Question

Освободите знаменатель от ирациональности 8/√32-⁴√2

Answer 1

$\frac{8}{\sqrt{32}-\sqrt[4]{2}}=\frac{8}{2^2\sqrt{2}-\sqrt[4]{2}}=\frac{8}{4\sqrt{2}-\sqrt[4]{2}}=\frac{8}{4\sqrt{2}-\sqrt[4]{2}}\cdot\frac{4\sqrt{2}+\sqrt[4]{2}}{4\sqrt{2}+\sqrt[4]{2}}=\frac{8(4\sqrt{2}+\sqrt[4]{2})}{(4\sqrt{2}-\sqrt[4]{2})\cdot(4\sqrt{2}+\sqrt[4]{2})}=\frac{32\sqrt{2}+8\sqrt[4]{2}}{16\cdot2-\sqrt{2}}=\frac{32\sqrt{2}+8\sqrt[4]{2}}{32-\sqrt{2}}=\frac{32\sqrt{2}+8\sqrt[4]{2}}{32-\sqrt{2}}\cdot\frac{32+\sqrt{2}}{32+\sqrt{2}}=$
$=\frac{(32\sqrt{2}+8\sqrt[4]{2})\cdot(32+\sqrt{2})}{(32-\sqrt{2})\cdot(32+\sqrt{2})}=\frac{1024\sqrt{2}+64+256\sqrt[4]{2}+8\sqrt[4]{2}\sqrt{2}}{1024-2}=\frac{1024\sqrt{2}+64+256\sqrt[4]{2}+8\sqrt[4]{2}\sqrt[4]{2^2}}{1022}=\frac{1024\sqrt{2}+64+256\sqrt[4]{2}+8\sqrt[4]{2\cdot2^2}}{1022}=\frac{1024\sqrt{2}+64+256\sqrt[4]{2}+8\sqrt[4]{2^3}}{1022}=\frac{1024\sqrt{2}+64+256\sqrt[4]{2}+8\sqrt[4]{8}}{1022}=\frac{2(512\sqrt{2}+32+128\sqrt[4]{2}+4\sqrt[4]{8})}{1022}=$
$=\frac{512\sqrt{2}+32+128\sqrt[4]{2}+4\sqrt[4]{8}}{511}$