已知函数$f(x)=\sin(2x+\dfrac{\mathrm{{\pi }}}{3})+\sin2x$，将$f(x)$的图象向右平移$\dfrac{\mathrm{{\pi }}}{6}$个单位长度，得到函数$g\left(x\right)$的图象，若$h\left(x\right)=f\left(x\right)+g\left(x\right)$，则$ $ $(\qquad)$ $ $

$f(x)$图象的对称轴的方程为$x=\\dfrac{k\\pi }{2}+\\dfrac{\\mathrm{{\\pi }}}{6}(k\\in {\\bf Z})换行符$

$g(x)=\\sqrt{3}\\cos\\left(2x+\\dfrac{\\mathrm{{\\pi }}}{3}\\right)$

$g(x)$在区间$\\left(\\dfrac{\\mathrm{{\\pi }}}{12},\\dfrac{\\mathrm{{\\pi }}}{3}\\right)$上单调递增

函数$h(x)$的零点为$\\dfrac{k\\mathrm{{\\pi }}}{2}(k\\in {\\bf Z})$

由题意得$f(x)=\sin\left(2x+\dfrac{\pi }{3}\right)+\sin2x=\sin2x\cos\dfrac{\pi}{3}+\cos2x\sin\dfrac{\pi}{3}+\sin2x=\dfrac{2}{3}\sin2x+\dfrac{\sqrt{3}}{2}\cos2x=$$\sqrt {3}$$\sin\left(2x+\dfrac{\pi}{6}\right)$,所以$f(x)$图象的对称轴的方程为$2x+\dfrac{\pi}{6}=k\pi +\dfrac{\pi}{2}(k$∈$\bf Z)$,即$x=k\dfrac{\pi}{2}+\dfrac{\pi}{6}(k$∈$\bf Z)$,故选项$\rm A$正确;将$f(x)$的图象向右平移$\dfrac{\pi}{6}$个单位长度,得到函数$g(x)=$$\sqrt {3}$$\sin[2\left(x-\dfrac{\pi}{6}\right)+\dfrac{\pi}{6}]=$$\sqrt {3}$$\sin\left(2x-\dfrac{\pi}{6}\right)=$$\sqrt {3}$$\cos\left(2x-\dfrac{\pi}{6}+\dfrac{\pi}{2}\right)=$$\sqrt {3}$$\cos\left(2x+\dfrac{\pi}{3}\right)$的图象,故选项$\rm B$正确;当$x$∈$(\dfrac{\pi}{12}$,$\dfrac{\pi}{3})$时,$2x+\dfrac{\pi}{3}$∈$(\dfrac{\pi}{2}$,$\pi )$,所以$g(x)$在区间$(\dfrac{\pi}{12}$,$\dfrac{\pi}{3})$上单调递减,故选项$\rm C$错误;$h(x)=f(x)+g(x)=$$\sqrt {3}$$\sin\left(2x+\dfrac{\pi}{6}\right)+$$\sqrt {3}$$\cos\left(2x+\dfrac{\pi}{3}\right)=$$\sqrt {3}$$\sin\left(2x+\dfrac{\pi}{6}\right)+$$\sqrt {3}$$\cos\left(2x+\dfrac{\pi}{2}-\dfrac{\pi}{6}\right)=$$\sqrt {3}$$\sin\left(2x+\dfrac{\pi}{6}\right)-$$\sqrt {3}$$\sin\left(2x+\dfrac{\pi}{6}\right)=0$,所以函数$h(x)$的零点为$k\dfrac{\pi}{2}(k$∈$\bf Z)$,故选项$\rm D$正确,故选$\rm ABD $.