求 [√(10+√1)+…+√(10+√99)]／[√(10－√1)+…+√(10－√99)] 的值

wintex

求 [√(10+√1)+…+√(10+√99)]／[√(10－√1)+…+√(10－√99)] 的值

Treenewbee

\[\sqrt{2}+1\]

wintex

Treenewbee 发表于 2023-5-30 23:01
\[\sqrt{2}+1\]

請問老師有過程嗎

王守恩

百度,好像有，可惜没看懂。
\(\frac{\big(\sqrt{2+\sqrt{1}}\big)+\big(\sqrt{2+\sqrt{2}}\big)+\big(\sqrt{2+\sqrt{3}}\big)}{\big(\sqrt{2-\sqrt{1}}\big)+\big(\sqrt{2-\sqrt{2}}\big)+\big(\sqrt{2-\sqrt{3}}\big)}=\sqrt{2}+1\)

\(\frac{\big(\sqrt{3+\sqrt{1}}\big)+\big(\sqrt{3+\sqrt{2}}\big)+\cdots+\big(\sqrt{3+\sqrt{8}}\big)}{\big(\sqrt{3-\sqrt{1}}\big)+\big(\sqrt{3-\sqrt{2}}\big)+\cdots+\big(\sqrt{3-\sqrt{8}}\big)}=\sqrt{2}+1\)

\(\frac{\big(\sqrt{4+\sqrt{1}}\big)+\big(\sqrt{4+\sqrt{2}}\big)+\cdots+\big(\sqrt{4+\sqrt{15}}\big)}{\big(\sqrt{4-\sqrt{1}}\big)+\big(\sqrt{4-\sqrt{2}}\big)+\cdots+\big(\sqrt{4-\sqrt{15}}\big)}=\sqrt{2}+1\)

\(\frac{\big(\sqrt{5+\sqrt{1}}\big)+\big(\sqrt{5+\sqrt{2}}\big)+\cdots+\big(\sqrt{5+\sqrt{24}}\big)}{\big(\sqrt{5-\sqrt{1}}\big)+\big(\sqrt{5-\sqrt{2}}\big)+\cdots+\big(\sqrt{5-\sqrt{24}}\big)}=\sqrt{2}+1\)

\(\frac{\big(\sqrt{6+\sqrt{1}}\big)+\big(\sqrt{6+\sqrt{2}}\big)+\cdots+\big(\sqrt{6+\sqrt{35}}\big)}{\big(\sqrt{6-\sqrt{1}}\big)+\big(\sqrt{6-\sqrt{2}}\big)+\cdots+\big(\sqrt{6-\sqrt{35}}\big)}=\sqrt{2}+1\)

cgl_74

证明表达太繁琐；我举个例子即可。这种无理化的化简太绕了。

luyuanhong

楼上 cgl_74 的解答已收藏。

王守恩

cgl_74 发表于 2023-6-1 11:54
证明表达太繁琐；我举个例子即可。这种无理化的化简太绕了。

谢谢 cgl_74 ！百度,好像有，可惜没看懂。谢谢 cgl_74 ！

\(\frac{\big(\sqrt{3+\sqrt{1}}\big)+\big(\sqrt{3+\sqrt{8}}\big)}{\big(\sqrt{3-\sqrt{1}}\big)+\big(\sqrt{3-\sqrt{8}}\big)}=\sqrt{2}+1\Rightarrow\frac{\big(\sqrt{n+\sqrt{k}}\big)+\big(\sqrt{n+\sqrt{n^2-k}}\big)}{\big(\sqrt{n-\sqrt{k}}\big)+\big(\sqrt{n-\sqrt{n^2-k}}\big)}=\sqrt{2}+1\)

\(\frac{\big(\sqrt{3+\sqrt{1}}\big)+\big(\sqrt{3+\sqrt{8}}\big)}{\big(\sqrt{3-\sqrt{1}}\big)+\big(\sqrt{3-\sqrt{8}}\big)}=\frac{\big(\sqrt{3+\sqrt{2}}\big)+\big(\sqrt{3+\sqrt{7}}\big)}{\big(\sqrt{3-\sqrt{2}}\big)+\big(\sqrt{3-\sqrt{7}}\big)}=\frac{\big(\sqrt{3+\sqrt{3}}\big)+\big(\sqrt{3+\sqrt{6}}\big)}{\big(\sqrt{3-\sqrt{3}}\big)+\big(\sqrt{3-\sqrt{6}}\big)}=\frac{\big(\sqrt{3+\sqrt{4}}\big)+\big(\sqrt{3+\sqrt{5}}\big)}{\big(\sqrt{3-\sqrt{4}}\big)+\big(\sqrt{3-\sqrt{5}}\big)}=\sqrt{2}+1\)

\(\frac{a+c+e+\cdots}{b+d+f+\cdots}=\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\cdots\Rightarrow\frac{\big(\sqrt{10+\sqrt{1}}\big)+\big(\sqrt{10+\sqrt{2}}\big)+\cdots+\big(\sqrt{10+\sqrt{99}}\big)}{\big(\sqrt{10-\sqrt{1}}\big)+\big(\sqrt{10-\sqrt{2}}\big)+\cdots+\big(\sqrt{10-\sqrt{99}}\big)}=\sqrt{2}+1\)