Câu hỏi 8

Xét hàm số: \(y = \dfrac{{{x^2} + x + 3}}{{x - 2}}\)  trên \(\left[ { - 2;\,\,1} \right]\) ta có:

\(y' = \dfrac{{\left( {2x + 1} \right)\left( {x - 2} \right) - {x^2} - x - 3}}{{{{\left( {x - 2} \right)}^2}}}\) \( = \dfrac{{2{x^2} - 3x - 2 - {x^2} - x - 3}}{{{{\left( {x - 2} \right)}^2}}}\)\( = \dfrac{{{x^2} - 4x - 5}}{{{{\left( {x - 2} \right)}^2}}}\)

\( \Rightarrow y' = 0 \Leftrightarrow {x^2} - 4x - 5 = 0\) \( \Leftrightarrow \left[ \begin{array}{l}x =  - 1 \in \left[ { - 2;\,\,1} \right]\\x = 5\,\, \notin \left[ { - 2;\,\,1} \right]\end{array} \right.\)

\( \Rightarrow \left\{ \begin{array}{l}y\left( { - 2} \right) =  - \dfrac{5}{4}\\y\left( { - 1} \right) =  - 1\\y\left( 1 \right) =  - 5\end{array} \right.\) \( \Rightarrow \left\{ \begin{array}{l}m = \mathop {Min}\limits_{\left[ { - 2;\,\,1} \right]} y =  - 5\\M = \mathop {Max}\limits_{\left[ { - 2;\,\,1} \right]} y =  - 1\end{array} \right.\) \( \Rightarrow M + m =  - 1 - 5 =  - 6.\)

Chọn C.