อ้างอิง:
ข้อความเดิมเขียนโดยคุณ นายสบาย
จงหาค่าของ $$ \frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\sqrt{10+\sqrt{3}}+...+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\sqrt{10-\sqrt{3}}+...+\sqrt{10-\sqrt{99}} } $$
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Very nice problem
Use the identity $\sqrt{a+\sqrt{a^2-b}}=\dfrac{1}{\sqrt{2}}\Big(\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}\Big)$.
$x= \dfrac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\sqrt{10+\sqrt{3}}+...+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\sqrt{10-\sqrt{3}}+...+\sqrt{10-\sqrt{99}} } $
$~=\dfrac{1}{\sqrt{2}}\dfrac{\big(\sqrt{10+\sqrt{99}}+\sqrt{10-\sqrt{99}}\big)+\cdots+\big(\sqrt{10+\sqrt{1}}+\sqrt{10-\sqrt{1}}\big)}{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\sqrt{10-\sqrt{3}}+...+\sqrt{10-\sqrt{99}} }$
$~=\dfrac{x+1}{\sqrt{2}}$
$\therefore x=\sqrt{2}+1$