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15. 计算：$(\frac{\sqrt{5}+1}{2})(\frac{\sqrt{5}-1}{2})$.
答案
16. 计算：$(\frac{\sqrt{3}+1}{2})^2+(\frac{\sqrt{3}-1}{2})^2$.
答案
17. 已知$x = \sqrt{3}+1$，$y = \sqrt{3}-1$，求下列各式的值.
（1）$x^2 + 2xy + y^2$；（2）$x^2 - y^2$.
（1）$x^2 + 2xy + y^2$；（2）$x^2 - y^2$.
答案
18. 阅读下列过程：
$\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{\sqrt{2}-1}{(\sqrt{2})^2 - 1^2}=\frac{\sqrt{2}-1}{2 - 1}=\sqrt{2}-1$；
$\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2}=\frac{\sqrt{3}-\sqrt{2}}{3 - 2}=\sqrt{3}-\sqrt{2}$.
仿照上述过程化去下列各式分母中的根号：
（1）$\frac{1}{\sqrt{5}-1}$； （2）$\frac{1}{\sqrt{5}+\sqrt{2}}$.
$\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{\sqrt{2}-1}{(\sqrt{2})^2 - 1^2}=\frac{\sqrt{2}-1}{2 - 1}=\sqrt{2}-1$；
$\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2}=\frac{\sqrt{3}-\sqrt{2}}{3 - 2}=\sqrt{3}-\sqrt{2}$.
仿照上述过程化去下列各式分母中的根号：
（1）$\frac{1}{\sqrt{5}-1}$； （2）$\frac{1}{\sqrt{5}+\sqrt{2}}$.
答案
