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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) \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}>\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{2007\cdot2008}\)
=> \(>\frac{1}{5}-\frac{1}{2008}>\frac{1}{5}\)
\(vậy:\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}>\frac{1}{5}\)
5A=\(\frac{1}{5}+\frac{2}{5^2}...+\frac{n}{5^n}...+\frac{11}{5^{11}}\)
=>4A=5A-A=\(\frac{1}{5}+\frac{1}{5^2}...+\frac{1}{5^{11}}-\frac{11}{5^{12}}\)
=>20A=\(1+\frac{1}{5}+...+\frac{1}{5^{10}}-\frac{11}{5^{11}}\)
=>16A=20A-4A=\(1-\frac{1}{5^{11}}+\frac{11}{5^{12}}-\frac{11}{5^{11}}\)
Mà \(1-\frac{1}{5^{11}}< 1\),\(\frac{11}{5^{12}}-\frac{11}{5^{11}}< 0\)
=>16A<1
Do đó: A<1/16(đpcm)
Đặt :
\(A=\frac{1}{5^2}+\frac{1}{6^2}+.........+\frac{1}{2007^2}\)
Ta thấy :
\(\frac{1}{5^2}>\frac{1}{5.6}\)
\(\frac{1}{6^2}>\frac{1}{6.7}\)
...........................
\(\frac{1}{2007^2}>\frac{1}{2007.2008}\)
\(\Leftrightarrow A>\frac{1}{5.6}+\frac{1}{6.7}+........+\frac{1}{2007.2008}\)
\(\Leftrightarrow A>\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+......+\frac{1}{2007}-\frac{1}{2008}\)
\(\Leftrightarrow A>\frac{1}{5}-\frac{1}{2008}>\frac{1}{5}\)
\(\Leftrightarrow A>\frac{1}{5}\)
Bài làm:
Xét: \(\frac{1}{5^2}>\frac{1}{5.6}\) ; \(\frac{1}{6^2}>\frac{1}{6.7}\) ; ... ; \(\frac{1}{100^2}>\frac{1}{100.101}\)
=> \(A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\) (1)
Lại có: \(\frac{1}{5^2}< \frac{1}{4.5}\) ; \(\frac{1}{6^2}< \frac{1}{5.6}\) ; ... ; \(\frac{1}{100^2}< \frac{1}{99.100}\)
=> \(A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\) (2)
Từ (1) và (2) => \(\frac{1}{6}< A< \frac{1}{4}\)
Ta có:
\(\frac{1}{5^2}<\frac{1}{4.5}\)
\(\frac{1}{6^2}<\frac{1}{5.6}\)
\(\frac{1}{7^2}<\frac{1}{6.7}\)
\(...\)
\(\frac{1}{2007^2}<\frac{1}{2006.2007}\)
\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}<\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{2006.2007}\)
\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}<\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{2006}-\frac{1}{2007}\)
\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}<\frac{1}{4}-\frac{1}{2007}\)
Mà \(\frac{1}{4}-\frac{1}{2007}<\frac{1}{4}\)
\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}<\frac{1}{4}\)
\(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{6}{13}\)
\(\Rightarrow\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)
\(\Rightarrow\frac{6}{2x+1}=\frac{6}{13}\Rightarrow2x+1=13\Rightarrow x=6\)
mình giải hơi gọn có gì ko hiểu thì hỏi nha !
#)Giải :
Ta có : \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{2007.2008}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{2007}-\frac{1}{2008}=\frac{1}{5}-\frac{1}{2008}=\frac{2003}{10004}>\frac{1}{5}\)
\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}>\frac{1}{5}\)