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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) \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}>\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{2007\cdot2008}\)
=> \(>\frac{1}{5}-\frac{1}{2008}>\frac{1}{5}\)
\(vậy:\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}>\frac{1}{5}\)
Ta có : 1/5^2 + 1/6^2 + 1/7^2 +....+ 1/2007^2 > 1/5.6 + 1/6.7 + 1/7.8 +...+ 1/2007.2008 = 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 +....+ 1/2007 - 1/2008 = 1/5 -1/2008 ko > 1/5
nhưng cái biểu thức nó cũng lớn hơn cái biểu thức bạn đưa ra nên ko thể chứng minh nó >\(\frac{1}{5}\)
Đặt :
\(A=\frac{1}{5^2}+\frac{1}{6^2}+.........+\frac{1}{2007^2}\)
Ta thấy :
\(\frac{1}{5^2}>\frac{1}{5.6}\)
\(\frac{1}{6^2}>\frac{1}{6.7}\)
...........................
\(\frac{1}{2007^2}>\frac{1}{2007.2008}\)
\(\Leftrightarrow A>\frac{1}{5.6}+\frac{1}{6.7}+........+\frac{1}{2007.2008}\)
\(\Leftrightarrow A>\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+......+\frac{1}{2007}-\frac{1}{2008}\)
\(\Leftrightarrow A>\frac{1}{5}-\frac{1}{2008}>\frac{1}{5}\)
\(\Leftrightarrow A>\frac{1}{5}\)
#)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 \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{...1}{2007^2}>\left(\frac{1}{5\cdot6}+\frac{1}{150}\right)+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{2007\cdot2008}\)
\(=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{2007\cdot2008}+\frac{1}{150}>\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{2007\cdot2008}+\frac{1}{2008}=\frac{1}{5}\left(đpcm\right)\)