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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
\(A=\dfrac{1}{2\times2}+\dfrac{1}{3\times3}+...+\dfrac{1}{100\times100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=1-\dfrac{1}{100}=\dfrac{99}{100}\)
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: \(\dfrac{99}{100}>\dfrac{3}{4}\left(\dfrac{396}{400}>\dfrac{300}{400}\right)\)
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
\(A>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)
\(A>\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2015-2014}{2014.2015}\)
\(A>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(A>1-\frac{1}{2015}\)
Mà \(\frac{1}{2015}< \frac{1}{4}\Rightarrow1-\frac{1}{2015}>1-\frac{1}{4}=\frac{3}{4}\Rightarrow A>\frac{3}{4}\)
Chứng minh:
C=\(\dfrac{1}{2x2}\)+\(\dfrac{1}{3x3}\)+\(\dfrac{1}{4x4}\)+.....+\(\dfrac{1}{100x100}\)<1
\(C=\dfrac{1}{2\times2}+\dfrac{1}{3\times3}+\dfrac{1}{4\times4}+...+\dfrac{1}{100\times100}\\ C< \dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{99\times100}\\ C< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ C< 1-\dfrac{1}{100}=\dfrac{99}{100}< 1\)
bài khó quá về mà hỏi đảng