（5分）若$\tan \theta =-2$，则$\dfrac{\sin \theta (1+\sin 2\theta )}{\sin \theta +\cos \theta }=$(　　)
A．$-\dfrac{6}{5}$              
B．$-\dfrac{2}{5}$            
C．$\dfrac{2}{5}$              
D．$\dfrac{6}{5}$
分析：由题意化简所给的三角函数式，然后利用齐次式的特征即可求得三角函数式的值．
解：由题意可得：$\dfrac{\sin \theta (1+\sin 2\theta )}{\sin \theta +\cos \theta }=\dfrac{\sin \theta ({\sin }^{2}\theta +{\cos }^{2}\theta +2\sin \theta \cos \theta )}{\sin \theta +\cos \theta }$
$=\dfrac{\sin \theta {(\sin \theta +\cos \theta )}^{2}}{\sin \theta +\cos \theta }=\sin \theta (\sin \theta +\cos \theta )$
$=\dfrac{{\sin }^{2}\theta +\sin \theta \cos \theta }{{\sin }^{2}\theta +{\cos }^{2}\theta }=\dfrac{{\tan }^{2}\theta +\tan \theta }{1+{\tan }^{2}\theta }$
$=\dfrac{4?2}{1+4}=\dfrac{2}{5}$．
故选：C．
点评：本题主要考查同角三角函数基本关系，三角函数式的求值等知识，属于中等题．