#)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
Ta có:(trội)\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}>\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{2007\cdot2008}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{2007}-\frac{1}{2008}=\frac{1}{5}-\frac{1}{2008}>\frac{1}{5}\left(đpcm\right)\)
Bao Pham [English club]
T.Ps
Cả 2 đều làm sai chõ cuối nhé
\(\frac{1}{5}-\frac{1}{2008}< \frac{1}{5}\)
Luôn luôn đúng
Đăt A= vế trái
=>5A=\(\frac{1}{5}+5.\left(\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}\right)\)
>\(\frac{1}{5}+5.\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{2007.2008}\right)\)
=\(\frac{1}{5}+5.\frac{1001}{6024}>1\)
=> A>1/5
=>dpcm
Mk nhầm!
\(\frac{1}{5}=\frac{1}{5}-\frac{1}{2007}+\frac{1}{2007}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{2006}-\frac{1}{2007}+\frac{1}{2007}\)
\(=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{2007\cdot2008}+\frac{1}{2008}\)
Mà \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{...1}{2007^2}>\left(\frac{1}{5\cdot6}+\frac{1}{150}\right)+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{2007\cdot2008}\)
\(=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{2007\cdot2008}+\frac{1}{150}>\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{2007\cdot2008}+\frac{1}{2008}=\frac{1}{5}\left(đpcm\right)\)
