Question

\(5\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}-\sqrt{\dfrac{5}{2}}\right)^2+\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}-\sqrt{\dfrac{3}{2}}\right)^2\)

Answer 1

Ta có: \(\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}-\sqrt{\dfrac{5}{2}}\right)^2+\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}-\sqrt{\dfrac{3}{2}}\right)^2\)

\(=\left(\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}+\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{2}}-\dfrac{\sqrt{5}}{\sqrt{2}}\right)^2+\left(\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}+\dfrac{\sqrt{6+2\sqrt{5}}}{\sqrt{2}}-\dfrac{\sqrt{3}}{\sqrt{2}}\right)^2\)

\(=\left(\dfrac{\sqrt{3}+1+\sqrt{5}-1-\sqrt{5}}{\sqrt{2}}\right)^2+\left(\dfrac{\sqrt{3}-1+\sqrt{5}+1-\sqrt{3}}{\sqrt{2}}\right)^2\)

\(=\dfrac{3}{2}+\dfrac{5}{2}=4\)