Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
`1)` Ta có `:` `1/sqrt1;1/sqrt2;1/sqrt3;…;1/sqrt99>1/sqrt100`
`=>` `1/sqrt1+1/sqrt2+1/sqrt3+…+1/sqrt99+1/sqrt100>100. 1/sqrt100=100/10=10`
`=>` `đpcm`
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:
\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(...............\)
\(\dfrac{1}{\sqrt{98}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
Cộng theo vế ta có:
\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{99}}>\dfrac{1}{10}+\dfrac{1}{10}+...+\dfrac{1}{10}=\dfrac{99}{10}\)
Lại có \(\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\) suy ra:
\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{100}}>\dfrac{1}{10}+\dfrac{1}{10}+...+\dfrac{1}{10}=\dfrac{100}{10}=10\)
Ta có:
1/√1>1/√100=1/10
1/√2>1/√100=1/10
........
1/√100=1/√100=1/10
Nên:
1/√1+1/√2+...+1/√100>1/10+1/10+...+1/10(100 phân số 1/10)
=1/√1+1/√2+..+1/√100>100/10
1/√1+1/√2+..+1/√100>10(đpcm)
Ta có :
\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}\\ \dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}\\ .........\\ \dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{100}}\)
\(\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{100}}>\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+....+\dfrac{1}{\sqrt{100}}\)( 100 phân số \(\dfrac{1}{\sqrt{100}}\) )
hay \(A>\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}+....+\dfrac{1}{10}\)(100 phân số \(\dfrac{1}{10}\) )
\(\Rightarrow A>\dfrac{100}{10}\\ \Rightarrow A>10\)
KL : Vậy ....
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)
refer
https://hoc24.vn/cau-hoi/cmr-1121sqrt-31sqrt-2500100.222352266174