[tex]\frac{4}{3+\sqrt{5}+\sqrt{2+2\sqrt{5}}}+\sqrt{\sqrt{5}-2}=1\Leftrightarrow \frac{4}{3+\sqrt{5}+\sqrt{2+2\sqrt{5}}}=1-\sqrt{\sqrt{5}-2}=\frac{3-\sqrt{5}}{1+\sqrt{\sqrt{5}-2}}\Leftrightarrow 4(1+\sqrt{\sqrt{5}-2})=(3-\sqrt{5})[(3+\sqrt{5})+\sqrt{2+2\sqrt{5}}]\Leftrightarrow 4+4\sqrt{\sqrt{5}-2}=4+(3-\sqrt{5})\sqrt{2+2\sqrt{5}}\Leftrightarrow 8\sqrt{\sqrt{5}-2}=(6-2\sqrt{5})\sqrt{2+2\sqrt{5}}\Leftrightarrow 8\sqrt{\sqrt{5}-2}=(\sqrt{5}-1)^2.\sqrt{2}.\sqrt{\sqrt{5}+1}=\sqrt{\sqrt{5}-1}.\sqrt{\sqrt{5}+1}.\sqrt{2}(\sqrt{\sqrt{5}-1})^3=2\sqrt{2}(\sqrt{\sqrt{5}-1})^3\Leftrightarrow 2\sqrt{2}.\sqrt{\sqrt{5}-2}=(\sqrt{\sqrt{5}-1})^3\Leftrightarrow 8(\sqrt{5}-2)=(\sqrt{5}-1)^3[/tex] (đúng)