Giải:
Ta có:
\(\sqrt{1}< \sqrt{100}\Leftrightarrow\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}\)
\(\sqrt{2}< \sqrt{100}\Leftrightarrow\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}\)
\(\sqrt{3}< \sqrt{100}\Leftrightarrow\dfrac{1}{\sqrt{3}}>\dfrac{1}{\sqrt{100}}\)
...
\(\sqrt{99}< \sqrt{100}\Leftrightarrow\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}\)
\(\sqrt{100}=\sqrt{100}\Leftrightarrow\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{100}}\)
Cộng vế theo vế, ta được:
\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...\dfrac{1}{\sqrt{100}}>\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+...+\dfrac{1}{\sqrt{100}}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...\dfrac{1}{\sqrt{100}}>\dfrac{100}{\sqrt{100}}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...\dfrac{1}{\sqrt{100}}>\dfrac{100}{10}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...\dfrac{1}{\sqrt{100}}>10\)
Vậy ...
