17. (本小题 8 分)计算.
(1) $(3+\sqrt{7})(3-\sqrt{7})+\sqrt{2}(2-\sqrt{2})$;
(2) $(\dfrac{3}{2}\sqrt{1\dfrac{2}{3}}-\sqrt{1\dfrac{1}{4}})^{2}$.

17. (1)
$\begin{aligned}&(3+\sqrt{7})(3 - \sqrt{7})+\sqrt{2}(2-\sqrt{2})\\=&3^{2}-(\sqrt{7})^{2}+2\sqrt{2}-2\\=&9 - 7+2\sqrt{2}-2\\=&2\sqrt{2}\end{aligned}$
(2)
$\begin{aligned}&(\frac{3}{2}\sqrt{1\frac{2}{3}}-\sqrt{1\frac{1}{4}})^{2}\\=&(\frac{3}{2}\sqrt{\frac{5}{3}}-\sqrt{\frac{5}{4}})^{2}\\=&(\frac{3}{2}×\frac{\sqrt{15}}{3}-\frac{\sqrt{5}}{2})^{2}\\=&(\frac{\sqrt{15}}{2}-\frac{\sqrt{5}}{2})^{2}\\=&\frac{15}{4}-2×\frac{\sqrt{15×5}}{4}+\frac{5}{4}\\=&\frac{15 + 5}{4}-\frac{5\sqrt{3}}{2}\\=&5-\frac{5\sqrt{3}}{2}\\=&\frac{10 - 5\sqrt{3}}{2}\end{aligned}$

18. (本小题 8 分)先化简,再求值:$(1-\dfrac{2}{a + 1})÷\dfrac{a^{2}-2a + 1}{a + 1}$,其中$a=\sqrt{2}+1$.

$(1-\dfrac{2}{a + 1})÷\dfrac{a^{2}-2a + 1}{a + 1}$
$=(\dfrac{a + 1}{a + 1}-\dfrac{2}{a + 1})÷\dfrac{(a - 1)^2}{a + 1}$
$=\dfrac{a + 1 - 2}{a + 1}×\dfrac{a + 1}{(a - 1)^2}$
$=\dfrac{a - 1}{a + 1}×\dfrac{a + 1}{(a - 1)^2}$
$=\dfrac{1}{a - 1}$
当$a = \sqrt{2} + 1$时，原式$=\dfrac{1}{\sqrt{2} + 1 - 1}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}$

19. (本小题 8 分)已知$x=\dfrac{\sqrt{3}-1}{2}$,$y=\dfrac{\sqrt{3}+1}{2}$.
(1) 求$x + y$,$xy$和$\dfrac{1}{x}+\dfrac{1}{y}$的值;
(2) 求$\sqrt{\dfrac{y}{x}}+\sqrt{\dfrac{x}{y}}$的值.

(1)
$x + y=\dfrac{\sqrt{3}-1}{2}+\dfrac{\sqrt{3}+1}{2}=\sqrt{3}$；
$xy=\dfrac{\sqrt{3}-1}{2}×\dfrac{\sqrt{3}+1}{2}=\dfrac{(\sqrt{3})^{2}-1^{2}}{4}=\dfrac{3 - 1}{4}=\dfrac{1}{2}$；
$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{y + x}{xy}=\dfrac{\sqrt{3}}{\dfrac{1}{2}} = 2\sqrt{3}$。
(2)
$\sqrt{\dfrac{y}{x}}+\sqrt{\dfrac{x}{y}}=\sqrt{\dfrac{y^{2}}{xy}}+\sqrt{\dfrac{x^{2}}{xy}}=\dfrac{x + y}{\sqrt{xy}}$
把$x + y=\sqrt{3}$，$xy=\dfrac{1}{2}$代入$\dfrac{x + y}{\sqrt{xy}}$可得：
$\dfrac{\sqrt{3}}{\sqrt{\dfrac{1}{2}}}=\dfrac{\sqrt{3}}{\dfrac{\sqrt{2}}{2}}=\sqrt{3}×\dfrac{2}{\sqrt{2}}=\sqrt{6}$。
综上，答案依次为：(1)$x + y=\sqrt{3}$，$xy=\dfrac{1}{2}$，$\dfrac{1}{x}+\dfrac{1}{y}=2\sqrt{3}$；(2)$\sqrt{6}$。