\(\dfrac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}=\dfrac{\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}.\sqrt{3+\sqrt{5}}}{\sqrt{10}+\sqrt{2}}\)
\(=\dfrac{\sqrt{9-5}.\sqrt{3+\sqrt{5}}}{\sqrt{10}+\sqrt[]{2}}=\dfrac{\sqrt{12+4\sqrt{5}}}{\sqrt{10}+\sqrt{2}}\)
\(\dfrac{\sqrt{\left(\sqrt{10}+\sqrt{2}\right)^2}}{\sqrt{10}+\sqrt{2}}=\dfrac{\sqrt{10}+\sqrt{2}}{\sqrt{10}+\sqrt{2}}=1\)
\(\dfrac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\left(\sqrt{5}+1\right)}\)
\(=\dfrac{2\sqrt{5}+2}{2\left(\sqrt{5}+1\right)}=1\)
A=\(\dfrac{\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
Nhân cả tử và mẫu với \(\sqrt{2}\) ta có:
\(A=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{20}+2}\)\(=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\left(\sqrt{5}+3\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\dfrac{\left(\sqrt{5}-1\right)\cdot\left(\sqrt{5}+3\right)}{2\left(\sqrt{5}+1\right)}=\dfrac{(\sqrt{5}-1)^2\cdot\left(\sqrt{5}+3\right)}{2\left(\sqrt{5}+1\right)\cdot\left(\sqrt{5}-1\right)}\)
vì \(\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\) do \(\sqrt{5}-1>0\)
\(=\dfrac{\left(6-2\sqrt{5}\right)\cdot\left(3+\sqrt{5}\right)}{2\cdot4}=1\)
