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b, \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>10\)
Ta có: \(1< 100\Rightarrow\sqrt{1}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{1}}< \frac{1}{\sqrt{100}}\)
\(2< 100\Rightarrow\sqrt{2}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{2}}< \frac{1}{\sqrt{100}}\)
\(3< 100\Rightarrow\sqrt{3}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{3}}< \frac{1}{\sqrt{100}}\)
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\(100=100\Rightarrow\sqrt{100}=\sqrt{100}\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\left(1\right)\)
Từ (1) suy ra:
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}\left(100sh\frac{1}{\sqrt{100}}\right)\)
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}.100\)
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>\frac{10}{\sqrt{100}}\)
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>10\left(ĐPCM\right)\)
b, \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
.............................................
Cộng với vế 99 của BĐT trên, ta được:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}>99.\frac{1}{10}=\frac{99}{10}\)
\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}>\frac{99}{10}=\frac{1}{10}=\frac{100}{10}=10\)
Wrecking Ball đã làm đúng
to ra kết quả giống cậu : Wrecking Ball
là đáp án đúng
tk nha ( chúc bn học gioi )
đề sai
chứng minh ngược lại C/m:>10
căn2<can3<can 4=>
1/căn2>1/căn3>1/căn4
1/căn2+1/can3+1/Căn4>3/can4=3/2
1/can5+....+1/can9>5.1/can9=5/3
1/can10+...+1/can16>7/can16=7/4
...
1/can81+...1/can100>18.1/can100= 19/10
A>B=1+3/2+5/3+7/4+...+19/10>10
Đề sai thật.
Xin phép sửa lại:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>10\)
Giải:
\(\sqrt{1}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}\)
\(\sqrt{2}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)
....
\(\sqrt{99}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\)
Cộng từng vế trên HĐT ta có:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}=\frac{100}{\sqrt{100}}=\frac{100}{10}\)
\(=10\)
Ta có:
\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)
\(.............\)
\(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}\)
Khi đó:
\(A=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{100}}\)
\(>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+.......+\frac{1}{\sqrt{100}}\left(100sohang\right)\)
\(=10\)
\(A=\frac{2!+\sqrt{3}}{2!}+\frac{3!+\sqrt{4}}{3!}+\frac{4!+\sqrt{5}}{4!}+....+\frac{2012!+\sqrt{2013}}{2012!}\)
\(=\frac{2!}{2!}+\frac{\sqrt{3}}{2!}+\frac{3!}{3!}+\frac{\sqrt{4}}{3!}+.....+\frac{2012!}{2012!}+\frac{\sqrt{2013}}{2012!}\)
\(=2012+\left(\frac{\sqrt{3}}{2!}+\frac{\sqrt{4}}{3!}+....+\frac{\sqrt{2011}}{2012!}\right)\)
Mà \(\frac{\sqrt{3}}{2!}+\frac{\sqrt{4}}{3!}+...+\frac{\sqrt{2013}}{2012!}>0\)
\(\Rightarrow A>2012+0=2012\)
Đề sai nên t sửa lại r nhé
\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
..............................
\(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
Cộng vế với vế của 99 bất đẳng thức trên ta được:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}>99\cdot\frac{1}{10}=\frac{99}{10}\)
=> A = \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}>\frac{99}{10}+\frac{1}{10}=\frac{100}{10}=10\)
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\)\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1 +\frac{x+349}{5}-4=0\)
\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
\(\Leftrightarrow\)\(x+329=0\) (vì 1/327 + 1/326 + 1/325 + 1/324 + 1/5 khác 0 )
\(\Leftrightarrow\)\(x=-329\)
Bài 1 :
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\)\(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Vì \(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)
\(\Rightarrow\)\(x+329=0\)
\(\Rightarrow\)\(x=-329\)
Vậy \(x=-329\)