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

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

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

đến đay bn tự tính nha

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

\(=9\left(\frac{1}{9.19}+\frac{1}{19.29}+...+\frac{1}{1999.2009}\right)=\frac{9}{10}\left(\frac{10}{9.19}+\frac{10}{19.29}+...+\frac{10}{1999.2009}\right)\)

\(=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{19}+\frac{1}{19}-\frac{1}{29}+...+\frac{1}{1999}-\frac{1}{2009}\right)=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{2009}\right)=\frac{9}{10}\cdot\frac{2000}{18081}=\frac{200}{2009}\)

Ta có: \(B=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)

\(B=9\left(\frac{1}{9.19}+\frac{1}{19.29}+...+\frac{1}{1999.2009}\right)\)

\(B=\frac{9}{10}\left(\frac{10}{9.19}+\frac{10}{19.29}+...+\frac{10}{1999.2009}\right)\)

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

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

\(B=\frac{9}{10}.\frac{2000}{18081}\)

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

Vậy \(B=\frac{200}{2009}\)

ta có\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)

\(=\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)

tách

\(B=\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)

\(2B=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)

\(2B-B=\frac{1}{2}-\frac{1}{1024}\)

thay vào B ta có 

\(\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)

\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{1024}=\frac{1}{1024}\)

\(A=\frac{1}{2}-\frac{1}{4}-\cdot\cdot\cdot-\frac{1}{1024}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\)

\(\Rightarrow2A=1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\)

\(\Rightarrow2A-A=\left(1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\right)-\left(\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\right)\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2^{10}}\)

\(\Rightarrow A=\frac{1}{2}+\frac{1}{2^{10}}\)

\(\Rightarrow A=\frac{2^9+1}{2^{10}}\)

\(\Rightarrow A=\frac{513}{1024}\)

\(\text{Ta có:}\frac{9}{9.19}+\frac{9}{19.29}+\frac{9}{29.39}+....+\frac{9}{2009.2019}\)

\(=\frac{9}{10}.\left(\frac{10}{9.19}+\frac{10}{19.29}+\frac{10}{29.39}+.....+\frac{10}{2009.2019}\right)\)

\(=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{2019}\right)\)

\(=\frac{9}{10}.\frac{670}{6057}\)

\(\frac{x+1}{x-1}=\frac{7}{3}\)

=> \(3.\left(x+1\right)=7.\left(x-1\right)\)

=> \(3x+3=7x-7\)

=> \(3x+10=7x\)

=> \(4x=10\)

=> \(x=\frac{10}{4}=\frac{5}{2}\)

Vậy \(x=\frac{5}{2}\)

\(\frac{x+1}{x-1}=\frac{7}{3}\)

\(\Rightarrow3\left(x+1\right)=7\left(x-1\right)\)

\(\Leftrightarrow3x+3=7x-7\)

\(\Leftrightarrow-4x=-10\)

\(\Leftrightarrow x=\frac{5}{2}\)

~~~!!!

\(\frac{x+1}{x-1}=\frac{7}{3}\)

\(\Leftrightarrow3x+3=7x-7\)

\(\Leftrightarrow3x=7x-7-3\)

\(\Leftrightarrow-4x=-10\)

\(\Rightarrow x=\frac{5}{2}\)

Phân số thứ 20 à , hơi khó đó

Nhưng kết quả là:\(\frac{1}{1599}\) 

Kết quả đúng là \(\frac{1}{1599}\)

mình làm rùi 

ủng hộ nha

toan lop 6 day a co ma lop 5 y ngu hay sao ma hoi