Ta có:
\(1+a^2+a^4=\left(a^2-a+1\right)\left(a^2+a+1\right)\)
Từ đây thì ta có:
\(A=\dfrac{1}{1+1^2+1^4}+\dfrac{2}{1+2^2+2^4}+...++\dfrac{2013}{1+2013^2+2013^4}\)
\(\Leftrightarrow2A=\dfrac{2}{\left(1^2-1+1\right)\left(1^2+1+1\right)}+\dfrac{4}{\left(2^2-2+1\right)\left(2^2+2+1\right)}+...+\dfrac{4026}{\left(2013^2-2013+1\right)\left(2013^2+2013+1\right)}\)
\(=\dfrac{2}{1.3}+\dfrac{4}{3.7}+...+\dfrac{4026}{4050157.4054183}\)
\(=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{7}+...+\dfrac{1}{4050157}-\dfrac{1}{4054183}=1-\dfrac{1}{4054183}=\dfrac{4054182}{4054183}\)
\(\Rightarrow A=\dfrac{2027091}{4054183}\)
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