Question

\(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1-1}}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}}+1+1}\)

Answer 1

Đúng như đã hứa nhé:

\(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}+1}}=\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3+1}}+1\right)}{\sqrt{3+1}-1}-\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3+1}}-1\right)}{\sqrt{3+1}-1}=\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3+1}}+1-\sqrt{\sqrt{3+1}}+1\right)}{2-1}=2\sqrt{3}\)

Answer 2

\(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}}+1}\) \(=\dfrac{\sqrt{3}}{\sqrt{\sqrt{4}}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{4}}+1}\)

\(=\dfrac{\sqrt{3}}{\sqrt{2}-1}-\dfrac{\sqrt{3}}{\sqrt{2}+1}\) \(=\dfrac{\sqrt{3}\left(\sqrt{2}+1\right)-\sqrt{3}\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}\)

\(=\dfrac{\sqrt{6}+\sqrt{3}-\sqrt{6}+\sqrt{3}}{\left(\sqrt{2}\right)^2+1^2}\) \(=\dfrac{2\sqrt{3}}{2-1}=\dfrac{2\sqrt{3}}{1}=2\sqrt{3}\)

Answer 3

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