1<1/3+1/8+1/15+1/24+....+1/360>

KO BIẾT ĐÚNG HAY KO NHÉ BẠN 

\(1\frac{1}{3}\)là hỗn số mà

\(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}.1\frac{1}{24}...1\frac{1}{360}\)

\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}...\frac{361}{360}\)

\(=\frac{2^2}{3}.\frac{3^2}{8}.\frac{4^2}{15}.\frac{5^2}{24}...\frac{19^2}{360}\)

\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}...\frac{19.19}{18.20}\)

\(=\left(\frac{2.3.4.5...19}{1.2.3.4...18}\right).\left(\frac{2.3.4.5...19}{3.4.5.6...20}\right)\)

\(=19.\frac{1}{10}\)

\(=\frac{19}{10}\)

\(A=1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}.1\frac{1}{24}.....1\frac{1}{360}\)

\(A=1+\left(\frac{1}{3}.\frac{1}{8}.\frac{1}{15}.\frac{1}{24}.....\frac{1}{360}\right)\)

Nếu đúng thì tk nha

\(1\dfrac{1}{3}.1\dfrac{1}{8}.1\dfrac{1}{15}......1\dfrac{1}{99}\)

\(=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.....\dfrac{100}{99}\)

\(=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.....\dfrac{10.10}{9.11}\)

\(=\dfrac{2.2.3.3.4.4.....10.10}{1.3.2.4.3.5.....9.11}\) ( Bước này bạn bỏ đi cũng được )

\(=\dfrac{\left(2.3.4.....10\right).\left(2.3.4.....10\right)}{\left(1.2.3.....9\right).\left(3.4.5.....11\right)}\)

\(=\dfrac{\left(2.3.4.....9\right).10.2.\left(3.4.5.....10\right)}{1.\left(2.3.4.....9\right).\left(3.4.5.....10\right).11}\)

\(=\dfrac{10.2}{1.11}=\dfrac{20}{11}=1\dfrac{9}{11}\)

\(1\dfrac{1}{3}.1\dfrac{1}{8}.1\dfrac{1}{15}.....1\dfrac{1}{99}\)

\(=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.....\dfrac{100}{99}\)

\(=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.....\dfrac{10.10}{9.11}\)

\(=\dfrac{2.2.3.3.4.4.....10.10}{1.3.2.4.3.5....9.11}\)

\(=\dfrac{2.3.4....10}{1.2.3....9}.\dfrac{2.3.4...10}{3.4.5....11}\)

\(=10.\dfrac{2}{11}=\dfrac{20}{11}\)

\(1\dfrac{1}{3}.1\dfrac{1}{8}.1\dfrac{1}{15}.....1\dfrac{1}{99}\)

\(=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.....\dfrac{100}{99}\)

\(=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.....\dfrac{10.10}{9.11}\)

\(=\dfrac{2.3.4.....10}{1.2.3.....9}.\dfrac{2.3.4.....10}{3.4.5.....11}\)

\(=10.\dfrac{2}{11}\)

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

1 1/9 x 1 1/10 x 1 1/11 x ... x 1 1/2011

=10/9 x 11/10 x 12/11 x ... x 2012/2011

khử

còn 2012/9

=\(\frac{10}{9}\)x\(\frac{11}{10}\)x\(\frac{12}{11}\)x.........x\(\frac{2012}{2011}\)

=\(\frac{2012}{9}\)

\(1\frac{1}{9}.1\frac{1}{10}.1\frac{1}{11}...1\frac{1}{2011}\)

= \(\frac{10}{9}.\frac{11}{10}.\frac{12}{11}...\frac{2012}{2011}\)

= \(\frac{2012}{9}\)

Bg

\(1\frac{1}{38}.1\frac{1}{39}.1\frac{1}{40}\)...\(.1\frac{1}{2013}\)

= \(\frac{39}{38}.\frac{40}{39}.\frac{41}{40}.\)...\(.\frac{2014}{2013}\)

= \(\frac{39.40.41.....2014}{38.39.40......2013}\)(39 trên, 39 dưới, 40 trên, 40 dưới,... 2013 trên, 2013 dưới, chịt tiêu hết)

= \(\frac{2014}{38}\)

= 53

cảm ơn bẹn nhe

\(1\frac{1}{38}.1\frac{1}{39}.1\frac{1}{40}...1\frac{1}{2013}=\frac{39}{38}.\frac{40}{39}.\frac{41}{40}...\frac{2014}{2013}=\frac{2014}{38}=53\)

Do thấy họ của bạn giống họ của tui nên giúp

\(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2016}\right)\left(1-\dfrac{1}{2017}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}...\dfrac{2015}{2016}.\dfrac{2016}{2017}=\dfrac{1}{2017}\)

Giải:

\(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2016}\right).\left(1-\dfrac{1}{2017}\right)\)

\(\Leftrightarrow A=\dfrac{1}{2}.\dfrac{2}{3}...\dfrac{2015}{2016}.\dfrac{2016}{2017}\)

\(\Leftrightarrow A=\dfrac{1.2...201.2016}{2.3...2016.2017}\)

\(\Leftrightarrow A=\dfrac{1.2.3...2015.2016}{2017.2.3...2015.2016.}\)

Rút gọ cả tử và mẫu với 2.3...2015.2016, ta được:

\(A=\dfrac{1}{2017}\)

Vậy \(A=\dfrac{1}{2017}\).

Chúc bạn học tốt!

Giải:

\(A=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2015}{2016}\cdot\dfrac{2016}{2017}\)

\(A=\dfrac{1}{2017}\)

Vậy, A= \(\dfrac{1}{2017}\)

Ta có \(1\frac{1}{3}=\frac{2^2}{3};1\frac{1}{8}=\frac{3^2}{8};.....\)

Nên thừa số thứ 98 là : \(1\frac{1}{9800}=\frac{99^2}{9800}\)

Ta có \(\frac{2^2}{3}.\frac{3^2}{8}......\frac{99^2}{9800}=\frac{2.2}{1.3}.\frac{3.3}{2.4}....\frac{99.99}{98.100}=\frac{2.2.3.3.....99.99}{1.3.2.4....98.100}\)

\(=\frac{\left(2.3.4...99\right).\left(2.3.4....99\right)}{\left(1.2.3....98\right).\left(3.4.5...100\right)}=\frac{99.2}{1.100}=\frac{198}{100}=\frac{99}{50}\)

giúp với !