\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}\right)\)

\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\cdot\dfrac{96}{97}=\dfrac{-4751}{9603}\)

a, \(A=-\dfrac{1}{20}-\left(\dfrac{1}{20\cdot19}+\dfrac{1}{19\cdot18}+...+\dfrac{1}{2\cdot1}\right)\\ \Rightarrow A=-\dfrac{1}{20}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\\ \Rightarrow A=-\dfrac{1}{20}-1+\dfrac{1}{20}=-1\)

b, \(B=\dfrac{1}{99}-\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{3\cdot1}\\ \Rightarrow B=\dfrac{1}{99}-\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}\right)\\ \Rightarrow B=\dfrac{1}{99}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\\ \Rightarrow B=\dfrac{1}{99}-\dfrac{1}{2}+\dfrac{1}{2\cdot99}=-\dfrac{16}{33}\)

\(=\dfrac{1}{99}-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}+\dfrac{1}{97\cdot99}\right)\\ =\dfrac{1}{99}-\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}+\dfrac{2}{97\cdot99}\right)\\ =\dfrac{1}{99}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\\ =\dfrac{1}{99}-\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)=\dfrac{1}{99}-\dfrac{1}{2}\cdot\dfrac{98}{99}\\ =\dfrac{1}{99}-\dfrac{49}{99}=-\dfrac{48}{99}=-\dfrac{16}{33}\)

\(\frac{1}{99.97}-\frac{1}{97.95}-...-\frac{1}{5.3}-\frac{1}{3.1}\)

\(=\frac{1}{97.99}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}\right)\)

\(=\frac{1}{97.99}-\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{95.97}\right)\)

\(=\frac{1}{97.99}-\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{97}\right)\)

\(=\frac{1}{9603}-\frac{1}{2}\left(1-\frac{1}{97}\right)\)

\(=\frac{1}{9603}-\frac{1}{2}.\frac{96}{97}\)

\(=\frac{-4751}{9603}\)

A=-(1/99.97+1/97.95+...+1/5.3+1/3.1)

2B=2/99.97+2/97.95+...+2/5.3+2/3.1

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

2B=1-1/99

2B=98/99

B=49/99

Suy ra A=-1/49/49

Mình giải đúng rồi bạn cứ yên tâm

 \(\frac{1}{99}-\frac{1}{99.97}-\frac{1}{97.95}-...-\frac{1}{5.3}-\frac{1}{3.1}\)

\(=\frac{1}{99}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}+\frac{1}{97.99}\right)\)

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

\(=\frac{1}{99}-\frac{1}{2}.\left(1-\frac{1}{99}\right)=\frac{1}{99}-\frac{1}{2}\cdot\frac{98}{99}=\frac{1}{99}-\frac{49}{99}=\frac{-48}{99}=\frac{-16}{33}\)

cảm on bạn két quả của mình cũng thế nhưng cách giải hơi khác bạn chút xíu

\(\frac{1}{99}-\frac{1}{99.97}-\frac{1}{97.95}-...-\frac{1}{5.3}-\frac{1}{3}\\ =\frac{1}{99}-\left(\frac{1}{99.97}+\frac{1}{97.95}+...+\frac{1}{5.3}+\frac{1}{3.1}\right)\\ =\frac{1}{99}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}+\frac{1}{97.99}\right)\\ =\frac{1}{99}-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\right)\\ =\frac{1}{99}-\frac{1}{2}.\left(1-\frac{1}{99}\right)\\ =\frac{1}{99}-\frac{1}{2}.\frac{98}{99}\\ =\frac{-16}{33}\)

Sai đề 

sai đề sao làm