Đặt [tex]A=\frac{\sqrt{\sqrt[4]{8}+\sqrt{\sqrt{2}-1}}+\sqrt{\sqrt[4]{8}-\sqrt{\sqrt{2}-1}}}{\sqrt{\sqrt[4]{8}+\sqrt{\sqrt{2}+1}}}>0\\\Rightarrow A^2=\frac{\sqrt[4]{8}+\sqrt{\sqrt{2}-1}+\sqrt[4]{8}-\sqrt{\sqrt{2}-1}+2\sqrt{\sqrt[4]{8}+\sqrt{\sqrt{2}-1}}.\sqrt{\sqrt[4]{8}-\sqrt{\sqrt{2}-1}}}{\sqrt[4]{8}+\sqrt{\sqrt{2}+1}}\\\Leftrightarrow A^2=\frac{2\sqrt[4]{8}+2\sqrt{\sqrt{8}-\sqrt{2}+1}}{\sqrt[4]{8}+\sqrt{\sqrt{2}+1}}\\\Leftrightarrow A^2=\frac{2(\sqrt[4]{8}+\sqrt{\sqrt{2}+1})}{\sqrt[4]{8}+\sqrt{\sqrt{2}+1}}=2\\\Rightarrow A=\sqrt{2}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
