\(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)\left(1-\dfrac{1}{15}\right)...\left(1-\dfrac{1}{780}\right)\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
TM
11 tháng 4 2017
\(A=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)......\left(1-\dfrac{1}{780}\right)\)
= \(\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.......\dfrac{779}{780}\)
= \(\dfrac{4}{6}.\dfrac{10}{12}.\dfrac{18}{20}.....\dfrac{1558}{1560}\)
= \(\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}......\dfrac{38.41}{39.40}\)
= \(\dfrac{1.4.2.5.3.6.....38.41}{2.3.3.4.4.5...39.40}\)
= \(\dfrac{\left(1.2.3....38\right)\left(4.5.6....41\right)}{\left(2.3.4....39\right)\left(3.4.5...40\right)}\)
= \(\dfrac{1}{39}.\dfrac{41}{3}\) = \(\dfrac{41}{117}\)
11 tháng 4 2017
\(B=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)
\(B=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}...\dfrac{779}{780}=\dfrac{4}{6}.\dfrac{10}{12}.\dfrac{18}{20}...\dfrac{1558}{1560}\)
\(B=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}.....\dfrac{38.41}{39.40}\)
\(B=\dfrac{\left(1.2.3.....38\right)\left(4.5.6.....41\right)}{\left(2.3.4.....39\right)\left(3.4.5.....40\right)}=\dfrac{1.41}{39.3}=\dfrac{41}{117}\)
BU
6 tháng 4 2022
\(\dfrac{\left(\dfrac{5}{30}+\dfrac{3}{30}+\dfrac{2}{30}\right):\left(\dfrac{5}{30}+\dfrac{3}{30}-\dfrac{2}{30}\right)}{\left(\dfrac{30}{60}-\dfrac{20}{60}+\dfrac{15}{60}-\dfrac{12}{60}\right):\left(\dfrac{3}{12}-\dfrac{2}{12}\right)}=\dfrac{\dfrac{1}{3}:\dfrac{1}{5}}{\dfrac{13}{60}:\dfrac{1}{12}}=\dfrac{\dfrac{1}{3}\times5}{\dfrac{13}{60}\times12}=\dfrac{\dfrac{5}{3}}{\dfrac{13}{5}}=\dfrac{25}{39}\)
30 tháng 11 2021
2: \(=\dfrac{203}{60}\cdot\dfrac{81}{1225}=\dfrac{783}{3500}\)
28 tháng 3 2018
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right).......\left(1-\dfrac{1}{10}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right).........\left(\dfrac{10}{10}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}......\dfrac{9}{10}\)
\(=\dfrac{1}{10}\)
8 tháng 4 2017
\(C=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)...\left(1-\dfrac{1}{210}\right)\)
\(=\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot\cdot\cdot\dfrac{209}{210}\)
14 tháng 3 2017
Ta có: \(A=\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{6}-1\right)\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{15}-1\right)\left(\dfrac{1}{21}-1\right)\left(\dfrac{1}{28}-1\right)\left(\dfrac{1}{36}-1\right)\)
\(=\dfrac{-2}{3}.\dfrac{-5}{6}.\dfrac{-9}{10}.\dfrac{-14}{15}.\dfrac{-20}{21}.\dfrac{-27}{28}.\dfrac{-35}{36}\)
\(=\dfrac{-2.\left(-5\right).3.\left(-3\right).2.\left(-7\right).\left(-4\right).5.\left(-3\right).9.5.\left(-7\right)}{3.2.3.2.5.3.5.3.7.4.7.4.9}\)
\(=\dfrac{-5}{3.4}=\dfrac{-5}{12}\)
Vậy \(A=\dfrac{-5}{12}.\)
AT
14 tháng 3 2017
\(C=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\)
\(2C=2\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\right)\)
\(2C=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{2015}}\)
\(2C-C=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2015}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2016}}\right)\)
\(C=2-\dfrac{1}{2^{2016}}\)
\(\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{6}\right).\left(1-\dfrac{1}{10}\right).\left(1-\dfrac{1}{15}\right)...\left(1-\dfrac{1}{780}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}...\dfrac{779}{780}\)
\(=\dfrac{4}{6}.\dfrac{10}{12}.\dfrac{18}{20}...\dfrac{1558}{1560}\)
\(=\dfrac{4.10.18...1558}{6.12.20...1560}\)
\(=\dfrac{41}{39}.3\)
\(=\dfrac{41}{11}\)