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27 tháng 6 2017

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

\(=-\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{100^2}\right)\) ( do có 99 cặp số )

\(=-\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)...\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)\)

\(=-\dfrac{1}{2}.\dfrac{3}{2}.\dfrac{2}{3}.\dfrac{4}{3}...\dfrac{99}{100}.\dfrac{101}{100}\)

\(=-\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)

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

Vậy \(A< \dfrac{-1}{2}\)

27 tháng 6 2017

Đề sai rồi kìa bn

Đúg ra phải là 1/100^2 -1 chứ

31 tháng 12 2023

a: \(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\)

\(=-\dfrac{1}{10}\)

9<10

=>1/9>1/10

=>\(-\dfrac{1}{9}< -\dfrac{1}{10}\)

=>\(A>-\dfrac{1}{9}\)

b: \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{10}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{11}{10}\)

\(=\dfrac{-1}{10}\cdot\dfrac{11}{2}=\dfrac{-11}{20}\)

20<21

=>\(\dfrac{11}{20}>\dfrac{11}{21}\)

=>\(-\dfrac{11}{20}< -\dfrac{11}{21}\)

=>\(B< -\dfrac{11}{21}\)

20 tháng 9 2023

\(B=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{100^2}\right)\)

\(B=\left(\dfrac{2^2}{2^2}-\dfrac{1}{2^2}\right)\cdot\left(\dfrac{3^2}{3^2}-\dfrac{1}{3^2}\right)....\left(\dfrac{100^2}{100^2}-\dfrac{1}{100^2}\right)\)

\(B=\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}....\cdot\dfrac{100^2-1}{100^2}\)

\(B=\dfrac{\left(2+1\right)\left(2-1\right)}{2^2}\cdot\dfrac{\left(3+1\right)\left(3-1\right)}{3^2}\cdot...\cdot\dfrac{\left(100+1\right)\left(100-1\right)}{100^2}\)

\(B=\dfrac{1\cdot3}{2^2}\cdot\dfrac{2\cdot4}{3^2}\cdot\dfrac{3\cdot5}{4^2}\cdot...\cdot\dfrac{99\cdot101}{100^2}\)

\(B=\dfrac{1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot101}{2^2\cdot3^2\cdot4^2\cdot5^2\cdot....\cdot100^2}\)

\(B=\dfrac{1\cdot101}{2\cdot3\cdot4\cdot5\cdot...\cdot100}\)

\(B=\dfrac{101}{2\cdot3\cdot4\cdot5\cdot...\cdot100}\)

Mà: \(\dfrac{1}{2}=\dfrac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot100}\) 

Ta có: \(101< 3\cdot4\cdot5\cdot...\cdot100\)

\(\Rightarrow\dfrac{101}{2\cdot3\cdot4\cdot5\cdot...\cdot100}< \dfrac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot100}\)

\(\Rightarrow B< \dfrac{1}{2}\)     

14 tháng 12 2017

Sửa đề:

\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)....\left(\dfrac{1}{100^2}-1\right)\)

\(A=\left(\dfrac{1}{2^2}-\dfrac{2^2}{2^2}\right)\left(\dfrac{1}{3^2}-\dfrac{3^2}{3^2}\right)\left(\dfrac{1}{4^2}-\dfrac{4^2}{4^2}\right)....\left(\dfrac{1}{100^2}-\dfrac{100^2}{100^2}\right)\)

\(A=\dfrac{\left(1-2^2\right)}{2^2}.\dfrac{\left(1-3^2\right)}{3^2}.\dfrac{\left(1-4^2\right)}{4^2}....\dfrac{\left(1-100^2\right)}{100^2}\)

\(A=\dfrac{\left(1-2\right)\left(1+2\right)}{2^2}.\dfrac{\left(1-3\right)\left(1+3\right)}{3^2}.\dfrac{\left(1-4\right)\left(1+4\right)}{4^2}......\dfrac{\left(1-100\right)\left(1+100\right)}{100^2}\)

\(A=\dfrac{-3}{2^2}.\dfrac{-8}{3^2}.\dfrac{-15}{4^2}....\dfrac{-9999}{100^2}\)

Ta xét từ \(2\) đến \(100\) có: \(\dfrac{\left(100-2\right)}{1}+1=99\)

\(50\) là số lẻ nên tích trên là số âm

Hay \(-A=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{99.101}{100.100}\)

\(-A=\dfrac{1.3.2.4.3.5....99.101}{2.2.3.3.4.4.....100.100}\)

\(-A=\dfrac{1.2.3....99}{2.3.4....100}.\dfrac{3.4.5....101}{2.3.4....100}\)

\(-A=\dfrac{1}{100}.\dfrac{101}{2}=\dfrac{101}{200}\)

\(A=-\dfrac{101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)

14 tháng 12 2017

Tập hợp Q các số hữu tỉ

18 tháng 8 2017

\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)..............\left(\dfrac{1}{100^2}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right).............\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)

\(=\dfrac{-1}{2}.\dfrac{3}{2}.\dfrac{-2}{3}.\dfrac{4}{3}.............\dfrac{-99}{100}.\dfrac{101}{100}\)

