19. 计算。
(1)$\sqrt{24} - \sqrt{0.5} - 2\sqrt{\dfrac{2}{3}} - (\sqrt{\dfrac{1}{8}} - \dfrac{1}{3}\sqrt{54})$；
(2)$9\sqrt{45} ÷ 3\sqrt{\dfrac{1}{5}} × \dfrac{3}{2}\sqrt{2\dfrac{2}{3}}$；
(3)$\sqrt{18} + \sqrt{12} - 2\sqrt{6} × \dfrac{\sqrt{3}}{4} - \sqrt{75}$。

(1) $\sqrt{24} - \sqrt{0.5} - 2\sqrt{\dfrac{2}{3}} - (\sqrt{\dfrac{1}{8}} - \dfrac{1}{3}\sqrt{54})$
$=2\sqrt{6}-\dfrac{\sqrt{2}}{2}-\dfrac{2\sqrt{6}}{3}-\dfrac{\sqrt{2}}{4}+\sqrt{6}$
$=(2\sqrt{6}-\dfrac{2\sqrt{6}}{3}+\sqrt{6})+(-\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{2}}{4})$
$=\dfrac{7\sqrt{6}}{3}-\dfrac{3\sqrt{2}}{4}$
(2) $9\sqrt{45} ÷ 3\sqrt{\dfrac{1}{5}} × \dfrac{3}{2}\sqrt{2\dfrac{2}{3}}$
$=9\sqrt{45}÷3\sqrt{\dfrac{1}{5}}×\dfrac{3}{2}\sqrt{\dfrac{8}{3}}$
$=(9÷3×\dfrac{3}{2})×\sqrt{45÷\dfrac{1}{5}×\dfrac{8}{3}}$
$=\dfrac{9}{2}×\sqrt{600}$
$=\dfrac{9}{2}×10\sqrt{6}$
$=45\sqrt{6}$
(3) $\sqrt{18} + \sqrt{12} - 2\sqrt{6} × \dfrac{\sqrt{3}}{4} - \sqrt{75}$
$=3\sqrt{2}+2\sqrt{3}-\dfrac{2\sqrt{18}}{4}-5\sqrt{3}$
$=3\sqrt{2}+2\sqrt{3}-\dfrac{3\sqrt{2}}{2}-5\sqrt{3}$
$=(3\sqrt{2}-\dfrac{3\sqrt{2}}{2})+(2\sqrt{3}-5\sqrt{3})$
$=\dfrac{3\sqrt{2}}{2}-3\sqrt{3}$

20. 已知$x = \sqrt{2} + 1$，$y = \sqrt{2} - 1$，求下列各式的值。
(1)$x^{2} - y^{2}$；
(2)$\dfrac{y}{x} + \dfrac{x}{y}$；
(3)$\dfrac{x^{2} - 2xy + y^{2}}{x^{2} - y^{2}} ÷ \dfrac{1}{2x + 2y}$。

4

(3)原式$=\dfrac{(x-y)^2}{(x-y)(x+y)}·2(x+y)=2(x-y)$
当$x=\sqrt{2}+1$，$y=\sqrt{2}-1$时，$x-y=(\sqrt{2}+1)-(\sqrt{2}-1)=2$，原式$=2×2=4$