Đặt tên bthuc là A

\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{19.21}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\)

\(2A=1-\frac{1}{21}=\frac{20}{21}\)

=>\(A=\frac{20}{21}:2=\frac{10}{21}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{17.19}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{17}-\frac{1}{19}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{19}\right)=\frac{1}{2}.\left(\frac{18}{19}\right)\)

\(=\frac{9}{19}\)

sr nhìn nhầm đề bài

Ta có:\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)=\frac{1}{2}\left(1-\frac{1}{21}\right)=\frac{1}{2}.\frac{20}{21}=\frac{10}{21}\)

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\)\(+...+\frac{1}{19.21}\)

=\(\frac{2}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\right)\)

=\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{19.21}\right)\)

=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{19}-\frac{1}{21}\right)\)

=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{21}\right)\)

=\(\frac{1}{2}.\frac{20}{21}\)

=\(\frac{20}{42}=\frac{10}{21}\)

Đặt :

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\)

\(\Leftrightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{19.21}\)

\(\Leftrightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{21}\)

\(\Leftrightarrow2A=1-\frac{1}{21}\)

\(\Leftrightarrow2A=\frac{20}{21}\)

\(\Leftrightarrow A=\frac{10}{21}\)

\(I=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2n+1\right).\left(2n+3\right)}\)

\(\Rightarrow I=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n+1\right).\left(2n+3\right)}\right)\)

\(\Rightarrow I=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n+1}-\frac{1}{2n+3}\right)\)

\(\Rightarrow I=\frac{1}{2}\left(1-\frac{1}{2n+3}\right)\)

\(\Rightarrow I=\frac{1}{2}.\frac{2n+2}{2n+3}\)

\(\Rightarrow I=\frac{n+1}{2n+3}\)

\(I=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2n+1\right)\left(2n+3\right)}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n+1\right)\left(2n+3\right)}\right)\)

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n+1}-\frac{1}{2n+3}=\frac{1}{1}-\frac{1}{2n+3}\)

\(=\frac{2n+3}{2n+3}-\frac{1}{2n+3}=\frac{2n+2}{2n+3}\)

thanks kiệt

Lồ

hơi khó đó tick mình nha Hoàng Thu Hà

gọi biểu thức là A

ta có :

A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}...\frac{1}{19.21}\)

=> 2A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}...\frac{2}{19.21}\)

2A = \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...-\frac{1}{21}\)

2A = 1 - \(\frac{1}{21}\)

2A = \(\frac{20}{21}\)

A = \(\frac{20}{21}:2=\frac{10}{21}\)

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{19\cdot21}-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{19\cdot21}\right)-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{1}{2}\left(1-\frac{1}{21}\right)-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{1}{2}\cdot\frac{20}{21}-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{10}{21}-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{x}{14}=\frac{10}{21}-\frac{2}{-7}\)

\(\frac{x}{14}=\frac{16}{21}\)

\(\Rightarrow x\cdot=21=14\cdot16\)

\(\Rightarrow x\cdot21=224\)

\(\Rightarrow x=\frac{224}{21}\)

\(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{\left(2n-1\right)\left(2n+1\right)}\)

\(=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)

\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)

\(=2.\left(1-\frac{1}{2n+1}\right)\)

\(=2.\left(\frac{2n}{2n+1}\right)\)

\(=\frac{4n}{2n+1}\)

Tham khảo nhé~