1/5^2< 1/4.5=1/4-1/5 
1/6^2<1/5.6=1/5-1/6 
.. 
1/99^2<1/98.99=1/98-1/99 
1/100^2<1/99.100=1/99-1/100 
Cộng vế theo vế

=> 1/5^2+1/6^2+...+1/100^2< 1/4 -1/100<1/4 

1/5^2> 1/5.6=1/5-1/6 
1/6^2>1/6.7=1/6-1/7 
.. 
1/99^2>1/99.100=1/99-1/100 
1/100^2>1/100.101=1/100-1/101 

Cộng vế theo vế
=> 1/5^2+1/6^2+...+1/100^2>1/5 -1/101=96/505>1/6 

Câu này có rồi

Đặt \(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...\frac{1}{100^2}\)

Ta có : 

\(A< \frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}+...+\frac{1}{99\times100}\)

\(\Rightarrow A< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)

Ta có : 

\(A>\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}+...+\frac{1}{100\times101}\)

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

Vậy \(\frac{1}{6}< A< \frac{1}{4}\left(đpcm\right)\)

CÂU HỎI TƯƠNG TỰ NHA BẠN

Dat A=1/5^2+1/6^2+1/7^2+............1/100^2<1/4.5+1/5.6+1/6.7+....+1/99.10=

1/4-1/5+1/5-1/6+1/6-1/7+.............1/99-1/100=

14-1/100=25/100-1/100=24/25/100=1/4(1)

A>1/5.6+1/6.7+1/7.8+....+1/100.101=

1/5-1/6+1/6-1/7+1/7-1/8 +...+1/100-1/101=

1/5-1/101>6 (2)

Tu 1 va 2 => dieu can chung minh

\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\)

\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}<\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}<\frac{1}{4}\)

Suy ra: điều cần chứng minh

đặt 1/5^2+1/6^2+,,,+1/100^2=A

*chứng minh A<1/4

ta có: \(\frac{1}{5^2}=\frac{1}{5.5}<\frac{1}{4.5}\)

\(\frac{1}{6^2}=\frac{1}{6.6}<\frac{1}{5.6}\)

...

\(\frac{1}{100^2}=\frac{1}{100.100}<\frac{1}{99.100}\)

\(=>A<\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)    
\(=>A<\frac{1}{4}-\frac{1}{100}<\frac{1}{4}=>A<\frac{1}{4}\left(1\right)\)

*chứng minh A>1/6

ta có \(\frac{1}{5^2}=\frac{1}{5.5}>\frac{1}{5.6}\)

\(\frac{1}{6^2}=\frac{1}{6.6}>\frac{1}{6.7}\)

...

\(\frac{1}{100^2}=\frac{1}{100.100}>\frac{1}{100.101}\)

\(=>A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)

\(=>A>\frac{1}{5}-\frac{1}{101}>\frac{1}{6}=>A>\frac{1}{6}\) (2)

từ (1) và (2)=>1/6<A<1/4 hay 1/6<1/5^2+...+1/100^2<1/4(đpcm)

tick nhé

1/4

Ta có:

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

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

.......

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

\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}\)                                                                                                                                                                               \(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)                                                                                                                             = \(\frac{1}{5}-\frac{1}{101}>\frac{1}{5}-\frac{1}{30}=\frac{1}{6}\)                                                                                  \(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{6}\)                              (1)

   Tương tự ta có:

\(\frac{1}{5^2}< \frac{1}{4.5}\)

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

......

\(\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)                                                                                                                                                                      \(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)

    \(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4}\)                                (2)

                     Từ (1) và (2)

                  \(\Rightarrow\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4}\)                  (đpcm)

     _Chúc_bạn_học_tốt_

thank you bn chúc bạn hok tot nhé

Ta có\(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}< \frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{101}< A< \frac{1}{4}-\frac{1}{100}\)(A là đề bài)

Mà \(\frac{1}{5}-\frac{1}{30}=\frac{1}{6}< \frac{1}{5}-\frac{1}{101}< A< \frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)

\(\Rightarrow\frac{1}{6}< A< \frac{1}{4}\left(ĐPCM\right)\)

Ta có: \(\frac{1}{5\cdot6}< \frac{1}{5^2}=\frac{1}{5\cdot5}< \frac{1}{4\cdot5}\)

           \(\frac{1}{6\cdot7}< \frac{1}{6^2}=\frac{1}{6\cdot6}< \frac{1}{5\cdot6}\)

            \(\frac{1}{7\cdot8}< \frac{1}{7^2}=\frac{1}{7\cdot7}< \frac{1}{6\cdot7}\)

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

            \(\frac{1}{100\cdot101}< \frac{1}{100^2}=\frac{1}{100\cdot100}< \frac{1}{99\cdot100}\)

Đặt \(A=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{100\cdot101}\)

          \(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{101}\)

          \(=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\)

        \(B=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{99\cdot100}\)

            \(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)

             \(=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)

\(=>\frac{1}{6}< A< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< B< \frac{1}{4}\)

\(=>\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(Đpcm\right)\)