\(\dfrac{1}{19}+\dfrac{9}{19\cdot29}+...+\dfrac{9}{1999\cdot2009}\)

\(=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{10}{19\cdot29}+...+\dfrac{10}{1999\cdot2009}\right)\)

\(=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{1}{19}-\dfrac{1}{2009}\right)\)

\(=\dfrac{1}{19}+\dfrac{1791}{38171}=\dfrac{200}{2009}\)

\(A=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+....+\frac{9}{1999.2009}\)

\(A=\frac{1}{19}+\left(\frac{9}{19.29}+\frac{9}{29.39}+.....+\frac{9}{1999.2009}\right)\)

\(A=\frac{1}{19}+\frac{9}{10}\left(\frac{10}{19.29}+\frac{10}{29.39}+....+\frac{10}{1999.2009}\right)\)

\(A=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+.....+\frac{1}{1999}-\frac{1}{2009}\right)\)

\(A=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)\)

\(A=\frac{1}{19}+\frac{9}{10}.\frac{1990}{38171}\)

\(A=\frac{1}{19}+\frac{1791}{38171}\)

\(A=\frac{200}{2009}\)

A = 200/2009

đúng 100%, mk thi r` nhưng làm biếng giải

mk đồng ý với bài của Asuka Kurashina

bài của bn ấy đúng 99%

còn 1% là mk ko biết

chúc bn học giỏi!

thanks

\(A=\frac{1}{19}+\frac{9}{10}\left(\frac{10}{19.29}+\frac{10}{29.39}+...+\frac{10}{1999.2009}\right)\)

\(A=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)

\(A=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{9}-\frac{1}{2009}\right)\)

\(A=\frac{1}{19}+\frac{9}{10}\left(\frac{2000}{18081}\right)\)

\(A=\frac{1}{19}+\frac{200}{2009}\)

\(A=\frac{5809}{38171}\)

MK ko chắc nhé =v ( mấy bước quy đồng lằng nhằng ko làm âu )

\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)

\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)=\frac{1}{19}+\frac{9}{10}\cdot\frac{1990}{38171}=\frac{1}{19}+\frac{1791}{38171}=\frac{200}{2009}\)

\(\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)

\(=\frac{1}{19}+\frac{9}{10}\left(\frac{10}{19.29}+\frac{10}{29.39}+...+\frac{10}{1999.2009}\right)\)

\(=\frac{1}{19}+\frac{9}{10}.\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)

\(=\frac{1}{19}+\frac{9}{10}.\left(\frac{1}{19}-\frac{1}{2009}\right)\)

b tự làm nốt nhé

\(\frac{1}{9.19}+\frac{1}{19.29}+\frac{1}{29.39}+...+\frac{1}{1999.2009}\)

\(=\frac{1}{10}\times\left(\frac{10}{9.19}+\frac{10}{19.29}+\frac{10}{29.39}+...+\frac{10}{1999.2009}\right)\)

\(=\frac{1}{10}\times\left(\frac{1}{9}-\frac{1}{19}+\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)

\(=\frac{1}{10}\times\left(\frac{1}{9}-\frac{1}{2009}\right)\)

\(=\frac{1}{10}\times\frac{2000}{18081}\)

\(=\frac{200}{18081}\)

_Chúc bạn học tốt_

Ta có:

\(\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)

\(=\frac{1}{19}+9\left(\frac{1}{19.29}+\frac{1}{29.39}+...+\frac{1}{1999.2009}\right)\)

\(=\frac{1}{19}+\frac{9}{10}\left(\frac{10}{19.29}+\frac{10}{29.39}+...+\frac{10}{1999.2009}\right)\)(Đây là dạng tổng đặc biệt bn nha)

\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)

\(=\frac{1}{9}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)\)

\(=\frac{1}{9}+\frac{9}{10}.\frac{1990}{38171}\)

\(=\frac{1}{9}+\frac{1791}{38171}\)

\(=0,1580...\approx0,16\)

\(A=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)

\(=\frac{1}{19}+\frac{9}{10}\left(\frac{10}{19.29}+\frac{10}{29.39}+...+\frac{10}{1999.2009}\right)\)

\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)

\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)\)

\(=\frac{1}{19}+\frac{1791}{38171}\)

\(=\frac{200}{2009}\)

Vậy \(A=\frac{200}{2009}\)

Ta có:

\(A=\dfrac{1}{19}+\dfrac{9}{19.29}+\dfrac{9}{29.39}+...+\) \(\dfrac{9}{1999.2009}\)

\(=\dfrac{1}{19}+\) \(\left(\dfrac{9}{19.29}+\dfrac{9}{29.39}+...+\dfrac{9}{1999.2009}\right)\)

\(=\dfrac{1}{19}\) \(+\) \(\dfrac{9}{10}\left(\dfrac{10}{19.29}+\dfrac{10}{29.39}+...+\dfrac{10}{1999.2009}\right)\)

\(=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{1}{19}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{39}+...+\dfrac{1}{1999}-\dfrac{1}{2009}\right)\)

\(=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{1}{19}-\dfrac{1}{2009}\right)=\dfrac{200}{2009}\)

Vậy \(A=\dfrac{200}{2009}\)

Ta có : \(A=\dfrac{1}{19}+\dfrac{9}{19.29}+\dfrac{9}{29.39}+...+\dfrac{9}{1999.2009}.\)\(\Rightarrow A=\dfrac{1}{19}+\left(\dfrac{9}{19.29}+\dfrac{9}{29.39}+...+\dfrac{9}{1999.2009}\right)\)\(\Rightarrow A=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{10}{19.29}+\dfrac{10}{29.39}+..+\dfrac{10}{1999}+\dfrac{10}{2009}\right)\)\(\Rightarrow A=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{1}{19}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{39}+...+\dfrac{1}{1999}-\dfrac{1}{2009}\right)\)\(\Rightarrow A=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{1}{19}-\dfrac{1}{2009}\right)\)

\(\Rightarrow A=\dfrac{1}{19}+\dfrac{9}{10}.\dfrac{1990}{38171}\)

\(\Rightarrow A=\dfrac{1}{19}+\dfrac{1791}{38171}\)

\(\Rightarrow A=\dfrac{200}{2009}\)

Vậy \(A=\dfrac{200}{2009}\)