Đáp án:
C2:
\(\left[ \begin{array}{l}
x \ge 2019\\
x \le 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
C1:\\
\mathop {\lim }\limits_{x \to - 2} \dfrac{{x + 2}}{{{x^2} - x - 6}} = \mathop {\lim }\limits_{x \to - 2} \dfrac{{x + 2}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to - 2} \dfrac{1}{{x - 3}} = \dfrac{1}{{ - 2 - 3}} = - \dfrac{1}{5}\\
C2:\\
y' = {x^2} - 2020x + 2019\\
y' \ge 0\\
\Leftrightarrow {x^2} - 2020x + 2019 \ge 0\\
\to \left( {x - 2019} \right)\left( {x - 1} \right) \ge 0\\
\to \left[ \begin{array}{l}
x \ge 2019\\
x \le 1
\end{array} \right.
\end{array}\)
