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3
AK
15 tháng 4 2018
Ta có :
\(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};...;\frac{1}{9^2}>\frac{1}{9.10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{1}{2}-\frac{1}{10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{5}{10}-\frac{1}{10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{4}{10}=\frac{2}{5}\left(1\right)\)
Ta có :
\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{9^2}< \frac{1}{8.9}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}< 1-\frac{1}{9}=\frac{8}{9}\left(2\right)\)
Từ ( 1 ) , ( 2 ) => ĐPCM
Chúc bạn học tốt !!!
AK
15 tháng 4 2018
Đề sai bạn nhé :
Đề đúng :
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}\)
CM : \(\frac{2}{5}< A< \frac{8}{9}\)
MM
14
5 tháng 8 2015
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
NB
5
M
14 tháng 4 2018
2. \(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2}{9}\)
\(\dfrac{2}{42}+\dfrac{2}{56}+\dfrac{2}{72}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2}{9}\)
\(\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2}{9}\)
\(2.\left(\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{x.\left(x+1\right)}\right)=\dfrac{2}{9}\)
\(2.\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)
\(2.\left(\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{2}{9}\right)\)
\(\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{2}{9}:2\)
\(\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\)
\(\dfrac{1}{x+1}=\dfrac{1}{18}\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=17\)
HT
1
14 tháng 4 2019
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
NM
1
24 tháng 6 2015
ta có A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\) < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
= \(1-\frac{1}{9}\)
= \(\frac{8}{9}\)
suy ra A < \(\frac{8}{9}\)
ta có A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)j> \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
= \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
= \(\frac{1}{2}-\frac{1}{10}\)
= \(\frac{2}{5}\)
suy ra A >\(\frac{2}{5}\)
\(\dfrac{1}{2^2}>\dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
\(\dfrac{1}{3^2}>\dfrac{1}{3\cdot4}=\dfrac{1}{3}-\dfrac{1}{4}\)
...
\(\dfrac{1}{9^2}>\dfrac{1}{9\cdot10}=\dfrac{1}{9}-\dfrac{1}{10}\)
Do đó: \(A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
...
\(\dfrac{1}{9^2}< \dfrac{1}{8\cdot9}=\dfrac{1}{8}-\dfrac{1}{9}\)
Do đó: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}=\dfrac{8}{9}\)
Suy ra: \(\dfrac{2}{5}< A< \dfrac{8}{9}\)