4}=\dfrac{3}{3}-\dfrac{23}{4}=1-\dfrac{23}{4}=\dfrac{4}{4}-\dfrac{23}{4}=\dfrac{-19}{4}\)
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