Cho biết

D

- Hướng dẫn giải

Ta có:  ${\sin }^2\alpha +{\cos }^2\alpha =1\Rightarrow {\cos }^2\alpha =1-{\sin }^2\alpha =\frac{8}{9}$

Mà $\frac{\mathrm{\pi }}{2}<\alpha <\mathrm{\pi }\Rightarrow \mathrm{cos\alpha }<0$ 

Suy ra: $\cos \alpha =-\frac{2\sqrt{2}}{3}$

$\sin \left(\alpha +\frac{\mathrm{\pi }}{6}\right)=\sin \alpha cos\frac{\mathrm{\pi }}{6}+\cos \alpha \sin \frac{\mathrm{\pi }}{6}=\frac{1}{3}.\frac{\sqrt{3}}{2}-\frac{2\sqrt{2}}{3}.\frac{1}{2}=\frac{\sqrt{3}-2\sqrt{2}}{6}$.

Chọn  D.