14. 已知$\sqrt{a - 3} + \sqrt{2 - b} = 0$，则$\dfrac{1}{\sqrt{a}} + \dfrac{\sqrt{6}}{\sqrt{b}} =$
$\frac{4}{3}\sqrt{3}$
.答案：$14.\frac{4}{3}\sqrt{3}$
解析：
因为$\sqrt{a - 3} \geq 0$，$\sqrt{2 - b} \geq 0$，且$\sqrt{a - 3} + \sqrt{2 - b} = 0$，所以$\sqrt{a - 3} = 0$，$\sqrt{2 - b} = 0$。
由$\sqrt{a - 3} = 0$，得$a - 3 = 0$，即$a = 3$。
由$\sqrt{2 - b} = 0$，得$2 - b = 0$，即$b = 2$。
将$a = 3$，$b = 2$代入$\dfrac{1}{\sqrt{a}} + \dfrac{\sqrt{6}}{\sqrt{b}}$，得：
$\begin{aligned}&\dfrac{1}{\sqrt{3}} + \dfrac{\sqrt{6}}{\sqrt{2}}\\=&\dfrac{\sqrt{3}}{3} + \sqrt{3}\\=&\dfrac{\sqrt{3}}{3} + \dfrac{3\sqrt{3}}{3}\\=&\dfrac{4\sqrt{3}}{3}\end{aligned}$
$\dfrac{4}{3}\sqrt{3}$
由$\sqrt{a - 3} = 0$，得$a - 3 = 0$，即$a = 3$。
由$\sqrt{2 - b} = 0$，得$2 - b = 0$，即$b = 2$。
将$a = 3$，$b = 2$代入$\dfrac{1}{\sqrt{a}} + \dfrac{\sqrt{6}}{\sqrt{b}}$，得：
$\begin{aligned}&\dfrac{1}{\sqrt{3}} + \dfrac{\sqrt{6}}{\sqrt{2}}\\=&\dfrac{\sqrt{3}}{3} + \sqrt{3}\\=&\dfrac{\sqrt{3}}{3} + \dfrac{3\sqrt{3}}{3}\\=&\dfrac{4\sqrt{3}}{3}\end{aligned}$
$\dfrac{4}{3}\sqrt{3}$
15. 已知$x = \sqrt{6} + \sqrt{2}$，则$x^{2} - 2\sqrt{2}x$的值是
4
.答案：15.4
解析：
$x = \sqrt{6} + \sqrt{2}$，则$x - \sqrt{2} = \sqrt{6}$，两边平方得$(x - \sqrt{2})^2 = (\sqrt{6})^2$，即$x^2 - 2\sqrt{2}x + 2 = 6$，所以$x^2 - 2\sqrt{2}x = 6 - 2 = 4$。
4
4
16. 如图，大正方形的边长为$\sqrt{15} + \sqrt{5}$，小正方形的边长为$\sqrt{15} - \sqrt{5}$，则图中阴影部分的面积为
$20\sqrt{3}$
.答案：$16.20\sqrt{3}$
解析：
阴影部分面积 = 大正方形面积 - 小正方形面积
大正方形面积 = $(\sqrt{15} + \sqrt{5})^2 = (\sqrt{15})^2 + 2\sqrt{15}·\sqrt{5} + (\sqrt{5})^2 = 15 + 2\sqrt{75} + 5 = 20 + 10\sqrt{3}$
小正方形面积 = $(\sqrt{15} - \sqrt{5})^2 = (\sqrt{15})^2 - 2\sqrt{15}·\sqrt{5} + (\sqrt{5})^2 = 15 - 2\sqrt{75} + 5 = 20 - 10\sqrt{3}$
阴影部分面积 = $(20 + 10\sqrt{3}) - (20 - 10\sqrt{3}) = 20\sqrt{3}$
$20\sqrt{3}$
大正方形面积 = $(\sqrt{15} + \sqrt{5})^2 = (\sqrt{15})^2 + 2\sqrt{15}·\sqrt{5} + (\sqrt{5})^2 = 15 + 2\sqrt{75} + 5 = 20 + 10\sqrt{3}$
小正方形面积 = $(\sqrt{15} - \sqrt{5})^2 = (\sqrt{15})^2 - 2\sqrt{15}·\sqrt{5} + (\sqrt{5})^2 = 15 - 2\sqrt{75} + 5 = 20 - 10\sqrt{3}$
阴影部分面积 = $(20 + 10\sqrt{3}) - (20 - 10\sqrt{3}) = 20\sqrt{3}$
$20\sqrt{3}$
17. (2024·启东期末)规定运算符号“$\Delta$”的意义如下：当$a ＞ b$时，$a\Delta b = a + b$；当$a\leqslant b$时，$a\Delta b = a - b$.计算$(\sqrt{3}\Delta\sqrt{2}) - (2\sqrt{3}\Delta3\sqrt{2})$的结果为
