Câu hỏi 28

Ta có:

\(\begin{array}{l}y' = \dfrac{{x + 1 - x + 1}}{{{{\left( {x + 1} \right)}^2}}} = \dfrac{2}{{{{\left( {x + 1} \right)}^2}}}\\ \Rightarrow y'' = \dfrac{{ - 2.2\left( {x + 1} \right)}}{{{{\left( {x + 1} \right)}^4}}} = \dfrac{{ - 4}}{{{{\left( {x + 1} \right)}^3}}} \Rightarrow y''\left( 0 \right) =  - 4\end{array}\)

Chọn B.