\(=\dfrac{-\left(1.2.3....99\right)}{2.3......100}.\dfrac{3.4...101}{2.3....100}\)

\(=\dfrac{-1}{100}.\dfrac{101}{2}\)

\(=\dfrac{-101}{200}< \dfrac{-1}{2}\)

\(\Leftrightarrow A< \dfrac{-1}{2}\)

29 tháng 7 2017

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

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)...\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)

\(=\dfrac{-1}{2}.\dfrac{3}{2}.\dfrac{-2}{3}.\dfrac{4}{3}...\dfrac{-99}{100}.\dfrac{101}{100}\)

\(=\dfrac{-\left(1.2...99\right)}{2.3...100}.\dfrac{3.4...101}{2.3...100}=\dfrac{-1}{100}.\dfrac{101}{2}\)

\(=\dfrac{-101}{200}< \dfrac{-1}{2}\)

\(\Rightarrow A< \dfrac{-1}{2}\)

19 tháng 10 2023

Ta có:

\(A=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{10}-1\right)\)

\(A=-\dfrac{1}{2}\cdot-\dfrac{2}{3}-\dfrac{3}{4}\cdot...\cdot-\dfrac{9}{10}\)

\(A=\dfrac{-1\cdot-2\cdot-3\cdot...\cdot-9}{2\cdot3\cdot4\cdot...\cdot10}\)

\(A=-\dfrac{1}{10}\)

Mà: \(10>9\)

\(\Rightarrow\dfrac{1}{10}< \dfrac{1}{9}\)

\(\Rightarrow-\dfrac{1}{10}>-\dfrac{1}{9}\)

\(\Rightarrow A>-\dfrac{1}{9}\)

19 tháng 9 2023

\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{2020^2}-1\right)\)

\(B=\left(\dfrac{1}{2^2}-\dfrac{2^2}{2^2}\right)\left(\dfrac{1}{3^2}-\dfrac{3^2}{3^2}\right)....\left(\dfrac{1}{2020^2}-\dfrac{2020^2}{2020^2}\right)\)

\(B=\left(\dfrac{1-2^2}{2^2}\right)\left(\dfrac{1-3^2}{3^2}\right)...\left(\dfrac{1-2020^2}{2020^2}\right)\)

\(B=\dfrac{\left(1-2\right)\left(1+2\right)}{2^2}\cdot\dfrac{\left(1-3\right)\left(1+3\right)}{3^2}....\cdot\dfrac{\left(2020-1\right)\left(2020+1\right)}{2020^2}\) 

\(B=\dfrac{-1\cdot3}{2^2}\cdot\dfrac{-2\cdot4}{3^2}\cdot\dfrac{-3\cdot5}{4^2}\cdot....\cdot\dfrac{-2019\cdot2021}{2020}\)

\(B=\dfrac{-1\cdot-2\cdot-3\cdot...\cdot-2019}{2\cdot3\cdot4\cdot....\cdot2020}\)

\(B=\dfrac{-1\cdot-1\cdot-1\cdot....\cdot-1}{1}\)

\(B=-1\) (2019 số -1) 

Mà: \(-1< \dfrac{1}{2}\)

\(\Rightarrow B< \dfrac{1}{2}\)

19 tháng 9 2023

 \(\dfrac{1}{2^2}\)\(\dfrac{1}{3^2}\);...;\(\dfrac{1}{2020^2}\) < 1 ⇒ 0 > \(\dfrac{1}{2^2}\) - 1 > \(\dfrac{1}{3^2}\) - 1 >..> \(\dfrac{1}{2020^2}\) - 1

Xét dãy số 2; 3; 4;...; 2020 dãy số này có số số hạng là:

        (2020 - 2):1 + 1 = 2019 (số hạng)

Vậy B là tích của 2019 số âm nên B < 0 ⇒ B < \(\dfrac{1}{2}\)

 

 

 

 

 

23 tháng 9 2021

\(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{19}\right)\left(1-\dfrac{1}{20}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{18}{19}.\dfrac{19}{20}=\dfrac{1}{20}>\dfrac{1}{21}\)

27 tháng 8 2017

\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right).....\left(\dfrac{1}{100^2}-1\right)\)

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

\(A=\dfrac{-\left(1.3\right)}{2.2}.\dfrac{-\left(2.4\right)}{3.3}.\dfrac{-\left(3.5\right)}{4.4}......\dfrac{-\left(99.101\right)}{100.100}\)

Từ \(-1\) đến \(-99\) có: \(\left(99-1\right):1+1=99\)

Cộng thêm số \(-101\) tất cả có \(100\) số hạng

\(A=\dfrac{1.3.2.4.3.5....99.101}{2.2.3.3.4.4.....100.100}\)

\(A=\dfrac{1.2.3....99}{2.3.4....100}.\dfrac{3.4.5....101}{2.3.4....100}\)

\(A=\dfrac{1}{100}.\dfrac{101}{2}\)

\(A=\dfrac{101}{200}\)

\(\)

27 tháng 8 2017

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