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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\Rightarrow B< \dfrac{1}{2}\)     

\(A=-\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{2014^2}\right)\)

\(A=\dfrac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2012\cdot2014\right)\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2013\cdot2013\right)\left(2014\cdot2014\right)}\)

\(A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot2012\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2014\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)}\)

\(A=\dfrac{1\cdot2015}{2014\cdot2}=\dfrac{2015}{4028}\)

Vì \(\dfrac{2015}{4028}>-\dfrac{1}{2}\) nên A > B

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

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

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

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

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

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

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

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

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

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

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

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

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: \(10>9\)

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

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

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

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

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

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

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

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

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

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

\(B=-1\) (2019 số -1) 

Mà: \(-1< \dfrac{1}{2}\)

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

 \(\dfrac{1}{2^2}\); \(\dfrac{1}{3^2}\);...;\(\dfrac{1}{2020^2}\) < 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 < \(\dfrac{1}{2}\)

b) Vì \(\left|x+\dfrac{1}{1.3}\right| \ge0;\left|x+\dfrac{1}{3.5}\right|\ge0;...;\left|x+\dfrac{1}{97.99}\right|\ge0\)

\(\Rightarrow50x\ge0\Rightarrow x\ge0\)

Khi đó: \(\left|x+\dfrac{1}{1.3}\right|=x+\dfrac{1}{1.3};\left|x+\dfrac{1}{3.5}\right|=x+\dfrac{1}{3.5};...;\left|x+\dfrac{1}{97.99}\right|=x+\dfrac{1}{97.99}\left(1\right)\)

Thay (1) vào đề bài:

\(x+\dfrac{1}{1.3}+x+\dfrac{1}{3.5}+...+x+\dfrac{1}{97.99}=50x\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right)=50x\)

\(\Rightarrow49x+\left[\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\right]=50x\)

\(\Rightarrow49x+\dfrac{16}{99}=50x\)

\(\Rightarrow x=\dfrac{16}{99}\)

Vậy \(x=\dfrac{16}{99}.\)

\(A=4.\dfrac{25}{16}+25.\left[\dfrac{9}{16}:\dfrac{125}{64}\right]:\dfrac{-27}{8}\)

\(=\dfrac{25}{16}+25.\dfrac{36}{125}:\dfrac{-27}{8}=-\dfrac{137}{240}\left(1\right)\)

\(B=125.\left[\dfrac{1}{25}+\dfrac{1}{64}:8\right]-64.\dfrac{1}{64}\)

\(=125.\dfrac{89}{1600}:8-64.\dfrac{1}{64}=\dfrac{-67}{512}\left(2\right)\)

Vì (2) > (1) => B > A