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Ta có:
\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{3}}>\dfrac{1}{10}\)
...
\(\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>100.\dfrac{1}{10}=10\).
Ta có:
\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)
\(.............\)
\(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}\)
Khi đó:
\(A=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{100}}\)
\(>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+.......+\frac{1}{\sqrt{100}}\left(100sohang\right)\)
\(=10\)
Ta có :
\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{`100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{3}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
........................................
\(\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+.......+\dfrac{1}{\sqrt{100}}>\dfrac{1}{10}+\dfrac{1}{10}+........+\dfrac{1}{10}=\dfrac{100}{10}=10\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+......+\dfrac{1}{\sqrt{100}}>10\left(đpcm\right)\)
Giải:
Ta thấy:
\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}.\)
\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}.\)
\(\dfrac{1}{\sqrt{3}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}.\)
...................................
\(\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}.\)
\(\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}.\)
\(\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}.\)
\(>\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+...+\dfrac{1}{\sqrt{100}}.\)
\(=\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}+...+\dfrac{1}{10}\) (100 số hạng \(\dfrac{1}{10}\)).
\(=\dfrac{100}{10}=10.\)
\(\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>10\left(đpcm\right).\)
Vậy..........
Ta có:
\(\sqrt{1}< \sqrt{100}\Rightarrow\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}\)
\(\sqrt{2}< \sqrt{100}\Rightarrow\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}\)
\(\sqrt{3}< \sqrt{100}\Rightarrow\dfrac{1}{\sqrt{3}}>\dfrac{1}{\sqrt{100}}\)
\(.............................\)
\(\sqrt{99}< \sqrt{100}\Rightarrow\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}\)
\(\sqrt{100}=\sqrt{100}\Rightarrow\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{100}}\)
Cộng từng vế của các BĐT trên ta được:
\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+...+\dfrac{1}{\sqrt{100}}\)
\(=\dfrac{100}{\sqrt{100}}=\dfrac{100}{10}=10\)
Vậy \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>10\) (Đpcm)
khos thees
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