Question

Answer 1

= \(2-\sqrt{3}+\dfrac{2\left(\sqrt{3}-1\right)}{\sqrt{3^2}-1^2}-6.\dfrac{4}{\sqrt{3}}\)

= \(2-\sqrt{3}+\dfrac{2\left(\sqrt{3}-1\right)}{2}-8\sqrt{3}\)

= \(2-\sqrt{3}+\sqrt{3}-1-8\sqrt{3}\)

= \(1-8\sqrt{3}\)

Answer 2

\(=2-\sqrt{3}+\sqrt{3}-1-8\sqrt{3}=-8\sqrt{3}+1\)