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

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

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

..........

..........

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

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

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

Ta có

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

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

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

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

=> \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}>\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}\)(100 số\(\frac{1}{10}\))  >10

M>10

cần cách làm ko?

\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{10}};\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{10}};...;\frac{1}{\sqrt{9}}>\frac{1}{\sqrt{10}};\frac{1}{\sqrt{10}}=\frac{1}{\sqrt{10}}\)

=>M>10

M>10 nha bạn

Hình như bạn hơi nhầm đề bài . Nếu B là 10 thì mình biết .

Nhận thấy : \(\frac{1}{\sqrt{1}}\)>\(\frac{1}{\sqrt{100}}\); \(\frac{1}{\sqrt{2}}\)>\(\frac{1}{\sqrt{100}}\);...: \(\frac{1}{\sqrt{100}}\)=\(\frac{1}{\sqrt{100}}\)

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

Hay : A > \(\frac{1}{\sqrt{100}}\).100

   <=> A > 10

<=> A>B

Nếu không đúng mong bạn thông cảm nhé !!

tui cm dc A> 18 do nha

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

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

CÓ BẠN NÀO BIẾT CẢ CÁCH LÀM HÔNG?

ta có A = 18,58960382

nên A>10

a)Ta có:\(\sqrt{17}>\sqrt{16}\)

             \(\sqrt{26}>\sqrt{25}\)

\(\implies\) \(\sqrt{17}+\sqrt{26}>\sqrt{16}+\sqrt{25}\)

\(\implies\) \(\sqrt{17}+\sqrt{26}+1>\sqrt{16}+\sqrt{25}+1=4+5+1=10\)

Mà \(\sqrt{100}=10\) \(\implies\) \(\sqrt{17}+\sqrt{26}+1>\sqrt{100}\)

Mà \(\sqrt{100}>\sqrt{99}\) \(\implies\) \(\sqrt{17}+\sqrt{26}+1>\sqrt{99}\)

b)Ta có:\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+....+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}=100.\frac{1}{\sqrt{100}}\)

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

\(\implies\) \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+....+\frac{1}{\sqrt{100}}>10\left(đpcm\right)\)

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

\(\Rightarrow A>\frac{100.1}{\sqrt{100}}=\frac{100}{10}=10\)

Vậy A > 10

ta có \(\frac{1}{\sqrt{1}}>\frac{1}{10}\)

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

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

\(\frac{1}{\sqrt{99}}>\frac{1}{10}\)

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

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

\(\Rightarrow A>\frac{100}{10}=10\)