Rút gọn
- \(\sqrt{5+\sqrt{21}}+\sqrt{5-\sqrt{21}}-2\sqrt{4\sqrt{7}}\)
- \(\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}\)
- \(\left(\sqrt{14}-\sqrt{10}\right)\left(6-\sqrt{35}\right)\sqrt{6+\sqrt{35}}\)
- \(\left(\sqrt{2}+1\right)\left(\sqrt{3}-1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)
- \(\sqrt{7-3\sqrt{5}}\left(7+3\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{10}\right)\)
1. \(=\sqrt{\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{3}{2}}\right)^2}+\sqrt{\left(\sqrt{\frac{7}{2}}-\sqrt{\frac{3}{2}}\right)^2}-2\sqrt{4\sqrt{7}}=\frac{7}{2}+\frac{3}{2}+\frac{7}{2}-\frac{3}{2}-2\sqrt{4\sqrt{7}}\)
\(=7-2\sqrt{4\sqrt{7}}\)
cho hỏi tại sao có số \(\frac{7}{2};\frac{3}{2}\)zậy chỉ với