Sửa đề: \(P=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(\Leftrightarrow\frac{P}{\sqrt{2}}=\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
\(\frac{P}{\sqrt{2}}=\frac{2+\sqrt{3}}{2+\left(\sqrt{3}+1\right)}+\frac{2-\sqrt{3}}{2-\left(\sqrt{3}-1\right)}\)
\(\frac{P}{\sqrt{2}}=\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{9-3}\)
\(\frac{P}{\sqrt{2}}=\frac{6-2\sqrt{3}+3\sqrt{3}-3+6+2\sqrt{3}-3\sqrt{3}}{6}\)
\(\frac{P}{\sqrt{2}}=\frac{6}{6}\)
\(\frac{P}{\sqrt{2}}=1\)
\(\Rightarrow P=1\cdot\sqrt{2}=\sqrt{2}\)
Vậy \(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\sqrt{2}\)
