Answer:
[tex]\huge\boxed{\text{Is inorrect. Correct answer is:}}\\\boxed{\dfrac{6-\sqrt3}{3}\neq\dfrac{1}{\sqrt3}}[/tex]
Step-by-step explanation:
[tex]\dfrac{\sqrt{12}-1}{\sqrt3}=\dfrac{\sqrt{12}}{\sqrt3}-\dfrac{1}{\sqrt3}=\sqrt{\dfrac{12}{3}}-\dfrac{1}{\sqrt3}=\sqrt4-\dfrac{1}{\sqrt3}\cdot\dfrac{\sqrt3}{\sqrt3}=2-\dfrac{\sqrt3}{3}\\\\=\dfrac{6}{3}-\dfrac{\sqrt3}{3}=\dfrac{6-\sqrt3}{3}\neq\dfrac{1}{\sqrt3}\\\\\dfrac{1}{\sqrt3}\cdot\dfrac{\sqrt3}{\sqrt3}=\dfrac{\sqrt3}{3}[/tex]
