Cho số M= 1+\(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+..+\dfrac{1}{\sqrt{10^6}}\)
Chứng minh rằng 1998<M<1999
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Khách
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YN
10 tháng 2 2023
Ta có:
\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{3}}>\dfrac{1}{10}\)
...
\(\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>100.\dfrac{1}{10}=10\).
10 tháng 3 2017
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:
\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{100}}>\dfrac{1}{10}+\dfrac{1}{10}+...+\dfrac{1}{10}=\dfrac{100}{10}=10\)
1 tháng 10 2017
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)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 9 2021
\(VT^2\le3\left(\dfrac{1}{2x^2+y^2+3}+\dfrac{1}{2y^2+z^2+3}+\dfrac{1}{2z^2+x^2+3}\right)\)
Mặt khác:
\(\dfrac{1}{2\left(x^2+1\right)+y^2+1}\le\dfrac{1}{4x+2y}=\dfrac{1}{2}\left(\dfrac{1}{x+x+y}\right)\le\dfrac{1}{18}\left(\dfrac{2}{x}+\dfrac{1}{y}\right)\)
\(\Rightarrow VT^2\le\dfrac{1}{6}\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)=\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{3}{2}\)
\(\Rightarrow VT\le\dfrac{\sqrt{6}}{2}\)
2 tháng 1 2018
Ta có :
\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{`100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{3}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
........................................
\(\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+.......+\dfrac{1}{\sqrt{100}}>\dfrac{1}{10}+\dfrac{1}{10}+........+\dfrac{1}{10}=\dfrac{100}{10}=10\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+......+\dfrac{1}{\sqrt{100}}>10\left(đpcm\right)\)
NN
2 tháng 1 2018
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..........
2 tháng 4 2017
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)
9 tháng 2 2021
a ĐKXĐ \(a\ge0,a\ne\dfrac{1}{4},a\ne1\)
\(\Rightarrow P=1+\left(\dfrac{\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\dfrac{\sqrt{a}\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}\right)\cdot\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
= \(1+\left(\dfrac{\left(-1\right)\left(2\sqrt{a}-1\right)}{\sqrt{a}-1}+\dfrac{\sqrt{a}\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{2\sqrt{a}-1}\)
= \(1+\left(-1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{a+\sqrt{a}+1}\right)\sqrt{a}\)
= \(1-\sqrt{a}+\dfrac{a\sqrt{a}+a}{a+\sqrt{a}+1}\) = \(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)+a\sqrt{a}+a}{a+\sqrt{a}+1}=\dfrac{1-a\sqrt{a}+a\sqrt{a}+a}{a+\sqrt{a}+1}=\dfrac{a+1}{a+\sqrt{a}+1}\)
b Xét hiệu \(P-\dfrac{2}{3}=\dfrac{a+1}{a+\sqrt{a}+1}-\dfrac{2}{3}=\dfrac{3a+3-2a-2\sqrt{a}-2}{a+\sqrt{a}+1}=\dfrac{a-2\sqrt{a}+1}{a+\sqrt{a}+1}=\dfrac{\left(\sqrt{a}-1\right)^2}{a+\sqrt{a}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}>0\) \(\Rightarrow P>\dfrac{2}{3}\)
c Ta có \(P=\dfrac{\sqrt{6}}{\sqrt{6}+1}\Rightarrow\dfrac{a+1}{a+\sqrt{a}+1}=\dfrac{\sqrt{6}}{\sqrt{6}+1}\) \(\Rightarrow\left(a+1\right)\left(\sqrt{6}+1\right)=\sqrt{6}\left(a+\sqrt{a}+1\right)\Leftrightarrow a\sqrt{6}+a+\sqrt{6}+1=a\sqrt{6}+\sqrt{6a}+\sqrt{6}\)
\(\Leftrightarrow a-\sqrt{6a}+1=0\Leftrightarrow a-\sqrt{6a}+\dfrac{6}{4}-\dfrac{2}{4}=0\Leftrightarrow\left(\sqrt{a}-\dfrac{\sqrt{6}}{2}\right)^2=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{a}=\dfrac{\sqrt{6}+1}{2}\\\sqrt{a}=\dfrac{1-\sqrt{6}}{2}\left(L\right)\end{matrix}\right.\) (Do \(\sqrt{a}\ge0\)) \(\Rightarrow a=\dfrac{\left(\sqrt{6}+1\right)^2}{4}=\dfrac{7+2\sqrt{6}}{4}\left(TM\right)\)
Vậy...
5 tháng 5 2018
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)
6 tháng 5 2018
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)
