A>1/2

Xin lỗi mình đang bận để lúc khác mình sẽ giải chi tiết

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

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

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

\(A=-\left(\frac{1}{100}\right).\left(\frac{101}{2}\right)\)

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

Thằng điên, có cái đầu bài cx chép sai thì làm ăn đếch j

Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)

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

               \(=\frac{1.2....18.19}{2.3...19.20}\)

               \(=\frac{1}{20}>\frac{1}{21}\)

Vậy A > 1/21

ta có: \(M=-\frac{3}{2^2}.-\frac{8}{3^2}.-\frac{15}{4}...-\frac{9999}{100^2}\)

M có 99 thừa số âm

=>M<0

\(-M=\frac{3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}....\frac{99.101}{100.100}=>-M=\frac{\left(2.3.4...99\right)\left(3.4.5...101\right)}{\left(2.3.4...100\right)\left(2.3.4...100\right)}=\frac{101}{100.2}=\frac{101}{200}\)

\(\frac{101}{200}>\frac{100}{200}=\frac{1}{2}=>-M>\frac{1}{2}=>M

A= -(3/4 . 8/9 . 15/16 ... 9999/10000)

 =-( (3.8.15....9999)/(4.9.16...10000))

 = - (( 1.3.2.4.3.5...99.101)/(2.2.3.3.4.4...100.100))

 = -( ( 1.2.3.4...99).(3.4.5..101) / (2.3.4...100) . (2.3.4..100))

 = -101/200< -100/200 = -1/2

Vậy A < -1/2