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5. 先化简，再求值：
$(x + 1)(x - 1)+(2x - 1)^{2}-2x(2x - 1)$，其中$x=\sqrt{2}+1$.
$(x + 1)(x - 1)+(2x - 1)^{2}-2x(2x - 1)$，其中$x=\sqrt{2}+1$.
答案
5. 解:原式$=x^{2}-1+4x^{2}-4x+1-4x^{2}+2x=x^{2}-2x$.
当$x=\sqrt{2}+1$时,
原式$=(\sqrt{2}+1)^{2}-2(\sqrt{2}+1)$
$=3+2\sqrt{2}-2\sqrt{2}-2$
$=1$.
当$x=\sqrt{2}+1$时,
原式$=(\sqrt{2}+1)^{2}-2(\sqrt{2}+1)$
$=3+2\sqrt{2}-2\sqrt{2}-2$
$=1$.
6. 已知$x=\sqrt{2}-1$，$y=\sqrt{2}+1$，求$\dfrac{x}{y}+\dfrac{y}{x}$的值.
答案
6. 解:原式$=\frac{x^{2}+y^{2}}{xy}=\frac{(x+y)^{2}-2xy}{xy}$.
$\because x+y=\sqrt{2}-1+\sqrt{2}+1=2\sqrt{2}$,
$xy=(\sqrt{2}-1)(\sqrt{2}+1)=1$,
$\therefore$原式$=(2\sqrt{2})^{2}-2=6$.
$\because x+y=\sqrt{2}-1+\sqrt{2}+1=2\sqrt{2}$,
$xy=(\sqrt{2}-1)(\sqrt{2}+1)=1$,
$\therefore$原式$=(2\sqrt{2})^{2}-2=6$.
7. 计算：
(1)$\sqrt{8}×\sqrt{2}-\sqrt[3]{64}-3\sqrt{\dfrac{1}{3}}-\sqrt{24}÷\sqrt{8}$；
(2)$\sqrt{48}÷\sqrt{3}-\sqrt{\dfrac{1}{2}}×\sqrt{12}+\sqrt{24}$；
(3)$(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})-\sqrt{(-5)^{2}}$；
(4)$(3\sqrt{2}+2\sqrt{3})(3\sqrt{2}-2\sqrt{3})-(\sqrt{3}-\sqrt{2})^{2}$.
(1)$\sqrt{8}×\sqrt{2}-\sqrt[3]{64}-3\sqrt{\dfrac{1}{3}}-\sqrt{24}÷\sqrt{8}$；
(2)$\sqrt{48}÷\sqrt{3}-\sqrt{\dfrac{1}{2}}×\sqrt{12}+\sqrt{24}$；
(3)$(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})-\sqrt{(-5)^{2}}$；
(4)$(3\sqrt{2}+2\sqrt{3})(3\sqrt{2}-2\sqrt{3})-(\sqrt{3}-\sqrt{2})^{2}$.
答案
7. (1)解:原式$=4-4-3×\frac{\sqrt{3}}{3}-\sqrt{3}$
$=-\sqrt{3}-\sqrt{3}=-2\sqrt{3}$
(2)解:原式$=4-\sqrt{6}+2\sqrt{6}=4+\sqrt{6}$
(3)解:原式$=7-5-5=-3$
(4)解:原式$=(3\sqrt{2})^{2}-(2\sqrt{3})^{2}-(5-2\sqrt{6})$
$=18-12-5+2\sqrt{6}=1+2\sqrt{6}$
$=-\sqrt{3}-\sqrt{3}=-2\sqrt{3}$
(2)解:原式$=4-\sqrt{6}+2\sqrt{6}=4+\sqrt{6}$
(3)解:原式$=7-5-5=-3$
(4)解:原式$=(3\sqrt{2})^{2}-(2\sqrt{3})^{2}-(5-2\sqrt{6})$
$=18-12-5+2\sqrt{6}=1+2\sqrt{6}$
1. 已知$a+\dfrac{1}{a}=\sqrt{7}$，求$a-\dfrac{1}{a}$的值.
答案
1. 解:$\because(a-\frac{1}{a})^{2}=(a+\frac{1}{a})^{2}-4=(\sqrt{7})^{2}-4=3$,
$\therefore a-\frac{1}{a}=\pm\sqrt{3}$.
$\therefore a-\frac{1}{a}=\pm\sqrt{3}$.
2. 计算：
(1)$(\sqrt{2}+1)(\sqrt{2}-1)=$；
(2)$(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})=$；
(3)$(2+\sqrt{3})(2-\sqrt{3})=$；
(4)$(\sqrt{5}+2)(\sqrt{5}-2)=$.
通过以上计算，观察规律，写出含$n$（$n$为正整数）的等式表示上面的规律：.
(1)$(\sqrt{2}+1)(\sqrt{2}-1)=$；
(2)$(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})=$；
(3)$(2+\sqrt{3})(2-\sqrt{3})=$；
(4)$(\sqrt{5}+2)(\sqrt{5}-2)=$.
通过以上计算，观察规律，写出含$n$（$n$为正整数）的等式表示上面的规律：.
答案
2. (1)1 (2)1 (3)1 (4)1
$(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}-\sqrt{n})=1$
$(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}-\sqrt{n})=1$
