Answer:
[tex]1-\dfrac{2}{3}\sqrt{3}[/tex]
Step-by-step explanation:
Maybe you want to simplify ...
[tex](1-\sqrt{3})\dfrac{1}{3+\sqrt{3}}[/tex]
Multiply numerator and denominator by the 'conjugate' of the denominator:
[tex](1-\sqrt{3})\dfrac{1}{3+\sqrt{3}}\cdot\dfrac{3-\sqrt{3}}{3-\sqrt{3}}=\dfrac{(1-\sqrt{3})(3-\sqrt{3})}{9-3}=\dfrac{3-4\sqrt{3}+3}{6}\\\\\boxed{1-\dfrac{2}{3}\sqrt{3}}[/tex]
