*Có : 5² < 5.6 => \(\frac{1}{5^2}>\frac{1}{5.6}\)

         6² < 6.7 =>\(\frac{1}{6^2}>\frac{1}{6.7}\)

   ....

         100² < 100 . 101 => \(\frac{1}{100^2}>\frac{1}{100.101}\)

Cộng từng vế có :

\(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}\)

\(A>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...\frac{1}{100}-\frac{1}{101}\)

\(A>\frac{1}{5}-\frac{1}{101}\)

Mà \(\frac{1}{5}-\frac{1}{101}=\frac{101-5}{105}=\frac{96}{505}\)

=> \(A>\frac{96}{505}\)

Mà \(\frac{1}{6}=\frac{96}{576}< \frac{96}{505}\)

=> \(A>\frac{1}{6}\)(1)

*Có 5² > 5.4 => \(\frac{1}{5^2}< \frac{1}{5.4}\)

.......

    100² > 100.99 => \(\frac{1}{100^2}< \frac{1}{100.99}\)

Cộng từng vế có :

........ => A < \(\frac{96}{400}\)

Có \(\frac{1}{4}=\frac{100}{400}>\frac{96}{400}\)

=> A < \(\frac{1}{4}\)(2)

Từ (1)(2) => đpcm

\(\text{Ta thấy :}\)

\(\frac{1}{5^2}>\frac{1}{5.6}\)

\(\frac{1}{6^2}>\frac{1}{6.7}\)

\(......................................\)

\(\frac{1}{100^2}>\frac{1}{100.101}\)

\(\Rightarrow A=\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}\)

\(\Rightarrow A>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...\frac{1}{100}-\frac{1}{101}\)

\(\Rightarrow A>\frac{1}{5}-\frac{1}{101}=\frac{101-5}{105}=\frac{96}{505}>\frac{96}{576}=\frac{1}{6}\)

\(\Rightarrow A>\frac{1}{6}\left(1\right)\)

\(\text{Lại thấy :}\)

\(\frac{1}{5^2}< \frac{1}{5.4}\)

\(\frac{1}{6^2}< \frac{1}{5.6}\)

\(..................................\)

\(\frac{1}{100^2}< \frac{1}{100.99}\)

\(\text{Tương tự như trên ta tính được }:\)

\(A< \frac{96}{400}< \frac{100}{400}=\frac{1}{4}\)

\(\Rightarrow A< \frac{1}{4}\left(2\right)\)

\(\text{Từ (1) và (2)}\Rightarrow\frac{1}{6}< A< \frac{1}{4}\)

Đặt \(A=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}\)

Ta có: \(\frac{1}{1+\sqrt{2}}>\frac{1}{2}\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}\right)\)

\(\frac{1}{\sqrt{3}+\sqrt{4}}>\frac{1}{2}\left(\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}\right)\)

...

\(\frac{1}{\sqrt{79}+\sqrt{80}}>\frac{1}{2}\left(\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\right)\)

Cộng các bất đẳng thức trên lại với nhau, ta được:

\(A>\frac{1}{2}\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{80}+\sqrt{81}}\right)\)

\(\Leftrightarrow A>\frac{1}{2}\left(\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{81}-\sqrt{80}}{81-80}\right)\)

\(\Leftrightarrow A>\frac{1}{2}\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{81}-\sqrt{80}\right)\)

\(\Leftrightarrow A>\frac{1}{2}\left(\sqrt{81}-1\right)=\frac{1}{2}\cdot\left(9-1\right)=\frac{1}{2}\cdot8=4\)

\(\Leftrightarrow A>4\)(đpcm)

\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{79}{80}\)

\(A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{80}{81}\)

\(A^2< \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}...\frac{79}{80}.\frac{80}{81}\)

\(A^2< \frac{1}{81}=\left(\frac{1}{9}\right)^2\)

=> \(A< \frac{1}{9}\left(đpcm\right)\)

Ta có:

\(\frac{1}{2}\)= 1- \(\frac{1}{2}\) < 1- \(\frac{1}{3}\)=\(\frac{2}{3}\)

\(\frac{3}{4}\)= 1- \(\frac{1}{4}\) < 1- \(\frac{1}{5}\) = \(\frac{4}{5}\)

...

\(\frac{79}{80}\) = 1- \(\frac{1}{80}\) < 1- \(\frac{1}{81}\)= \(\frac{80}{81}\)

Từ trên, ta có:

A= \(\frac{1}{2}\). \(\frac{3}{4}\). \(\frac{5}{6}\)...\(\frac{79}{80}\)< \(\frac{2}{3}\). \(\frac{4}{5}\). \(\frac{6}{7}\)...\(\frac{80}{81}\)

A² <  \(\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{80}{81}\right)\). \(\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{79}{80}\right)\)

A² < \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{79}{80}.\frac{80}{81}\)

A² <\(\frac{1.\left(2.3.4...79.80\right)}{\left(2.3.4...79.80\right).81}\)

A² < \(\frac{1}{81}\) =\(\left(\frac{1}{9}\right)^2\)

A  ^<  \(\frac{1}{9}\)  (đpcm)

Vậy A< \(\frac{1}{9}\)

\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{79}{80}\)

\(A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{80}{81}\)

\(A^2< \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}....\frac{79}{80}.\frac{80}{81}\)

\(A^2< \frac{1}{81}=\left(\frac{1}{9}\right)^2\)

\(\Rightarrow A< \frac{1}{9}\left(\text{đ}pcm\right)\)

Ta có \(A

A=1+4+4²+...+4⁷⁹

A=(1+4)+(4²+4³)+...+(4⁷⁸+4⁷⁹)

A=1.(1+4)+4².(1+4)+...+4⁷⁸.(1+4)

A=1.5+4².5+...+4⁷⁸.5

A=5.(1+4²+...+4⁷⁸)

=>A chia hết cho 5

A=1+4+4²+4³+4⁴+4⁵⁺...+4⁵⁸⁺4⁵⁹

b, A chia hết cho21

A = 2 + 2² + 2³ + ...+ 2³⁰

A = ( 2 + 2² ) + ( 2³ + 2⁴ ) + ....+ ( 2²⁹ + 2³⁰ )

A = 2(1+2) + 2³(1+2) + ....+ 2²⁹(1+2)

A = 2.3 + 2³ . 3 + ...+ 2²⁹.3

A = 3(2+2³ + ...+ 2²⁹) \(⋮\) 3

Vậy  A chia hết cho 3