Question

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

Answer 1

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

Vậy \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}=4\)