Quy đồng ta có:
$P=\dfrac{(2+\sqrt{5})(\sqrt{2}-\sqrt{3-\sqrt{5}})+(2-\sqrt{5})(\sqrt{2}+\sqrt{3-\sqrt{5}})}{(\sqrt{2}-\sqrt{3+\sqrt{5}})(\sqrt{2}+\sqrt{3+\sqrt{5}})}
\\=\dfrac{2\sqrt{2}-2\sqrt{3-\sqrt{5}}+\sqrt{10}-\sqrt{15-5\sqrt{5}}+..}{(\sqrt{2}-\sqrt{3-\sqrt{5}})(\sqrt{2}+\sqrt{3+\sqrt{5}})}
\\=\dfrac{4\sqrt{2}+2(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}})-(\sqrt{15-5\sqrt{5}}+\sqrt{15+5\sqrt{5}})}{\sqrt{5}-1}$
Đặt $A=\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}} \Rightarrow A^2=6-2\sqrt{4}=2 \Rightarrow A=\sqrt{2}$
Đặt $B=\sqrt{15-5\sqrt{5}}+\sqrt{15+5\sqrt{5}} \Rightarrow B^2=30+2\sqrt{100}=50 \Rightarrow B=\sqrt{50}$
Tới đây dễ rồi