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 \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{2}}>...>\frac{1}{\sqrt{100}}\)

\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>100.\frac{1}{\sqrt{100}}\) 

\(=\sqrt{100}=10\RightarrowĐPCM\)

( 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:

\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}\Rightarrow\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}=100.\frac{1}{\sqrt{100}}=10\left(đpcm\right)\)

Dấu "⇒" mình đánh nhầm....Bạn Lê Phương Uyên Nhi chuyển thành dấu ">" nhé!!!

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)

\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}\)

Ta có \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}\)

       \(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)

 ... 

      \(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}\)

=> \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}\)

                                         (có 100 số hạng \(\frac{1}{\sqrt{100}}\))

=>\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}\)\(>\frac{1}{\sqrt{100}}.100\)

=>\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}\)\(\ge\frac{1}{10}.100=10\)

Vậy \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>10\)

Học tốt

Ta có: \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}\)

\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)

\(\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{100}}\)

 \(..............\)

\(\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\)

\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}\)

\(\Rightarrow>100.\frac{1}{\sqrt{100}}=10\)

\(\Rightarrowđpcm\)

Làm tắt thôi, hôm trước vừa làm câu nỳ xong, hiểu thì tự trình bày nhé~

Đặt biểu thức trên là A.Ta có:

\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}\)

\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)

             ...

\(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}\)

\(\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\)

\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>100.\frac{1}{\sqrt{100}}\)

                                                                                          \(=10\)

Vậy \(A>10\left(đpcm\right)\)

Vì 

\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

.............................

\(\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}=\frac{1}{10}\)

Cộng vế với vế ta được :

\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{100}}>\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+....+\frac{1}{10}\) ( có 100 số \(\frac{1}{10}\) )

\(\Leftrightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{100}}>\frac{100}{10}=10\) (đpcm)

con cặc 

 ddddddddddd