Question

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

Answer 1

Answer 2

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

\(=\dfrac{\sqrt{3}}{\sqrt{\sqrt{4}}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{4}}+1}\)

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

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

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