Chứng minh liên hợp ta được:
\(\left\{{}\begin{matrix}\sqrt{1^3+2^3}=1+2\\\sqrt{1^3+2^3+3^3}=1+2+3\\...........\\\sqrt{1^3+2^3+...+n^3}=1+2+3+...+n\end{matrix}\right.\)(về cách liên hợp gg tìm hiểu nhé)
\(A=\dfrac{1}{\sqrt{1^3}}+\dfrac{1}{\sqrt{1^3+2^3}}+...+\dfrac{1}{\sqrt{1^3+2^3+...+2018^3}}\)
\(A=1+\dfrac{1}{1+2}+...+\dfrac{1}{1+2+3+...+2018}\)
\(A=1+\dfrac{1}{\dfrac{2.3}{2}}+...+\dfrac{1}{\dfrac{2018.2019}{2}}\)
\(A=2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2018.2019}\right)\)
\(A=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2018}-\dfrac{1}{2019}\right)\)
\(A=2\left(1-\dfrac{1}{2019}\right)=\dfrac{4036}{2019}\)
