\(\dfrac{1}{3-\sqrt{5}}-\dfrac{1}{\sqrt{5}+1}=\dfrac{3+\sqrt{5}}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}-\dfrac{\sqrt{5}-1}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
\(=\dfrac{3+\sqrt{5}}{9-5}-\dfrac{\sqrt{5}-1}{5-1}=\dfrac{3+\sqrt{5}}{4}-\dfrac{\sqrt{5}-1}{4}=\dfrac{4}{4}=1\)
Ta có: \(A=\dfrac{1}{3-\sqrt{5}}-\dfrac{1}{\sqrt{5}+1}\)
\(=\dfrac{3+\sqrt{5}-\sqrt{5}+1}{4}\)
=1