Question

Chứng minh đẳng thức : \(\dfrac{\sqrt{6+2\sqrt{5}}}{\sqrt{5}+1}=\dfrac{\sqrt{5-2\sqrt{6}}}{\sqrt{3}-\sqrt{2}}\)

Answer 1

Ta có: \(\dfrac{\sqrt{6+2\sqrt{5}}}{\sqrt{5}+1}=\dfrac{\sqrt{5+2\cdot\sqrt{5}\cdot1+1}}{\sqrt{5}+1}=\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{5}+1}\)

\(=\dfrac{\sqrt{5}+1}{\sqrt{5}+1}=1\) (1)

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

Từ (1)(2) \(\Rightarrow\dfrac{\sqrt{6+2\sqrt{5}}}{\sqrt{5}+1}=\dfrac{\sqrt{5-2\sqrt{6}}}{\sqrt{3}-\sqrt{2}}\) (đpcm)