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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:
\(A=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{100}}\)
\(>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+.......+\frac{1}{\sqrt{100}}\left(100sohang\right)\)
\(=10\)
Có : \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
..................
\(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
=> \(\frac{1}{\sqrt{1}}\)+ \(\frac{1}{\sqrt{2}}\)+ ......... + \(\frac{1}{\sqrt{100}}\)> 1/10 + 1/10 + ...... +1/10 ( có 100 phân số 1/10 )
= 100/10 = 10
=> ĐPCM
Tk mk nha
Do \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{2}}>...>\frac{1}{\sqrt{100}}\)
\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>100.\frac{1}{\sqrt{100}}\)
\(=\sqrt{100}=10\RightarrowĐPCM\)
Ta có:\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}};\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}};...;\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}\) và \(\frac{1}{\sqrt{100}}=\frac{1}{10}\)
=>\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}>\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}\) (100 số hạng 1/10)
\(=100.\frac{1}{10}\)
\(=10\) (đpcm)
Ta có :
\(\frac{1}{\sqrt{1}}>\frac{1}{10}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{10}\)
.....
\(\frac{1}{\sqrt{99}}>\frac{1}{10}\)
\(\frac{1}{\sqrt{100}}=\frac{1}{10}\)
Cộng vế theo vế ta có :\(\frac{1}{\sqrt{1}}\frac{1}{\sqrt{2}}+......+\frac{1}{\sqrt{99}}+\frac{1}{100}>100.\frac{1}{10}=10\)
b, \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>10\)
Ta có: \(1< 100\Rightarrow\sqrt{1}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{1}}< \frac{1}{\sqrt{100}}\)
\(2< 100\Rightarrow\sqrt{2}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{2}}< \frac{1}{\sqrt{100}}\)
\(3< 100\Rightarrow\sqrt{3}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{3}}< \frac{1}{\sqrt{100}}\)
______________________________________________
\(100=100\Rightarrow\sqrt{100}=\sqrt{100}\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\left(1\right)\)
Từ (1) suy ra:
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}\left(100sh\frac{1}{\sqrt{100}}\right)\)
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}.100\)
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>\frac{10}{\sqrt{100}}\)
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>10\left(ĐPCM\right)\)
( Bạn đặt A = (...) biểu thức đã cho )
Ta có :
\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{100}}\)
\(............\)
\(\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\)
\(\Rightarrow\)\(A=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}\)
\(\Rightarrow\)\(A>100.\frac{1}{\sqrt{100}}=\frac{100}{\sqrt{100}}=\frac{100}{10}=10\)
\(\Rightarrow\)\(A>10\) ( đpcm )
Vậy \(A>10\)
Chúc bạn học tốt ~
Ta có:\(\frac{1}{\sqrt{1}}>\frac{1}{10}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{10}\)
\(\frac{1}{\sqrt{3}}>\frac{1}{10}\)
...........
\(\frac{1}{\sqrt{100}}=\frac{1}{10}\)
\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}\)
=\(100.\frac{1}{10}=10\)
=> đpcm ( Tự KL nhé)
Ta có: \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{3}}>...>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{100}}>100.\frac{1}{\sqrt{100}}=\frac{100.1}{10}=10\)
Vậy \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{100}}>10\)
cảm ơn bạn nhiều