Lời giải:
Đặt \(\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}=a; \sqrt{2+\sqrt{3}-\sqrt{2-\sqrt{3}}}=b\)
Có:
\(a^2+b^2=(2+\sqrt{3}+\sqrt{2-\sqrt{3}})+(2+\sqrt{3}-\sqrt{2-\sqrt{3}})=2(2+\sqrt{3})\)
\(=4+2\sqrt{3}=3+1+2\sqrt{3.1}=(\sqrt{3}+1)^2\)
\(ab=\sqrt{(2+\sqrt{3}+\sqrt{2-\sqrt{3}})(2+\sqrt{3}-\sqrt{2-\sqrt{3}})}\)
\(=\sqrt{(2+\sqrt{3})^2-(2-\sqrt{3})}=\sqrt{5+5\sqrt{3}}\)
Như vậy:
\(\frac{\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}}{\sqrt{2+\sqrt{3}-\sqrt{2-\sqrt{3}}}}+\frac{\sqrt{2+\sqrt{3}-\sqrt{2-\sqrt{3}}}}{\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}}=\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}\)
\(=\frac{(\sqrt{3}+1)^2}{\sqrt{5+5\sqrt{3}}}=\frac{(\sqrt{3}+1)^2}{\sqrt{5}.\sqrt{\sqrt{3}+1}}=\frac{(\sqrt{3}+1)^{1.5}}{\sqrt{5}}\)
