[polsk133]

ให้ $A=4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{5^3}$

$\sqrt[4]{5}A=4\sqrt[4]{5}-3\sqrt{5}+2\sqrt[4]{5^3}-5$

$(\sqrt[4]{5}+1)A=\sqrt[4]{5}-\sqrt{5}-1+\sqrt[4]{5^3}=\sqrt[4]{5}(\sqrt{5}+1)-(\sqrt{5}+1)=(\sqrt[4]{5}-1)(\sqrt{5}+1)$

$A=\frac{(\sqrt[4]{5}-1)(\sqrt{5}+1)}{\sqrt[4]{5}+1}=\frac{(\sqrt[4]{5}-1)^2(\sqrt{5}+1)}{\sqrt{5}-1}=\frac{(\sqrt[4]{5}-1)^2(\sqrt{5}+1)^2}{4}$

ดังนั้น $\frac{2}{\sqrt{A}}=\frac{4}{(\sqrt[4]{5}-1)(\sqrt{5}+1)}=\frac{4(\sqrt[4]{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}=1+\sqrt[4]{5}$