Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt \(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{18.19}+\frac{2}{19.20}\)
\(A=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(A=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(A=2\left(1-\frac{1}{20}\right)\)
\(A=2.\frac{19}{20}=\frac{19}{10}\)
Vậy ...
=2.(\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+......+\(\frac{1}{19.20}\))
=2.( 1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+..........+\(\frac{1}{19}\)-\(\frac{1}{20}\))
=2.(1-\(\frac{1}{20}\))
=2.\(\frac{19}{20}\)
= \(\frac{19}{10}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\)
=1-1/20
=19/20
\(A=\frac{1^2.2^2.3^2...1000^2}{1.2^2.3^2.4^2...1000^2.1001}=\frac{1}{1001}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{99.100}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2.\left(1-\frac{1}{100}\right)\)
\(=2.\frac{99}{100}=\frac{99}{50}\)
=\(2\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{99.100}\right)\)
=\(2\left(1-\frac{1}{100}\right)\)
=\(2\cdot\frac{99}{100}=\frac{99}{50}\)
Đặt tổng trên là A , ta có :
\(\frac{A}{2}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(\frac{A}{2}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{98}-\frac{1}{99}\right)+\left(\frac{1}{99}-\frac{1}{100}\right)\)
\(\frac{A}{2}=\left(1-\frac{1}{100}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+...+\left(\frac{1}{98}-\frac{1}{98}\right)+\left(\frac{1}{99}-\frac{1}{99}\right)\)\(\frac{A}{2}=\frac{99}{100}\)
\(A=\frac{99}{100}.2\)
\(A=\frac{99}{50}\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....++\frac{1}{9}-\frac{1}{10}\)
\(B=1-\frac{1}{10}=\frac{9}{10}\)
\(C=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(C=1-\frac{1}{100}\)
\(C=\frac{99}{100}\)