Question

\(\dfrac{2}{\sqrt{3}-1}-\sqrt{4-2\sqrt{3}}\)

Answer 1

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

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

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

Answer 2

\(\dfrac{2}{\sqrt{3}-1}-\sqrt{\left(\sqrt{3}-1\right)^2}\)

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

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

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