Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
\(B=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{81}\right).\left(1-\frac{1}{100}\right)\)
\(B=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{80}{81}.\frac{99}{100}\)
\(B=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{8.10}{9.9}.\frac{9.11}{10.10}\)
\(B=\frac{1.2.3...8.9}{2.3.4...9.10}.\frac{3.4.5...10.11}{2.3.4...9.10}\)
\(B=\frac{1}{10}.\frac{11}{2}\)
\(B=\frac{11}{20}>\frac{11}{21}\)
Ta có :
\(M=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{99}{100}=\frac{3.8.15.....99}{4.9.16.....100}=\frac{1.3.2.4.3.5.....9.11}{2.2.3.3.4.4.....10.10}\)\(=\frac{1.2.3...9}{2.3...10}.\frac{3.4...11}{2.3...10}=\frac{1}{10}.\frac{11}{2}=\frac{11}{20}< \frac{11}{19}\)
ta có M = (1- 1/4) (1- 1/9)... ( 1- 1/100)
= 3/2^2.8/3^2 ... 99/10^2
= 1.3/2^2 . 2.4/3^2 ... 9.11/10^ 2
= 1.2.3...9/ 2.3.4...10 . 3.4.5... 11/ 2.3.4... 10
= 1/10 . 11/2 = 11/20 < 11/19
Vậy M < 11/19
\(A=\left(\frac{1-2^2}{2^2}\right)\left(\frac{1-3^2}{3^2}\right)....\left(\frac{1-10^2}{10^2}\right)\)
\(A=\frac{\left(1+2\right)\left(1-2\right)}{2^2}.\frac{\left(1-3\right)\left(1+3\right)}{3^2}.......\frac{\left(1-10\right)\left(1+10\right)}{10^2}\)
\(A=\frac{3.\left(-1\right)}{2^2}.\frac{\left(-2\right).4}{3^2}........\frac{\left(-9\right).11}{10^2}=-\left(\frac{1.3}{2^2}.\frac{2.4}{3^2}....\frac{9.11}{10^2}\right)\)
\(=-\left(\frac{1.2....9}{2.3....10}.\frac{3.4....11}{2.3.4...10}\right)=-\left(\frac{1}{10}.\frac{11}{2}\right)=\frac{-11}{20}\)
\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)....\left(\frac{1}{81}-1\right)\left(\frac{1}{100}-1\right)\)
\(=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}....\frac{-80}{81}.\frac{-99}{100}\)
\(=\left[\left(-1\right).\left(-1\right)...\left(-1\right)\left(9\text{số (-1)}\right)\right].\frac{3}{4}.\frac{8}{9}....\frac{99}{100}\)
\(=\left(-1\right).\frac{1.3}{2.2}.\frac{2.4}{3.3}....\frac{9.11}{10.10}\)
\(=-\frac{1.11}{2.10}=-\frac{11}{10}\)
\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{81}-1\right)\left(\frac{1}{100}-1\right)\)
\(=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}.....\frac{-80}{81}.\frac{-99}{100}\)
\(=\left[\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right)\right].\frac{3}{4}.\frac{8}{9}.....\frac{99}{100}\)
\(=\left(-1\right).\frac{1.3}{2.2}.\frac{2.4}{3.3}....\frac{9.11}{10.10}\)
\(=-\frac{1.11}{2.10}=-\frac{11}{10}\)
A>1/2
Xin lỗi mình đang bận để lúc khác mình sẽ giải chi tiết
