\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac{1}{3}.\left(1-\frac{1}{99}\right)\)

\(=\frac{1}{3}.\frac{98}{99}\)

\(=\frac{98}{297}\)

Chuc bn học tốt

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{99}\)

\(=1-\frac{1}{99}\)

\(=\frac{98}{99}\)

Đặt tổng là M

Ta có

\(M=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)\)

\(\Rightarrow M=\frac{1}{2}\left(1-\frac{1}{99}\right)\)

\(\Rightarrow M=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)

S=2(1-1/3+1/3-1/5+...+1/97-1/99)
  =2(1-1/99)
  =2(98/99)
  =196/99
 

2S=2/1*3+2/3*5+...+2/97*99

2S=1/1-1/3+1/3-1/5+...+1/97-1/99

2S=1-1/99

2S=98/99

S=49/99

giải sai 

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(\Leftrightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)

\(\Leftrightarrow2A=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{99-97}{97.99}\)

\(\Leftrightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

\(\Leftrightarrow2A=1-\frac{1}{99}\)

\(\Leftrightarrow2A=\frac{99}{99}-\frac{1}{99}\)

\(\Leftrightarrow2A=\frac{98}{99}\)

\(\Leftrightarrow A=\frac{98}{99}\div2\)

\(\Leftrightarrow A=\frac{49}{99}\)

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97+99}\)

\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\)

\(A=\left(1-\frac{1}{99}\right)+\left(-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{95}-\frac{1}{97}\right)\)

\(A=\frac{98}{99}+0\)

\(A=\frac{98}{99}\)

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(A=\frac{1}{2}.\left(1-\frac{1}{99}\right)\)

\(A=\frac{1}{2}.\frac{98}{99}\)

\(A=\frac{98}{198}=\frac{49}{99}\)

2A=2-1/3+1/3-1/5+...+1/97-1/99

2A=2-1/99

2A=197/99

A=197/198

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{97\cdot99}\)

\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{97\cdot99}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{99}\right)\)

\(=\frac{1}{2}\cdot\frac{98}{99}\)

\(=\frac{49}{99}\)

=))

\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{99}\right)\)

\(=\frac{1}{2}.\frac{98}{99}\)

\(=\frac{49}{99}\)

\(=\frac{2}{1.3.2}+\frac{2}{3.5.2}+\frac{2}{5.7.2}+...+\frac{2}{97.99.2}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{99}\right)=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)

= 1-1/3+1/3-1/5+1/5-1/7+...+1/97-1/99

= 1 - 1/99

= 98/99

=49/99 NHA

HT

k cho mình nha

@@@@@@@@@@@@@@@@@@

\(P=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)

Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)

2A=\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)

2A=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\)

2A=\(1-\frac{1}{99}\)

A=\(\frac{49}{99}\)

Chúc bạn học tốt

HYC-30/1/2022

\(\frac{1}{x}-\frac{1}{9999}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(\frac{1}{x}-\frac{1}{9999}=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(\frac{1}{x}-\frac{1}{9999}=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{99}\right)\)

\(\frac{1}{x}-\frac{1}{999}=\frac{1}{2}.\frac{98}{99}\)

\(\frac{1}{x}-\frac{1}{9999}=\frac{49}{99}\)

\(\frac{1}{x}=\frac{49}{99}+\frac{1}{9999}\)

\(\frac{1}{x}=\frac{50}{101}\)

\(x=1:\frac{50}{101}\)

\(x=\frac{101}{50}\)

Vậy \(x=\frac{101}{50}\)