Cho hai số thực \(a\) và \(b\) thỏa \[\mathop {\lim }\limits_{x \to - 5} \frac{{4{x^2} + \left( {a + 10} \right)x + b}}{{{x^2} + 5x}} = \frac{{23}}{5}\,.\] Khi đó giá trị \(2024a + 2023b\) bằng                                                                                          

Chọn B

+ Vì \[\mathop {\lim }\limits_{x \to - 5} \frac{{4{x^2} + \left( {a + 10} \right)x + b}}{{{x^2} + 5x}} = \frac{{23}}{5}\, \Rightarrow 4{\left( { - 5} \right)^2} + \left( {a + 10} \right)\left( { - 5} \right) + b = 0 \Leftrightarrow b = 5a - 50\]

Với \(b = 5a - 50\)  ta có

\[\begin{array}{l}\mathop {\lim }\limits_{x \to - 5} \frac{{4{x^2} + \left( {a + 10} \right)x + b}}{{{x^2} + 5x}} = \mathop {\lim }\limits_{x \to - 5} \frac{{4{x^2} + \left( {a + 10} \right)x + 5a - 50}}{{{x^2} + 5x}}\\ = \mathop {\lim }\limits_{x \to - 5} \frac{{\left( {x + 5} \right)\left( {4x + a - 10} \right)}}{{\left( {x + 5} \right)x}} = \mathop {\lim }\limits_{x \to - 5} \frac{{4x + a - 10}}{x} = \frac{{a - 30}}{{ - 5}} = \frac{{23}}{5}\\ \Leftrightarrow a = 7 \Rightarrow b = - 15\, \Rightarrow 2024a + 2023b = - 16117.\end{array}\]