+ \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)-n}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Do đó : \(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)

\(=1-\frac{1}{10}=\frac{9}{10}\)

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

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

 . .. . . .  .

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

\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}\)( 100 phân số \(\frac{1}{\sqrt{100}}\)) . Mà:

  \(\sqrt{100}=10\RightarrowĐPCM\)

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)

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:

\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}=\frac{100}{\sqrt{100}}=\frac{100}{10}\)

\(=10\)

chào hen 

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 :\(\frac{1}{\sqrt{1}}\frac{1}{\sqrt{2}}+......+\frac{1}{\sqrt{99}}+\frac{1}{100}>100.\frac{1}{10}=10\)

( 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:

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

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 \(A>10\left(đpcm\right)\)