Ta có S=1/2^2+1/3^2+1/4^2+...+1/9^2 
           <1/2²+1/2*3+1/3*4+....+1/8*9 
           =1/2²+1/2-1/3+1/3-1/4+....+1/8-1/9 
           =1/4+1/2-1/9=23/36<32/36=8/9 (♪) 
Ta lại có S=1/2^2+1/3^2+1/4^2+...+1/9^2 
                >1/2²+1/3*4+1/4*5+....+1/9*10 
                =1/2²+1/3-1/4+1/4-1/5+........+1/9-1/10 
                =1/2²+1/3-1/10 
                =19/20>8/20=2/5 ( ♫) 
                Từ (♪)( ♫) cho ta đpcm

Đpcm là j thế bạn

Điều phải chứng minh

Ta có:\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)

            \(=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)

Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)

     \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)

    \(=1-\frac{1}{9}\)   

      \(=\frac{8}{9}\)

Lại có \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)

\(=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

Mà        \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}\)

\(=\frac{2}{5}\)

Vậy \(\frac{2}{5}< S< \frac{8}{9}\)

S< 1/1.2+1/2.3+1/3.4+...+1/8.9 = 1/1-1/2+1/2-1/3+1/3-1/4+...+1/8-1/9=1-1/9=8/9

=> S < 8/9

S> 1/2.3+1/3.4+1/4.5+...+1/9.10=1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10=1/2-1/10=4/10=2/5

=> S > 2/5

Đs: 2/5 < S < 8/9

copy à

câu nào cũng trả lời.trốn học à

mình hỏi giúp bạn mình thôi

\(\frac{1}{2\cdot2}< \frac{1}{1\cdot2};...;\frac{1}{9\cdot9}< \frac{1}{8\cdot9}\)

\(\Rightarrow S=\frac{1}{2^2}+...+\frac{1}{9^2}< \frac{1}{1\cdot2}+...+\frac{1}{8\cdot9}=1-\frac{1}{2}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\left(1\right)\)

\(\frac{1}{2\cdot2}>\frac{1}{2\cdot3};...;\frac{1}{9\cdot9}>\frac{1}{9\cdot10}\)

\(\Rightarrow S=\frac{1}{2^2}+...+\frac{1}{9^2}>\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}=\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\left(2\right)\)

Từ (1)(2) => đpcm

S<1/2^2 + 1/2.3 + 1/3.4 +...+ 1/8.9

S<1/4 + 1/2 - 1/3 + 1/3 - 1/4+...+1/8 - 1/9

S<1/4 + 1/2 - 1/9

S<23/36<8/9 (1)

Mặt khác: S>1/2^2 + 1/3.4 + ...+ 1/9*10

S>1/4 + 1/3 - 1/4 + ... + 1/9 - 1/10

S>1/4 + 1/3 - 1/10

S>29/60>2/5 (2)

Từ (1),(2)

=> 2/5<S<8/9