Chứng minh: \(17< \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+.....+\dfrac{1}{\sqrt{100}}\)
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Chứng minh: 17 < \(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{100}}\) < 18
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 \(x< \dfrac{1}{2}\)
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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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 đẳng thức sau:
\(\dfrac{1}{1+\sqrt{2}}\)+\(\dfrac{1}{\sqrt{2}+\sqrt{3}}\)+...+\(\dfrac{1}{\sqrt{99}+\sqrt{100}}\)=9
VT tương đương với \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\dfrac{\sqrt{1}-\sqrt{2}}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+...+\dfrac{\sqrt{99}-\sqrt{100}}{99-100}\)
\(=\sqrt{100}-\sqrt{99}+\sqrt{99}-....-\sqrt{3}+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}\) (kiểu do mẫu số nó có kết quả âm nên đảo lại phép)
\(=10-1=9=VP\)
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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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Bài 1 : chứng minh. \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}+\dfrac{1}{\sqrt{100}}>10\)
Chứng minh rằng: \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>10\)
Ta có :
\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{`100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{3}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
........................................
\(\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+.......+\dfrac{1}{\sqrt{100}}>\dfrac{1}{10}+\dfrac{1}{10}+........+\dfrac{1}{10}=\dfrac{100}{10}=10\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+......+\dfrac{1}{\sqrt{100}}>10\left(đpcm\right)\)
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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..........
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Ta có :
1√1>1√‘100=11011>1‘100=110
1√2>1√100=11012>1100=110
1√3>1√100=11013>1100=110
........................................
1√99>1√100=110199>1100=110
1√100=1√100=1101100=1100=110
⇔1√1+1√2+.......+1√100>110+110+........+110=10010=10⇔11+12+.......+1100>110+110+........+110=10010=10
⇔1√1+1√2+......+1√100>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:
\(\sqrt{1}< \sqrt{100}\Rightarrow\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}\)
\(\sqrt{2}< \sqrt{100}\Rightarrow\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}\)
\(\sqrt{3}< \sqrt{100}\Rightarrow\dfrac{1}{\sqrt{3}}>\dfrac{1}{\sqrt{100}}\)
\(.............................\)
\(\sqrt{99}< \sqrt{100}\Rightarrow\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}\)
\(\sqrt{100}=\sqrt{100}\Rightarrow\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{100}}\)
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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