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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做分母有理化.
类似以上这种化简的步骤叫做分母有理化.
答案
(1) $\boldsymbol{\frac{\sqrt{6}}{2}}$，$\boldsymbol{\frac{\sqrt{6}}{3}}$，$\boldsymbol{\sqrt{5}-\sqrt{3}}$；
(2) $\boldsymbol{2025}$。
(2) $\boldsymbol{2025}$。
解析
解：
(1)
$\frac{3}{\sqrt{6}}=\frac{3×\sqrt{6}}{\sqrt{6}×\sqrt{6}}=\frac{3\sqrt{6}}{6}=\frac{\sqrt{6}}{2}$
$\sqrt{\frac{2}{3}}=\sqrt{\frac{2×3}{3×3}}=\frac{\sqrt{6}}{3}$
$\frac{2}{\sqrt{5}+\sqrt{3}}=\frac{2×(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}=\frac{2(\sqrt{5}-\sqrt{3})}{5-3}=\sqrt{5}-\sqrt{3}$
(2) 对括号内每一项分别分母有理化：
$\frac{1}{\sqrt{2}+\sqrt{1}}=\sqrt{2}-1$
$\frac{1}{\sqrt{3}+\sqrt{2}}=\sqrt{3}-\sqrt{2}$
$\dots$
$\frac{1}{\sqrt{2026}+\sqrt{2025}}=\sqrt{2026}-\sqrt{2025}$
将上述各式相加：
$\begin{aligned}&\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\dots+\frac{1}{\sqrt{2026}+\sqrt{2025}}\\=&(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+\dots+(\sqrt{2026}-\sqrt{2025})\\=&\sqrt{2026}-1\end{aligned}$
因此原式$=(\sqrt{2026}-1)×(\sqrt{2026}+1)=(\sqrt{2026})^2 -1^2=2026-1=2025$
(1)
$\frac{3}{\sqrt{6}}=\frac{3×\sqrt{6}}{\sqrt{6}×\sqrt{6}}=\frac{3\sqrt{6}}{6}=\frac{\sqrt{6}}{2}$
$\sqrt{\frac{2}{3}}=\sqrt{\frac{2×3}{3×3}}=\frac{\sqrt{6}}{3}$
$\frac{2}{\sqrt{5}+\sqrt{3}}=\frac{2×(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}=\frac{2(\sqrt{5}-\sqrt{3})}{5-3}=\sqrt{5}-\sqrt{3}$
(2) 对括号内每一项分别分母有理化：
$\frac{1}{\sqrt{2}+\sqrt{1}}=\sqrt{2}-1$
$\frac{1}{\sqrt{3}+\sqrt{2}}=\sqrt{3}-\sqrt{2}$
$\dots$
$\frac{1}{\sqrt{2026}+\sqrt{2025}}=\sqrt{2026}-\sqrt{2025}$
将上述各式相加：
$\begin{aligned}&\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\dots+\frac{1}{\sqrt{2026}+\sqrt{2025}}\\=&(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+\dots+(\sqrt{2026}-\sqrt{2025})\\=&\sqrt{2026}-1\end{aligned}$
因此原式$=(\sqrt{2026}-1)×(\sqrt{2026}+1)=(\sqrt{2026})^2 -1^2=2026-1=2025$
