Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{2013^2-1}{2013^2}.\frac{2014^2-1}{2014^2}\)
\(A=\frac{1.3.2.4.3.5....2012.2014.2013.2015}{2^2.3^2.4^2...2013^2.2014^2}\)
\(A=\frac{\left(1.2.3...2012.2013\right).\left(3.4.5...2014.2015\right)}{\left(2.3.4...2013.2014\right).\left(2.3.4...2013.2014\right)}\)(nhóm từng số ở trước và sau vào 2 nhóm khác nhau)
\(A=\frac{3.2015}{2014.2}\)
\(A=\frac{6045}{4028}\)
\(A=\frac{6045}{4028}\),nha bạn ,chúc bạn hok tốt ,love bạn nhìu ,cách làm giống như Monozono Nanami nha
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1000}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{999}{1000}\)
\(=\frac{1}{1000}\)
chúc
bn
hk
tốt
\(C=\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.........\frac{2499}{2500}\)
\(=\frac{2.4}{3^2}.\frac{3.5}{4^2}.\frac{4.6}{5^2}......\frac{49.51}{50^2}\)
\(=\frac{2.3.4....49}{3.4.5....50}.\frac{4.5.6....51}{3.4.5....50}\)
\(=\frac{1}{25}.17=\frac{17}{25}\)
\(a)\) \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{1000}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{999}{1000}\)
\(A=\frac{1.2.3.....999}{2.3.4.....1000}\)
\(A=\frac{1}{1000}.\frac{2.3.4.....999}{2.3.4.....999}\)
\(A=\frac{1}{1000}\)
Vậy \(A=\frac{1}{1000}\)
b) \(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}+\frac{1}{2015}\)
\(B=1-\frac{1}{2015}\)
\(B=\frac{2014}{2015}\)
a) \(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)
\(=\frac{1}{100}\)
b)\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(=1-\frac{1}{2015}\)
\(=\frac{2014}{2015}\)
còn lại tự giải nha gần giống như phần b thôi cũng thú vị.
ủng hộ nha
A = \(\left(1-\frac{1}{2^2}\right)\)x \(\left(1-\frac{1}{3^2}\right)\)x \(\left(1-\frac{1}{4^2}\right)\)x . . . x \(\left(1-\frac{1}{100^2}\right)\)
A=\(\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{100^2-1}{100^2}=\frac{\left(2-1\right)\left(2+1\right)}{2^2}.\frac{\left(3-1\right)\left(3+1\right)}{3^2}...\frac{\left(100-1\right)\left(100+1\right)}{100^2}\)
A=\(\frac{1.3.2.4.3.5....98.100.99.101}{2^2.3^2....100^2}=\frac{101}{2.100}>\frac{1}{2}\)
Ta có :
\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(A=\left(\frac{1}{2^2}-\frac{2^2}{2^2}\right).\left(\frac{1}{3^2}-\frac{3^2}{3^2}\right).\left(\frac{1}{4^2}-\frac{4^2}{4^2}\right)...\left(\frac{1}{100^2}-\frac{100^2}{100^2}\right)\)
\(A=\left(-\frac{3}{2^2}\right).\left(-\frac{8}{3^2}\right).\left(-\frac{15}{4^2}\right)...\left(-\frac{99}{100^2}\right)\)
\(A=-\left(\frac{1.3.2.4.3.5.....9.11}{2.2.3.3.4.4....10.10}\right)\)
\(A=-\left(\frac{1.2.3....9}{2.3.4....10}.\frac{3.4.5.....11}{2.3.4....10}\right)\)
\(A=-\left(\frac{1}{10}.\frac{11}{2}\right)=-\frac{11}{20}=\frac{-11}{20}\)
Lại có : \(\frac{-1}{2}=\frac{-1.10}{2.10}=\frac{-10}{20}\)
Vì \(-11< -10\)nên \(\frac{-11}{20}< \frac{-10}{20}\)hay \(A< \frac{-1}{2}\)
Mk mới học bài này xong,nhớ ủng hộ mk nha !!! ^_^