Câu hỏi 24

Ta có:

\(\begin{array}{l}y = \dfrac{{{{\sin }^3}x + {{\cos }^3}x}}{{1 - \sin x\cos x}} = \dfrac{{\left( {\sin x + \cos x} \right)\left( {{{\sin }^2}x - \sin x\cos x + {{\cos }^2}x} \right)}}{{1 - \sin x\cos x}}\\ = \dfrac{{\left( {\sin x + \cos x} \right)\left( {1 - \sin x\cos x} \right)}}{{1 - \sin x\cos x}} = \sin x + \cos x\end{array}\)

\(\begin{array}{l} \Rightarrow y' = \cos x - \sin x,\,\,y'' =  - \sin x - \cos x =  - y\\ \Rightarrow y'' + y = 0\end{array}\)

Chọn D.