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Ta có \[A=\sum\limits_{k=1}^{2018}{\dfrac{{{\varepsilon }_{k}}}{{{\varepsilon }_{k}}+4}}=\sum\limits_{k=1}^{2018}{\left( 1-\dfrac{4}{{{\varepsilon }_{k}}+4} \right)}=2018-4\sum\limits_{k=1}^{2018}{\dfrac{1}{{{\varepsilon }_{k}}+4}}.\]
Xét \[P(z) = {z^{2018}} - 2018 = \prod\limits_{k = 1}^{2018} {(z - {\varepsilon _k})} \Rightarrow P'(z) = P(z)\left( {\sum\limits_{k = 1}^{2018} {\dfrac{1}{{z - {\varepsilon _k}}}} } \right).\]
Suy ra \[\sum\limits_{k=1}^{2018}{\dfrac{1}{z-{{\varepsilon }_{k}}}}=\dfrac{{P}'(z)}{P(z)}=\dfrac{2018{{z}^{2017}}}{{{z}^{2018}}-2018}.\]
Do đó \[\sum\limits_{k=1}^{2018}{\dfrac{1}{-4-{{\varepsilon }_{k}}}}=\dfrac{2018{{(-4)}^{2017}}}{{{(-4)}^{2018}}-2018}\Rightarrow \sum\limits_{k=1}^{2018}{\dfrac{1}{{{\varepsilon }_{k}}+4}}=\dfrac{{{2018.4}^{2017}}}{{{4}^{2018}}-2018}.\]
Vậy \[A=2018-4\dfrac{{{2018.4}^{2017}}}{{{4}^{2018}}-2018}=-\dfrac{{{2018}^{2}}}{{{4}^{2018}}-2018}.\]
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