11. 已知$y=\sqrt{1-4x}+\sqrt{4x-1}+\frac{1}{2}$,求$(\sqrt{2x}+\sqrt{y})^{2}-(\sqrt{2x}-\sqrt{y})^{2}$的值.
答案：11. 解：由二次根式的非负性，得$1-4x≥0$且$4x-1≥0$，
解得$x=\dfrac{1}{4}$，则$y=\dfrac{1}{2}$.
原式$=2x+2\sqrt{2xy}+y-(2x-2\sqrt{2xy}+y)=4\sqrt{2xy}$.
当$x=\dfrac{1}{4}$，$y=\dfrac{1}{2}$时，原式$=4\sqrt{2xy}=4\sqrt{2×\dfrac{1}{4}×\dfrac{1}{2}}=4×$
$\dfrac{1}{2}=2$.
解得$x=\dfrac{1}{4}$，则$y=\dfrac{1}{2}$.
原式$=2x+2\sqrt{2xy}+y-(2x-2\sqrt{2xy}+y)=4\sqrt{2xy}$.
当$x=\dfrac{1}{4}$，$y=\dfrac{1}{2}$时，原式$=4\sqrt{2xy}=4\sqrt{2×\dfrac{1}{4}×\dfrac{1}{2}}=4×$
$\dfrac{1}{2}=2$.
12. 已知$a=\frac{\sqrt{b-4}+\sqrt{4-b}}{\sqrt{b^{4}}}+4b$,求$\frac{4}{\sqrt{a}}-\frac{\sqrt{a}}{b}+4$的值.
答案：12. 解：由二次根式的非负性，得$b-4≥0$，$4-b≥0$，解得$b=$
$4$.$\therefore a=4b=4×4=16$.
原式$=\dfrac{4}{\sqrt{16}}-\dfrac{\sqrt{16}}{4}+4=\dfrac{4}{4}-\dfrac{4}{4}+4=4$.
$4$.$\therefore a=4b=4×4=16$.
原式$=\dfrac{4}{\sqrt{16}}-\dfrac{\sqrt{16}}{4}+4=\dfrac{4}{4}-\dfrac{4}{4}+4=4$.
13. (2025·咸宁期末)阅读理解:
$\frac{1}{\sqrt{2}+1}=\frac{1×(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=\sqrt{2}-1$,$\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{1×(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\sqrt{3}-\sqrt{2}$.
应用计算:
(1)$\frac{1}{\sqrt{7}+\sqrt{6}}=$
(2)$\frac{1}{\sqrt{n+1}+\sqrt{n}}$n为正整数$=$
归纳拓展:
$\frac{1}{\sqrt{2}+1}=\frac{1×(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=\sqrt{2}-1$,$\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{1×(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\sqrt{3}-\sqrt{2}$.
应用计算:
(1)$\frac{1}{\sqrt{7}+\sqrt{6}}=$
$\sqrt{7}-\sqrt{6}$
;(2)$\frac{1}{\sqrt{n+1}+\sqrt{n}}$n为正整数$=$
$\sqrt{n+1}-\sqrt{n}$
;归纳拓展:
答案：13. (1)$\sqrt{7}-\sqrt{6}$ (2)$\sqrt{n+1}-\sqrt{n}$
(3)解：原式$=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\dots+\sqrt{2025}$
$-\sqrt{2024}=\sqrt{2025}-1=45-1=44$.
(3)解：原式$=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\dots+\sqrt{2025}$
$-\sqrt{2024}=\sqrt{2025}-1=45-1=44$.