Question by Dao Trong Luan

Dao Trong Luan Coordinator

07/04/2018 at 05:07

\(A=\dfrac{\dfrac{\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{1}{2017}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}}}{\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2017}}{\dfrac{1}{2016}+\dfrac{2}{2015}+...+\dfrac{2016}{1}}}\)

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Lê Quốc Trần Anh Coordinator 07/04/2018 at 11:38

Calculate(v), not "calculator"(n) or "calculating"(v-ing)
Unknown account 07/04/2018 at 11:31

Calculating not "caculator" :)

Dao Trong Luan Coordinator

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