Chứng minh rằng: \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>10\)
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Chứng minh rằng : \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>10\)
Ta có:
\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{3}}>\dfrac{1}{10}\)
...
\(\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>100.\dfrac{1}{10}=10\).
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Chứng minh rằng :\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}+\dfrac{1}{\sqrt{100}}\)>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:
\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{100}}>\dfrac{1}{10}+\dfrac{1}{10}+...+\dfrac{1}{10}=\dfrac{100}{10}=10\)
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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)
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Ta có:
1√1>1√100=11011>1100=110
1√2>1√100=11012>1100=110
..............................
1√98>1√100=110198>1100=110
1√99>1√100=110199>1100=110
Cộng theo vế ta có:
1√1+1√2+...+1√99>110+110+...+110=991011+12+...+199>110+110+...+110=9910
Lại có 1√100=1101100=110 suy ra:
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Chứng minh rằng: \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+....+\dfrac{1}{\sqrt{100}}>10\)
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)
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Bài 40: Chứng minh rằng:
a) \(A=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}=9\)
b) \(B=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}=\dfrac{9}{10}\)
chứng minh rằng
\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+........+\dfrac{1}{\sqrt{100}}>10\)
Ta có:
1/ căn 1> 1/10
1/ căn 2> 1/10
...
1/ căn 99> 1/10
1/ căn 100 = 1/10
=> 1/ căn 1 + 1/ căn 2 + ... + 1/ căn 99 + 1/ căn 100 > 100 . 1/10 = 10 (đpcm)
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1/√1 > 1/10
1/√2 > 1/10
1/√3 > 1/10
....................
1/√99 > 1/10
1/√100 = 1/10
Cộng từng vế ta có:
1/√1 + 1/√2 + 1/√3 + ... + 1/√100 >100.1/0 = 10 (Đpcm)
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\(\dfrac{1}{\sqrt{1}}\) > \(\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{2}}\)> \(\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{3}}\) > \(\dfrac{1}{10}\)
....................
\(\dfrac{1}{\sqrt{99}}\) > \(\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{100}}\) = \(\dfrac{1}{10}\)
Cộng theo vế ta có:
\(\dfrac{1}{\sqrt{1}}\)+\(\dfrac{1}{\sqrt{2}}\)+\(\dfrac{1}{\sqrt{3}}\)+ ... +\(\dfrac{1}{\sqrt{100}}\) >100.\(\dfrac{1}{10}\) = 10 (Đpcm)
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Chứng minh rằng :
\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+....+\dfrac{1}{\sqrt{100}}\)> 10 .
Ta có :
\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
.........................................
\(\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+..........+\dfrac{1}{\sqrt{100}}>\dfrac{1}{10}+\dfrac{1}{10}+....+\dfrac{1}{10}=\dfrac{1}{10}.100=10\left(đpcm\right)\)
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Ta có:
1/√1 > 1/10
1/√2 > 1/10
1/√3 > 1/10
....................
1/√99 > 1/10
1/√100 = 1/10
Cộng từng vế ta có:
1/√1 + 1/√2 + 1/√3 + ... + 1/√100 >100.1/0 = 10 (Đpcm)
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Bài 1 : chứng minh. \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}+\dfrac{1}{\sqrt{100}}>10\)
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Chứng minh rằng : \(\dfrac{1}{\sqrt{1}}\)+\(\dfrac{1}{\sqrt{2}}\)+\(\dfrac{1}{\sqrt{3}}\)+.....+\(\dfrac{1}{\sqrt{100}}\)>10
Ta có :
\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
.............................
\(\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+.........+\dfrac{1}{\sqrt{99}}+\dfrac{1}{\sqrt{100}}>\dfrac{1}{10}+\dfrac{1}{10}+.....+\dfrac{1}{10}=\dfrac{100}{10}=10\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+......+\dfrac{1}{\sqrt{100}}>10\left(đpcm\right)\)
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Cho \(x=\dfrac{\sqrt{2}-\sqrt{1}}{1+\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{2+\sqrt{3}}+\dfrac{\sqrt{4}-\sqrt{3}}{3+4}+...+\dfrac{\sqrt{100}-\sqrt{99}}{99+100}\). Chứng minh \(x< \dfrac{1}{2}\)