$- \sqrt{3} + 4\sqrt{2}$
.答案：$17.- \sqrt{3} + 4\sqrt{2}$
解析：
$\because \sqrt{3} ＞ \sqrt{2}$，$\therefore \sqrt{3}\Delta\sqrt{2} = \sqrt{3} + \sqrt{2}$；
$\because 2\sqrt{3} = \sqrt{12}$，$3\sqrt{2} = \sqrt{18}$，$\sqrt{12} ＜ \sqrt{18}$，即$2\sqrt{3} ＜ 3\sqrt{2}$，$\therefore 2\sqrt{3}\Delta3\sqrt{2} = 2\sqrt{3} - 3\sqrt{2}$；
$\therefore (\sqrt{3}\Delta\sqrt{2}) - (2\sqrt{3}\Delta3\sqrt{2}) = (\sqrt{3} + \sqrt{2}) - (2\sqrt{3} - 3\sqrt{2}) = \sqrt{3} + \sqrt{2} - 2\sqrt{3} + 3\sqrt{2} = -\sqrt{3} + 4\sqrt{2}$。
$\because 2\sqrt{3} = \sqrt{12}$，$3\sqrt{2} = \sqrt{18}$，$\sqrt{12} ＜ \sqrt{18}$，即$2\sqrt{3} ＜ 3\sqrt{2}$，$\therefore 2\sqrt{3}\Delta3\sqrt{2} = 2\sqrt{3} - 3\sqrt{2}$；
$\therefore (\sqrt{3}\Delta\sqrt{2}) - (2\sqrt{3}\Delta3\sqrt{2}) = (\sqrt{3} + \sqrt{2}) - (2\sqrt{3} - 3\sqrt{2}) = \sqrt{3} + \sqrt{2} - 2\sqrt{3} + 3\sqrt{2} = -\sqrt{3} + 4\sqrt{2}$。
18. 计算：
(1)$\sqrt{8} + 2\sqrt{3} - (\sqrt{27} + \sqrt{2})$；
(2)$\sqrt{54} - (\sqrt{\dfrac{2}{3}} + 2\sqrt{\dfrac{1}{2}} - \sqrt{32})$；
(3)$-3\sqrt{\dfrac{8}{27}}÷\sqrt{\dfrac{3}{4}}×\sqrt{27}$；
(4)$(2\sqrt{3} - 1)^{2} + (\sqrt{3} + 2)(\sqrt{3} - 2)$；
(5)$(3\sqrt{2} - 2\sqrt{6})×(-3\sqrt{2} - 2\sqrt{6})$；
(6)$(2\sqrt{7} + 5\sqrt{2})^{2} - (2\sqrt{7} - 5\sqrt{2})^{2}$.
(1)$\sqrt{8} + 2\sqrt{3} - (\sqrt{27} + \sqrt{2})$；
(2)$\sqrt{54} - (\sqrt{\dfrac{2}{3}} + 2\sqrt{\dfrac{1}{2}} - \sqrt{32})$；
(3)$-3\sqrt{\dfrac{8}{27}}÷\sqrt{\dfrac{3}{4}}×\sqrt{27}$；
(4)$(2\sqrt{3} - 1)^{2} + (\sqrt{3} + 2)(\sqrt{3} - 2)$；
(5)$(3\sqrt{2} - 2\sqrt{6})×(-3\sqrt{2} - 2\sqrt{6})$；
(6)$(2\sqrt{7} + 5\sqrt{2})^{2} - (2\sqrt{7} - 5\sqrt{2})^{2}$.
答案：$18.(1)\sqrt{2} - \sqrt{3} (2)\frac{8\sqrt{6}}{3} + 3\sqrt{2} (3)- 4\sqrt{6} (4)12 - 4\sqrt{3}$
$(5)6 (6)40\sqrt{14}$
$(5)6 (6)40\sqrt{14}$
19. 已知$x = \sqrt{3} + \sqrt{7}$，$y = \sqrt{3} - \sqrt{7}$，求$4x^{2} - 7xy + 4y^{2}$的值.
答案：19.
∵$x = \sqrt{3} + \sqrt{7}，$$y = \sqrt{3} - \sqrt{7}，$
∴$x - y = 2\sqrt{7}，$xy = - 4.
∴原
式$=4(x - y)^2 + xy = 4×(2\sqrt{7})^2 + (- 4)=108$
∵$x = \sqrt{3} + \sqrt{7}，$$y = \sqrt{3} - \sqrt{7}，$
∴$x - y = 2\sqrt{7}，$xy = - 4.
∴原
式$=4(x - y)^2 + xy = 4×(2\sqrt{7})^2 + (- 4)=108$