[tex]A= (\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}})/ (\frac{2}{\sqrt{3}+1}-\frac{2}{\sqrt{3}-1})[/tex]
\[\begin{align}
& \frac{\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}}{\frac{2}{\sqrt{3}+1}-\frac{2}{\sqrt{3}-1}} \\
& \frac{2}{\sqrt{3}+1}-\frac{2}{\sqrt{3}-1}=\frac{2\sqrt{3}-2-2\sqrt{3}-2}{3-1}=\frac{-4}{2}=-2 \\
& \sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}} \\
& =\frac{\sqrt{2}}{2}(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}) \\
& =\frac{\sqrt{2}}{2}(\sqrt{{{(\sqrt{5}+1)}^{2}}}-\sqrt{{{(\sqrt{5}-1)}^{2}}}) \\
& =\frac{\sqrt{2}}{2}(\sqrt{5}+1-\sqrt{5}+1) \\
& =\sqrt{2} \\
& \Rightarrow \frac{\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}}{\frac{2}{\sqrt{3}+1}-\frac{2}{\sqrt{3}-1}}=\frac{-\sqrt{2}}{2} \\
\end{align}\]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
