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3 tháng 8 2016

\(A=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{99!}+\frac{1}{100!}\)

\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(A< 1+1-\frac{1}{100}\)

\(A< 2-\frac{1}{100}< 2\)

4 tháng 8 2016

\(A=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{99!}+\frac{1}{100!}\)

\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(A< 1+1-\frac{1}{100}\)

\(A< 2-\frac{1}{100}< 2\)

7 tháng 3 2017

A<2 k mk nha

19 tháng 4 2017

A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

A < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

A < \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

A < 1 - \(\frac{1.}{100}\)

A < \(\frac{99}{100}< \frac{199}{100}\)

=> A < \(\frac{199}{100}\)

b,

S = \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{99}{10^2}\)

S = \(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{9.11}{10.10}\)

S = \(\frac{1.3.2.4.3.5.4.6.5.7...9.11}{2.2.3.3.4.4...10.10}\)

S = \(\frac{1.2.3^2.4^2.5^2...9^2.10.11}{2^2.3^3.4^2...10^2}\)

S = \(\frac{1.11}{2.10}\)

S = \(\frac{11}{20}\)

14 tháng 6 2017

Đặt \(B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

+ Xét : \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)

...

\(\frac{1}{100^2}< \frac{1}{99\cdot100}\)

\(\Rightarrow B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}< 2\)

\(\Leftrightarrow A< B< 2\left(đpcm\right)\)

8 tháng 6 2018

Ta có :

\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right)....\left(\frac{1}{100^2}-1\right)\)

\(A< \left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)...\left(\frac{1}{100}-1\right)\)

           \(\Rightarrow A< \left(\frac{-1}{2}\right).\left(\frac{-2}{3}\right)....\left(\frac{-99}{100}\right)\)

\(\Rightarrow A< -\left(\frac{1}{2}.\frac{2}{3}...\frac{99}{100}\right)\)

\(A< -\left(\frac{1.2....99}{2.3...100}\right)=\frac{-1}{100}\)

\(\)Mà \(\frac{-1}{100}>\frac{-1}{2}\)

\(\Rightarrow A>\frac{-1}{2}\)

8 tháng 6 2018

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Tham khảo ở link này nha bạn.

12 tháng 5 2019

câu hỏi của Lê Vũ Anh Thư nhé, vừa đc đăng lên

chúc hok tốt!

13 tháng 5 2019

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)

\(\Rightarrow A=\left(\frac{1}{2^2}-\frac{4}{2^2}\right)\left(\frac{1}{3^2}-\frac{9}{3^2}\right)...\left(\frac{1}{100^2}-\frac{10000}{100^2}\right)\)

\(\Rightarrow A=\frac{-3}{2^2}.\frac{-8}{3^2}...\frac{-9999}{100^2}\)

\(\Rightarrow A=-\frac{3}{2^2}.\frac{8}{3^2}...\frac{9999}{100^2}\)

\(\Rightarrow A=-\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{99.101}{100.100}\)

\(\Rightarrow A=-\frac{\left(1.2...99\right)\left(3.4...101\right)}{\left(2.3...100\right)\left(2.3...100\right)}\)

\(\Rightarrow A=-\frac{101}{100.2}=\frac{-101}{200}< \frac{-100}{200}=\frac{-1}{2}\)

Vậy \(A< \frac{-1}{2}\)

3 tháng 5 2017

Ta có: \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)

\(=\left(-\frac{1.3}{2.2}\right).\left(-\frac{2.4}{3.3}\right)...\left(-\frac{99.101}{100.100}\right)\)

\(=-\frac{1}{2}.\frac{101}{100}=-\frac{101}{200}< -\frac{100}{200}=-\frac{1}{2}\)

Vậy \(A< -\frac{1}{2}\)

29 tháng 5 2017

\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)...\left(\frac{1}{10000}-1\right)\)

\(=\frac{-3}{4}\cdot\frac{-8}{9}\cdot\frac{-15}{16}\cdot...\cdot\frac{-9999}{10000}\)

\(=\frac{-1\cdot3}{2\cdot2}\cdot\frac{-2\cdot4}{3\cdot3}\cdot...\cdot\frac{-99\cdot111}{100.100}\)

\(=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot...\cdot\frac{99\cdot111}{100\cdot100}\)

\(=\frac{\left(1\cdot2\cdot3\cdot4\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot6\cdot...\cdot111\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot100\right)^2}\)

\(=\frac{101}{2\cdot100}\)

\(=\frac{101}{200}>\frac{1}{2}\)