\(A=\dfrac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\dfrac{3-\sqrt{5}}{\sqrt{10}-\sqrt{3+\sqrt{5}}}\)
\(\dfrac{A}{\sqrt{2}}=\dfrac{3+\sqrt{5}}{2\sqrt{5}+\sqrt{6+2\sqrt{5}}}-\dfrac{3-\sqrt{5}}{2\sqrt{5}-\sqrt{6+2\sqrt{5}}}\)
\(\dfrac{A}{\sqrt{2}}=\dfrac{3+\sqrt{5}}{2\sqrt{5}+\sqrt{\left(\sqrt{5}+1\right)^2}}-\dfrac{3-\sqrt{5}}{2\sqrt{5}-\sqrt{\left(\sqrt{5}+1\right)^2}}\)
\(\dfrac{A}{\sqrt{2}}=\dfrac{3+\sqrt{5}}{2\sqrt{5}+\sqrt{5}+1}-\dfrac{3-\sqrt{5}}{2\sqrt{5}-\sqrt{5}-1}\)
\(\dfrac{A}{\sqrt{2}}=\dfrac{3+\sqrt{5}}{3\sqrt{5}+1}-\dfrac{3-\sqrt{5}}{\sqrt{5}-1}\)
\(\dfrac{A}{\sqrt{2}}=\dfrac{\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)-\left(3-\sqrt{5}\right)\left(3\sqrt{5}+1\right)}{14-2\sqrt{5}}\)
\(\dfrac{A}{\sqrt{2}}=\dfrac{2+2\sqrt{5}+12-8\sqrt{5}}{14-2\sqrt{5}}\)
\(\dfrac{A}{\sqrt{2}}=\dfrac{14-6\sqrt{5}}{14-2\sqrt{5}}\)
Bạn kiểm tra lại cái đề bài nhé
