CMR: \(1+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{2500}}< 100\)
K
Khách
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PP
0
GH
2
26 tháng 5 2018
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 ....
SM
1
HT
30 tháng 9 2018
Đặt A=\(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{100}}\)
\(\Leftrightarrow A=\dfrac{2}{2\sqrt{2}}+\dfrac{2}{2\sqrt{3}}+....+\dfrac{2}{2\sqrt{100}}\)
\(\Leftrightarrow A=\dfrac{2}{\sqrt{2}+\sqrt{2}}+\dfrac{2}{\sqrt{3}+\sqrt{3}}+....+\dfrac{2}{\sqrt{99}+\sqrt{99}}+\dfrac{2}{\sqrt{100}+\sqrt{100}}\)
\(\Leftrightarrow A=2\left(\dfrac{1}{\sqrt{2}+\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{99}}+\dfrac{1}{\sqrt{100}+\sqrt{100}}\right)\)
Ta có:
\(\dfrac{1}{\sqrt{2}+\sqrt{2}}< \dfrac{1}{1+\sqrt{2}};\dfrac{1}{\sqrt{3}+\sqrt{3}}< \dfrac{1}{\sqrt{2}+\sqrt{3}}\)
Tường tự, ta có:
\(\dfrac{A}{2}< \dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(A< 2\left(\dfrac{1-\sqrt{2}}{-1}+\dfrac{\sqrt{2}-\sqrt{3}}{-1}+\dfrac{\sqrt{99}-\sqrt{100}}{-1}\right)\)
\(A< -2\left(1-\sqrt{2}+\sqrt{2}-\sqrt{3}+...-\sqrt{99}+\sqrt{99}-\sqrt{100}\right)\)
\(A< -2\left(1-\sqrt{100}\right)\)
\(A< 18\)
Vậy\(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{100}}< 18\)
BS
1
15 tháng 10 2017
ta có :
\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}\)
\(=\dfrac{2}{2\sqrt{1}}+\dfrac{2}{2\sqrt{2}}+\dfrac{2}{2\sqrt{3}}+...+\dfrac{2}{2\sqrt{100}}\)
\(>\dfrac{2}{\sqrt{1}+\sqrt{2}}+\dfrac{2}{\sqrt{2}+\sqrt{3}}+\dfrac{2}{\sqrt{3}+\sqrt{4}}+...+\dfrac{2}{\sqrt{100}+\sqrt{101}}\)
\(=2\left(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{100}+\sqrt{101}}\right)\)
\(=2\left(\dfrac{\sqrt{1}-\sqrt{2}}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+\dfrac{\sqrt{3}+\sqrt{4}}{3-4}+...+\dfrac{\sqrt{100}-\sqrt{101}}{100-101}\right)\)
\(=2\left(\dfrac{\sqrt{1}-\sqrt{101}}{-1}\right)=2\left(\sqrt{101}-\sqrt{1}\right)=18,1\)
\(>18\)
\(\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>18\)
3 tháng 10 2017
\(linh=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+....+\dfrac{1}{\sqrt{99}}+\dfrac{1}{\sqrt{100}}\)
\(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}\\\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}\\.............\\\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}\end{matrix}\right.\)
Suy ra:
\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+....+\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+...+\dfrac{1}{\sqrt{100}}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{99}}>\dfrac{99}{\sqrt{100}}\)
\(linh=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+.....+\dfrac{1}{\sqrt{99}}+\dfrac{1}{\sqrt{100}}>\dfrac{99}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}\)
\(\)\(linh>10\left(đpcm\right)\)
Bài này ko phải 100 nhé
AH
Akai Haruma
Giáo viên
1 tháng 3 2021
Bạn xem lời giải tại đây:
cho 100 STN \(a_1,a_2,...,a_{100}\) thỏa mãn: \(\dfrac{1}{\sqrt{a_1}} \dfrac{1}{\sqrt{a_2}} ... \dfrac{1}{\sqrt{a_{100}... - Hoc24
1 tháng 8 2018
Ta có : \(\sqrt{n+1}-\sqrt{n}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n+1}+\sqrt{n}}=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}< \dfrac{1}{\sqrt{n}+\sqrt{n}}=\dfrac{1}{2\sqrt{n}}\) ⇒ \(2\left(\sqrt{n+1}-\sqrt{n}\right)< \dfrac{1}{\sqrt{n}}\left(1\right)\)
\(\sqrt{n}-\sqrt{n-1}=\dfrac{\left(\sqrt{n}-\sqrt{n-1}\right)\left(\sqrt{n}+\sqrt{n+1}\right)}{\sqrt{n}+\sqrt{n-1}}=\dfrac{1}{\sqrt{n}+\sqrt{n-1}}>\dfrac{1}{\sqrt{n}+\sqrt{n}}=\dfrac{1}{2\sqrt{n}}\) ⇒ \(2\left(\sqrt{n+1}-\sqrt{n}\right)>\dfrac{1}{\sqrt{n}}\left(2\right)\)
Từ \(\left(1;2\right)\text{⇒ }đpcm\)
nè
`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`