Ta có :

\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{19}\right).\left(1-\frac{1}{20}\right)\)

\(=\)\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{18}{19}.\frac{19}{20}\)

\(=\)\(\frac{1.2.3.....18.19}{2.3.4.....19.20}\)

\(=\)\(\frac{1}{20}\)

Vì \(\frac{1}{20}>\frac{1}{21}\)nên \(A>\frac{1}{21}\)

Vậy \(A>\frac{1}{21}\)

bít kq nhưng ko thích giải

cậu ko giúp cậu ấy thì thôi đừng bảo như thế

\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{81}\right)\left(1-\frac{1}{100}\right)\)

\(B=\frac{3}{4}\cdot\frac{8}{9}\cdot...\cdot\frac{80}{81}\cdot\frac{99}{100}\)

\(B=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot...\cdot\frac{8.10}{9.9}\cdot\frac{9.11}{10.10}\)

\(B=\frac{\left(1\cdot2\cdot...\cdot8\cdot9\right).\left(3\cdot4\cdot...\cdot10\cdot11\right)}{\left(2\cdot3\cdot..\cdot9\cdot10\right).\left(2\cdot3\cdot...\cdot9\cdot10\right)}\)

\(B=\frac{1\cdot2\cdot...\cdot8\cdot9}{2\cdot3\cdot...\cdot9\cdot10}\cdot\frac{3\cdot4\cdot...\cdot10\cdot11}{2\cdot3\cdot...\cdot9\cdot10}\)

\(B=\frac{1}{10}\cdot\frac{11}{2}=\frac{11}{20}\)

Vì 20 < 21 nên 11/20 > 11/21

Vậy ..... 

bạn vào link này nè:https://olm.vn/hoi-dap/question/980572.html

S > 1/3

ta thấy \(\frac{1}{20}\)<\(\frac{1}{3}\)

thì \(\frac{1}{20}\)+...+\(\frac{1}{29}\)<\(\frac{1}{20}\)+...+\(\frac{1}{20}\)<\(\frac{1}{3}\)

vậy \(\frac{1}{20}\)+...+\(\frac{1}{29}\)<\(\frac{1}{3}\)

Ta có: \(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

\(\frac{1}{4^2}<\frac{1}{3\cdot4}=\frac13-\frac14\)

...

\(\frac{1}{20^2}<\frac{1}{19\cdot20}=\frac{1}{19}-\frac{1}{20}\)

Do đó: \(\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{20^2}<\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{19}-\frac{1}{20}<\frac12\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{20^2}<\frac14+\frac12\)

=>\(A<\frac34\)

a: Ta có: \(A=\frac59+\left(-\frac57\right)+\left(-\frac{20}{48}\right)+\frac{8}{12}+\left(-\frac{21}{48}\right)\)

\(=\frac59-\frac57-\frac{41}{48}+\frac{32}{48}\)

\(=\frac{35-45}{63}-\frac{9}{48}=\frac{-10}{63}-\frac{3}{16}=\frac{-160-189}{63\cdot16}=\frac{-349}{1008}\)

b: \(B=\left(-\frac59\right)+\frac{8}{15}+\left(-\frac{2}{11}\right)+\left(\frac{4}{-9}\right)+\frac{2}{45}\)

\(=\left(-\frac59-\frac49\right)+\frac{8}{15}+\frac{2}{45}-\frac{2}{11}\)

\(=-1-\frac{2}{11}+\frac{24}{45}+\frac{2}{45}=-\frac{13}{11}+\frac{26}{45}=\frac{-13\cdot45+26\cdot11}{11\cdot45}=\frac{-299}{495}\)

c: \(\frac{1}{11}>\frac{1}{20};\frac{1}{12}>\frac{1}{20};\ldots;\frac{1}{20}=\frac{1}{20}\)

Do đó: \(\frac{1}{11}+\frac{1}{12}+\cdots+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+\cdots+\frac{1}{20}\)

=>S>10/20

=>S>1/2