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

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

Ta có:

\(\frac{1}{5^2}

thuỳ dung đúng đấy

ME TOO

Đặt \(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+....+\frac{1}{2007^2}\)\(A< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{2006.2007}\)

\(=\frac{5-4}{4.5}+\frac{6-5}{6.5}+....+\frac{2007-2006}{2006.2007}\)

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

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

\(\Leftrightarrow A< \frac{1}{4}-\frac{1}{2007}< \frac{1}{4}\)

vậy đpcm

\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{2006.2007}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{2006}-\frac{1}{2007}=\frac{1}{4}-\frac{1}{2007}< \frac{1}{4}\)