计算：$\dfrac {1}{ \sqrt {3}+ \sqrt {2}}-2^{-1}+(\cos 60 ^{\circ} + \dfrac {1}{2})^{0}-3^{ \frac {1}{2}}$．

1. 对第一项进行分母有理化

$\dfrac {1}{ \sqrt {3}+ \sqrt {2}} = \dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})} = \sqrt{3}-\sqrt{2}$

2. 计算负整数指数幂

$2^{-1} = \dfrac{1}{2}$

3. 计算含有特殊角三角函数值的零指数幂

因为 $\cos 60^{\circ} = \dfrac{1}{2}$，所以 $(\cos 60 ^{\circ} + \dfrac {1}{2})^{0} = \left(\dfrac{1}{2}+\dfrac{1}{2}\right)^{0} = 1^{0} = 1$

4. 化简分数指数幂

$3^{\frac{1}{2}} = \sqrt{3}$

5. 合并各项并求出最终化简结果

原式 $= (\sqrt{3} - \sqrt{2}) - \dfrac{1}{2} + 1 - \sqrt{3}$
$= \dfrac{1}{2} - \sqrt{2}$