\(\frac{1}{5^2}+\frac{1}{6^2}+......+\frac{1}{2007^2}>\frac{1}{5}\)

Có \(\frac{1}{5^2}>\frac{1}{4.5}\)

    \(\frac{1}{6^2}>\frac{1}{5.6}\)

     \(........\)

     \(\frac{1}{2007^2}=\frac{1}{2006.2007}\)

\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+.......+\frac{1}{2007^2}< \frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{2006.2007}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+.....+\frac{1}{2006}-\frac{1}{2007}\)

\(=\frac{1}{4}-\frac{1}{2007}\)

\(=\frac{2003}{8028}>\frac{1}{5}\)

có thể tham khảo phương pháp giải ở đây https://hoc24.vn/hoi-dap/question/205816.html

cầu xin các bạn mở lòng từ bi giúp tớ bài này nhé 

a) \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}<\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{2006\cdot2007}\)

=>              \(<\frac{1}{4}-\frac{1}{2007}<\frac{1}{4}\)

\(vậy:\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{2007^2}<\frac{1}{4}\)

b) \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}>\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{2007\cdot2008}\)

=>    \(>\frac{1}{5}-\frac{1}{2008}>\frac{1}{5}\)

\(vậy:\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}>\frac{1}{5}\)

cảm ơn bạn nha

#)Giải :

Ta có : \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{2007.2008}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{2007}-\frac{1}{2008}=\frac{1}{5}-\frac{1}{2008}=\frac{2003}{10004}>\frac{1}{5}\)

\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}>\frac{1}{5}\)

\(\frac{1}{5}-\frac{1}{2018}>\frac{1}{5}????\)

#)Góp ý :

Chết ! máy tính lỗi rùi :v xin lỗi bn, mk tính nhầm, ph là \(\frac{2003}{10040}>\frac{1}{5}\) nhé @@ sai òi

15: A= 1/3-3/4+3/5+1/2007-1/36+1/15-2/9

Sửa đề: 

A=-3/4-2/9-1/36+1/3+3/5+1/15+1/2007

=-27/36-8/36-1/36+5/15+9/15+1/15+1/2007

=-1+1+1/2007=1/2007

16:

\(A=\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}-\dfrac{3}{4}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{64}\)

\(=\dfrac{5+9+1}{15}+\dfrac{-27-8-1}{36}+\dfrac{1}{64}\)

=1/64

17:

=1/2-1/2+2/3-2/3+3/4-3/4+4/5-4/5+5/6-5/6-6/7

=-6/7