A. \(4037\)
B. \(4035\)
C. \( - 4035\)
D. \(4033\)
B
\(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^{2018}} + x - 2}}{{{x^{2017}} + x - 2}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^{2018}} - 1 + x - 1}}{{{x^{2017}} - 1 + x - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {{x^{2017}} + {x^{2016}} + ... + 1} \right) + x - 1}}{{\left( {x - 1} \right)\left( {{x^{2016}} + {x^{2015}} + ... + 1} \right) + x - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {{x^{2017}} + {x^{2016}} + ... + 1 + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^{2016}} + {x^{2015}} + ... + 1 + 1} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^{2017}} + {x^{2016}} + ... + 1 + 1}}{{{x^{2016}} + {x^{2015}} + ... + 1 + 1}} = \dfrac{{2018}}{{2017}}\\ \Rightarrow a = 2018,\,\,b = 2017\\ \Rightarrow {a^2} - {b^2} = {2018^2} - {2017^2} = \left( {2018 - 2017} \right)\left( {2018 + 2017} \right) = 4035\end{array}\)
Chọn B.
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