已知函数$f\left( x \right)=\sin x\cos x-\sqrt{3}{{\cos }^{2}}x+\dfrac{\sqrt{3}}{2}$．

$(1)$求函数$f\left( x \right)$的最小正周期及单调递减区间；

$(2)$求函数$f\left( x \right)$在$\left[ -\dfrac{\text{ }\!\!\pi}{12},\dfrac{2\pi}{3} \right]$上的最值．

$(1)$最小正周期为$\\pi\\!\\!\\text{ }$，单调递减区间为$\\left[ \\dfrac{\\text{5 }\\!\\!\\pi\\!\\!\\text{ }}{12}+k\\pi,\\dfrac{\\text{11 }\\!\\!\\pi\\!\\!\\text{ }}{12}+k\\pi \\right]\\left( k\\in \\bf{Z} \\right)$

$(2)$$f{{(x)}_{\\min }}=-1$，$f(x)_{\\max }=1$

$(1)$由函数$f\left( x \right)=\sin x\cos x-\sqrt{3}{{\cos }^{2}}x+\dfrac{\sqrt{3}}{2}=\dfrac{1}{2}\times 2\sin x\cos x-\dfrac{\sqrt{3}}{2}\left( 2{{\cos }^{2}}x-1 \right)$

$=\dfrac{1}{2}\sin 2x-\dfrac{\sqrt{3}}{2}\cos 2x=\sin \left( 2x-\dfrac{\pi}{3} \right)$，

$\therefore f\left( x \right)$的最小正周期为$T=\dfrac{2\pi}{2}=\pi$，

令$\dfrac{\pi}{2}+2k\pi\leqslant 2x-\dfrac{\pi}{3}\leqslant \dfrac{\text{3 }\!\!\pi}{2}+2k\pi\!\!\text{ ,}k\in \bf{Z}$，可得$\dfrac{\text{5 }\!\!\pi}{12}+k\pi\leqslant x\leqslant \dfrac{\text{11 }\!\!\pi}{12}+k\pi\!\!\text{ ,}k\in \bf{Z}$，

$\therefore f\left( x \right)$的单调减区间为$\left[ \dfrac{\text{5 }\!\!\pi}{12}+k\pi,\dfrac{\text{11 }\!\!\pi}{12}+k\pi \right]\left( k\in \bf{Z} \right)$；

$(2)$由$(1)$知，函数的单调递增区间为$\left[ k\text{ }\!\!\pi-\dfrac{\pi}{12},k\text{ }\!\!\pi+\dfrac{5\pi}{12} \right],k\in \bf{Z}$，

$\because x\in \left[ -\dfrac{\pi}{12},\dfrac{2\pi}{3} \right]$，

$\therefore {{f}({x})}$在$\left[ -\dfrac{\pi}{12},\dfrac{5\pi}{12} \right]$上单调递增，在$\left[ \dfrac{5\pi}{12},\dfrac{2\pi}{3} \right]$上单调递减，

且$f\left(-\dfrac{\pi}{12}\right)=-1$，$f\left(\dfrac{5\pi}{12}\right)=1$，$f\left(\dfrac{2\pi}{3}\right)=\text{0}$，

$\therefore f{{(x)}_{\min }}=-1$，$f(x)_{\max }=1$．