Câu hỏi 19

Ta có

\(\begin{array}{l}\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{2\left( {\sqrt {x + 3} - 2} \right)}}{{{x^2} - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{2\left( {x - 1} \right)}}{{\left( {\sqrt {x + 3} + 2} \right)\left( {x - 1} \right)\left( {x + 1} \right)}} = \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{2}{{\left( {\sqrt {x + 3} + 2} \right)\left( {x + 1} \right)}} = \dfrac{1}{4}\\\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {a{x^2} + bx + \dfrac{1}{4}} \right) = a + b + \frac{1}{4}\\f\left( 1 \right) = a - b - \dfrac{7}{4}\end{array}\)

Hàm số liên tục tại x = 1 nên ta có 

\(\begin{array}{l}\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = f\left( 1 \right) \Leftrightarrow \dfrac{1}{4} = a + b + \dfrac{1}{4} = a - b - \dfrac{7}{4} \Leftrightarrow \left\{ \begin{array}{l}a + b = 0\\a - b = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 1\\b = - 1\end{array} \right.\\ \Rightarrow A = 2018a + b = 2018 - 1 = 2017.\end{array}\)

Chọn B.