ra to lắm

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

\(=-\left(-\frac{1}{99.97}+\frac{1}{97.95}+.........+\frac{1}{5.3}+\frac{1}{3.1}\right)\)

\(=-\left(-\frac{1}{99.97}+\frac{1}{97.95}+.......+\frac{1}{5.3}+\frac{1}{3.1}\right).\frac{2}{2}\)

\(=-\left(-\frac{2}{99.97}+\frac{2}{97.95}+......+\frac{2}{5.3}+\frac{2}{3.1}\right).\frac{1}{2}\)

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

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

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

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

Đặt: \(A=\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-...-\frac{1}{5.3}-\frac{1}{3.1}\)

\(=\frac{1}{99.97}-\left(\frac{1}{97.95}+\frac{1}{95.93}+...+\frac{1}{3.1}\right)\)

\(=\frac{1}{2}\left(\frac{1}{97}-\frac{1}{99}\right)-\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{93}-\frac{1}{95}+\frac{1}{95}-\frac{1}{97}\right)\)

\(=\frac{1}{2}\left(\frac{1}{97}-\frac{1}{99}\right)-\frac{1}{2}\left(1-\frac{1}{97}\right)\)

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

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

Vậy ....

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

\(=\frac{1}{99.97}-\left(\frac{1}{97.95}+...+\frac{1}{5.3}+\frac{1}{3.1}\right)\)

\(=\frac{1}{99.97}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}\right)\left(1\right).\)

Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}\)

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

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

\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{97}\right)\)

\(\Rightarrow A=\frac{1}{2}.\frac{96}{97}\)

\(\Rightarrow A=\frac{48}{97}.\)

+ Thay A vào \(\left(1\right)\) ta được:

\(\frac{1}{99.97}-\frac{48}{97}\)

\(=\frac{1}{99.97}-\frac{48.99}{99.97}\)

\(=\frac{1-48.99}{99.97}\)

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

Vậy \(\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-...-\frac{1}{5.3}-\frac{1}{3.1}=-\frac{4751}{9603}.\)

Chúc bạn học tốt!

Đặt \(B=\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-...-\frac{1}{5.3}-\frac{1}{3.1}\)

\(\Rightarrow B=\frac{1}{99.97}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}\right)\)

Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}\)

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

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

\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{97}\right)\)

\(\Rightarrow A=\frac{1}{2}.\frac{96}{97}\)

\(\Rightarrow A=\frac{48}{97}\)

\(\Rightarrow B=\frac{1}{99.97}-\frac{48.99}{97.99}\)

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

rễ thế k cho mình trả lời cho

kết quả : ~-0.4

-0,4

k nha

\(\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-...-\frac{1}{5.3}-\frac{1}{3.1}.\)

\(=\frac{1}{99.97}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{93.95}+\frac{1}{95.97}\right)\)

\(=\frac{1}{99.97}-\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{93.95}+\frac{2}{95.97}\right)\)

\(=\frac{1}{99.97}-\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...-\frac{1}{95}+\frac{1}{95}-\frac{1}{97}\right)\)

\(=\frac{1}{99.97}-\frac{1}{2}\left(1-\frac{1}{97}\right)=\frac{1}{99.97}-\frac{1}{2}.\frac{96}{97}\)

\(=\frac{1}{99.97}-\frac{48}{97}\)

chúc bạn học tốt

       \(\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-...-\frac{1}{5.3}-\frac{1}{3.1}\)

\(=\frac{1}{99.97}-\left(\frac{1}{97.95}+\frac{1}{95.93}+...+\frac{1}{5.3}+\frac{1}{3.1}\right)\)

\(=\frac{1}{99.97}-\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{93}-\frac{1}{95}+\frac{1}{95}-\frac{1}{97}\right)\)

\(=\frac{1}{99.97}-\frac{1}{2}\left(1-\frac{1}{97}\right)\)

\(=\frac{1}{99.97}-\frac{48}{97}\)

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

ko chắc nha

1/2(-1/99+1/97-1/97+....+1)=1/2(1-1/99)=49/99

hình như làm nhầm r xin lỗi nha! làm lại 

1/2(1/(99*97))-1/2(-1/97+1/95-1/95+1/93...+1)=1/2(1/(99*97))-1/2(-1/97+1)=-9503/19206

lần này hi vọng ko nhầm

đặt tổng là A

=>A=\(\frac{-1}{3}\left(\frac{1}{99}-\frac{1}{97}+\frac{1}{97}-\frac{1}{96}+....+\frac{1}{5}-\frac{1}{3}\right)\)

=>A=\(\frac{-1}{3}\left(\frac{1}{99}-\frac{1}{3}\right)=\frac{32}{297}\)

A=\(\frac{1}{99.97}\)-\(\frac{1}{97.95}\)-\(\frac{1}{95.93}\)-...-\(\frac{1}{5.3}\)-\(\frac{1}{3.1}\)

A=\(\frac{1}{2}\).(\(\frac{1}{99}\)+\(\frac{1}{97}\)-\(\frac{1}{97}\)+\(\frac{1}{95}\)-\(\frac{1}{95}\)+\(\frac{1}{93}\)-...-\(\frac{1}{5}\)+\(\frac{1}{3}\)-\(\frac{1}{3}\)+1)

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

A=\(\frac{1}{2}\).\(\frac{100}{99}\)=\(\frac{50}{99}\)

tớ nghĩ là bằng \(\frac{-4751}{9603}\) chứ bạn Kayoko Suzue