Question

Tính:
\(\sqrt{3+\sqrt{5}}\) - \(\sqrt{3-\sqrt{5}}\) - \(\sqrt{2}\)

Answer 1

`\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}`

`=(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}})/(\sqrt{2})-\sqrt{2}`

`=(\sqrt{(\sqrt{5}+1)^{2}}-\sqrt{(\sqrt{5}-1)^{2}})/(\sqrt{2})-\sqrt{2}`

`=(|\sqrt{5}+1|-|\sqrt{5}-1|)/(\sqrt{2})-\sqrt{2}`

`=(\sqrt{5}+1-(\sqrt{5}-1))/(\sqrt{2})-\sqrt{2}`

`=(2)/(\sqrt{2})-\sqrt{2}`

`=\sqrt{2}-\sqrt{2}=0`

Answer 2

\(=\dfrac{\sqrt{2}\sqrt{3+\sqrt{5}}-\sqrt{2}\sqrt{3-\sqrt{5}}-2}{\sqrt{2}}\\ =\dfrac{\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}-2}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}-2}{\sqrt{2}}\\ =\dfrac{\sqrt{5}+1-\sqrt{5}+1-2}{\sqrt{2}}\\ =\dfrac{2-2}{\sqrt{2}}=\dfrac{0}{\sqrt{2}}\)

Answer 3

Đặt \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}=A\)

\(\Rightarrow A\sqrt{2}=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}-\sqrt{4}\)

\(A\sqrt{2}=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}-2\)

\(A\sqrt{2}=\sqrt{5}+1-\sqrt{5}+1-2\left(\sqrt{5}>1\right)\)

\(A\sqrt{2}=2-2=0\Rightarrow A=0\)