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\(E=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{49.49}\)
Ta có \(\frac{1}{2.2}>\frac{1}{2.3}\)
\(\frac{1}{3.3}>\frac{1}{3.4}\)
...
\(\frac{1}{49.49}>\frac{1}{49.50}\)
=> \(E=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{49.49}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=\frac{1}{2}-\frac{1}{50}=\frac{24}{50}=\frac{12}{25}=F\)
=> E > F
đặt \(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+\frac{1}{7.7}+\frac{1}{8.8}=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
\(A<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=1-\frac{1}{8}<1\)
vậy \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+\frac{1}{7.7}+\frac{1}{8.8}<1\)
\(A=\dfrac{1}{2\times2}+\dfrac{1}{3\times3}+...+\dfrac{1}{100\times100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=1-\dfrac{1}{100}=\dfrac{99}{100}\)
Quy đồng 99/100 với 3/4, ta có:
\(\dfrac{99}{100}=\dfrac{396}{400};\dfrac{3}{4}=\dfrac{300}{400}\)
So sánh A với 3/4: \(\dfrac{99}{100}>\dfrac{3}{4}\left(\dfrac{396}{400}>\dfrac{300}{400}\right)\)
Ta có:
\(\frac{1}{2x2}<\frac{1}{1.2}\)
\(\frac{1}{3x3}<\frac{1}{2.3}\)
\(...\)
\(\frac{1}{2015x2015}<\frac{1}{2014x2015}\)
\(\Rightarrow\frac{1}{2x2}+\frac{1}{3x3}+...+\frac{1}{2015x2015}<\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{2014x2015}\)
\(\Rightarrow\frac{1}{2x2}+\frac{1}{3x3}+...+\frac{1}{2015x2015}<1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(\Rightarrow\frac{1}{2x2}+\frac{1}{3x3}+...+\frac{1}{2015x2015}<1-\frac{1}{2015}<1\)
\(\Rightarrow\)Đpcm
a, \(A=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{2011\cdot2011}\)
có :
\(\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
\(\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)
...
\(\frac{1}{2011\cdot2011}< \frac{1}{2010\cdot2011}\)
nên :
\(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2010\cdot2011}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(\Rightarrow A< 1-\frac{1}{2011}\)
\(\Rightarrow A< \frac{2010}{2011}< 1\)
b, \(A=\frac{2010}{2011}=1-\frac{1}{2011}\)
\(\frac{3}{4}=1-\frac{1}{4}\)
\(\frac{1}{4}>\frac{1}{2011}\)
nên :
\(A>\frac{3}{4}\)
\(A>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)
\(A>\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2015-2014}{2014.2015}\)
\(A>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(A>1-\frac{1}{2015}\)
Mà \(\frac{1}{2015}< \frac{1}{4}\Rightarrow1-\frac{1}{2015}>1-\frac{1}{4}=\frac{3}{4}\Rightarrow A>\frac{3}{4}\)
bài khó quá về mà hỏi đảng