Đặt \(A=\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\ge0\)
\(\to A^2=\dfrac{2+\sqrt{3}}{2-\sqrt{3}}+\dfrac{2-\sqrt{3}}{2+\sqrt{3}}+2\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}\cdot\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\\ \to A^2=\dfrac{\left(2+\sqrt{3}\right)^2+\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+2\\ \to A^2=\dfrac{7+4\sqrt{3}+7-4\sqrt{3}}{4-3}+2\\ \to A^2=14+2=16\\ \to A=4\)
\(\left(\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\right)+\left(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\right)\\ =\dfrac{\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{2}.\sqrt{2-\sqrt{3}}}+\dfrac{\sqrt{2}.\sqrt{2-\sqrt{3}}}{\sqrt{2}.\sqrt{2+\sqrt{3}}}\\ =\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{4-2\sqrt{3}}}+\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{4+2\sqrt{3}}}\\ =\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{\left(\sqrt{3}-1\right)^2}}+\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{\left(\sqrt{3}+1\right)^2}}\\ =\dfrac{\left|\sqrt{3}+1\right|}{\left|\sqrt{3}-1\right|}+\dfrac{\left|\sqrt{3}-1\right|}{\left|\sqrt{3}+1\right|}\)
\(=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}+\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\\ =\dfrac{\left(\sqrt{3}+1\right)^2+\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\\ =\dfrac{4+2\sqrt{3}+4-2\sqrt{3}}{\sqrt{3^2}-1}\\ =\dfrac{8}{2}\\ =4\)
