Question

Answer 1

Tách ra đi cậu, nhiều vậy không ai làm đâu ạ ☕

Answer 2

1: \(=\dfrac{1}{\sqrt{2}}\cdot\sqrt{4-2\sqrt{3}}=\dfrac{1}{\sqrt{2}}\left(\sqrt{3}-1\right)\)

2: \(=\dfrac{1}{\sqrt{2}}\cdot\sqrt{8+2\sqrt{15}}=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\)

3: \(=\dfrac{1}{\sqrt{2}}\cdot\sqrt{10-2\sqrt{21}}=\dfrac{\sqrt{7}-\sqrt{3}}{\sqrt{2}}\)

4: \(=\dfrac{1}{\sqrt{2}}\cdot\sqrt{12-2\sqrt{35}}=\dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{2}}\)

5: \(=\dfrac{1}{\sqrt{2}}\cdot\sqrt{4+2\sqrt{3}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}\)

6: \(=\dfrac{1}{\sqrt{2}}\cdot\sqrt{8-2\sqrt{15}}=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\)

7: \(=\dfrac{1}{\sqrt{2}}\cdot\sqrt{10+2\sqrt{21}}=\dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{2}}\)

8: \(=\dfrac{1}{\sqrt{2}}\cdot\sqrt{12+2\sqrt{35}}=\dfrac{\sqrt{7}+\sqrt{5}}{\sqrt{2}}\)