\(\left(2+\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\cdot\left(2-\dfrac{5+\sqrt{5}}{\sqrt{5}+1}\right)\\ =\left(2+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\cdot\left(2-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\\ =\left(2+\sqrt{5}\right)\cdot\left(2-\sqrt{5}\right)\\ =2^2-\sqrt{5}^2=4-5=-1\)
Ta có: \(\left(2+\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\cdot\left(2-\dfrac{5+\sqrt{5}}{\sqrt{5}+1}\right)\)
\(=\left(2+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\cdot\left(2-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\)
\(=\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)\)
=4-5=-1
