11.阅读下列材料,回答问题.
在进行二次根式化简时,我们有时会碰上如$\frac{2}{\sqrt{3}},\sqrt{\frac{2}{5}},\frac{2}{\sqrt{3}+1}$这样的式子,其实我们还可以将其进一步化简,如:
(1)$\frac{2}{\sqrt{3}}=\frac{2×\sqrt{3}}{\sqrt{3}×\sqrt{3}}=\frac{2\sqrt{3}}{3}$;(2)$\sqrt{\frac{2}{5}}=\sqrt{\frac{2×5}{5×5}}=\frac{\sqrt{10}}{5}$;
(3)$\frac{2}{\sqrt{3}+1}=\frac{2×(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{2×(\sqrt{3}-1)}{(\sqrt{3})^2 -1^2}=\frac{2(\sqrt{3}-1)}{2}=\sqrt{3}-1$.
类似以上这种化简的步骤叫作分母有理化.
(1)化简:$\frac{3}{\sqrt{6}}=$,$\sqrt{\frac{2}{3}}=$,$\frac{2}{\sqrt{5}+\sqrt{3}}=$;
(2)已知$x=\frac{\sqrt{2}-1}{\sqrt{2}+1},y=\frac{\sqrt{2}+1}{\sqrt{2}-1}$,求$(x+y)^2$的值;
(3)计算:$(\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\dots+\frac{1}{\sqrt{2026}+\sqrt{2025}})×(\sqrt{2026}+1)$.

11. (1) $\frac{\sqrt{6}}{2}$ $\frac{\sqrt{6}}{3}$ $\sqrt{5}-\sqrt{3}$
(2)$x=\frac{\sqrt{2}-1}{\sqrt{2}+1}=\frac{(\sqrt{2}-1)^2}{(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{3-2\sqrt{2}}{2-1}=\frac{3-2\sqrt{2}}{1}= 3-2\sqrt{2}$,
$y=\frac{\sqrt{2}+1}{\sqrt{2}-1}=\frac{(\sqrt{2}+1)^2}{(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{3+2\sqrt{2}}{2-1}=\frac{3+2\sqrt{2}}{1}=3+2\sqrt{2}$,
$(x+y)^2=(3-2\sqrt{2}+3+2\sqrt{2})^2=6^2=36$.
(3) 原式$=[\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)} + \frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})} + \dots + \frac{\sqrt{2026}-\sqrt{2025}}{(\sqrt{2026}+\sqrt{2025})(\sqrt{2026}-\sqrt{2025})}] × (\sqrt{2026}+1)$
$= (\frac{\sqrt{2}-1}{2-1} + \frac{\sqrt{3}-\sqrt{2}}{3-2} + \dots + \frac{\sqrt{2026}-\sqrt{2025}}{2026-2025}) × (\sqrt{2026}+1)$
$=(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\dots+\sqrt{2026}-\sqrt{2025})×(\sqrt{2026}+1)$
$=(\sqrt{2026}-1)(\sqrt{2026}+1)$
$=2026-1$
$=2025$