13.「★★☆」定义:已知$(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=(\sqrt{a})^2-(\sqrt{b})^2=a-b$,可以去掉根号,我们称$\sqrt{a}+\sqrt{b}$与$\sqrt{a}-\sqrt{b}$为一对“对偶式”.若$\sqrt{18-x}-\sqrt{11-x}=1$,则$\sqrt{18-x}+\sqrt{11-x}=$.

答案 $7$
解析 由题意可知$(\sqrt{18-x}-\sqrt{11-x})(\sqrt{18-x}+\sqrt{11-x})=18-x-(11-x)=7$,
$\because \sqrt{18-x}-\sqrt{11-x}=1$,
$\therefore \sqrt{18-x}+\sqrt{11-x}=7÷1=7$.

14.「★★☆」对于任意不相等的两个数$a,b$,定义一种运算※如下：
$a※b=\dfrac{\sqrt{a+b}}{a-b}$,如$5※4=\dfrac{\sqrt{5+4}}{5-4}=3$,那么$(2-\sqrt{3})※(7※5)=$.

答案 $-\dfrac{\sqrt{2}+\sqrt{6}}{4}$
解析 $(2-\sqrt{3})※(7※5)=(2-\sqrt{3})※\dfrac{\sqrt{7+5}}{7-5}=(2-\sqrt{3})※\dfrac{\sqrt{12}}{2}=(2-\sqrt{3})※\dfrac{2\sqrt{3}}{2}=(2-\sqrt{3})※\sqrt{3}=\dfrac{\sqrt{2-\sqrt{3}+\sqrt{3}}}{2-\sqrt{3}-\sqrt{3}}=\dfrac{\sqrt{2}}{2-2\sqrt{3}}=\dfrac{\sqrt{2}×(2+2\sqrt{3})}{(2-2\sqrt{3})×(2+2\sqrt{3})}=\dfrac{2\sqrt{2}+2\sqrt{6}}{4-12}=\dfrac{2\sqrt{2}+2\sqrt{6}}{-8}=-\dfrac{\sqrt{2}+\sqrt{6}}{4}$.

15.「2026湖南长沙期中，★☆」先化简,再求值:$(a-\sqrt{2})(a+\sqrt{2})+a(a-\dfrac{1}{a})$,其中$a=-\sqrt{5}$.

$(a-\sqrt{2})(a+\sqrt{2})+a(a-\dfrac{1}{a})=a^2-2+a^2-1=2a^2-3$,把$a=-\sqrt{5}$代入得,原式$=2×(-\sqrt{5})^2-3=2×5-3=7$.

16.「2026 浙江绍兴期末改编，★☆」已知 $x=\sqrt{2}-\sqrt{3}$，$y=\sqrt{2}+\sqrt{3}$，求下列各式的值：
(1) $x^2 - 2xy + y^2$.
(2) $x^2 - y^2$.

(1)$x^2-2xy+y^2=(x-y)^2$,把$x=\sqrt{2}-\sqrt{3}$,$y=\sqrt{2}+\sqrt{3}$代入得,原式$=[(\sqrt{2}-\sqrt{3})-(\sqrt{2}+\sqrt{3})]^2=12$.
(2)$x^2-y^2=(x+y)(x-y)=[(\sqrt{2}-\sqrt{3})+(\sqrt{2}+\sqrt{3})][(\sqrt{2}-\sqrt{3})-(\sqrt{2}+\sqrt{3})]=-2\sqrt{2}×2\sqrt{3}=-4\sqrt{6}$.