Bài làm:

\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{99.100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\right)\)

\(=\frac{1}{99.100}-\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{98-97}{97.98}+\frac{99-98}{98.99}\right)\)

\(=\frac{1}{99.100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\right)\)

\(=\frac{1}{99.100}-\left(1-\frac{1}{99}\right)\)

\(=\frac{1}{99.100}-\frac{98}{99}\)

\(=\frac{1-98.100}{99.100}=\frac{1-9800}{9900}=-\frac{9799}{9900}\)

Học tốt!!!!

\(\left(\frac{1}{100.99}\right)-\left(\frac{1}{99.98}\right)-\left(\frac{1}{98.97}\right)-...-\left(\frac{1}{3.2}\right)-\left(\frac{1}{2.1}\right)\)

\(=\frac{1}{100.99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{2.1}\right)\)

\(=\frac{1}{99}-\frac{1}{100}-\left(\frac{1}{98}-\frac{1}{99}+\frac{1}{97}-\frac{1}{98}+...+1+\frac{1}{2}\right)\)

\(=\frac{1}{99}-\frac{1}{100}-\left(1-\frac{1}{99}\right)\)

\(=\frac{1}{99}-\frac{1}{100}-1+\frac{1}{99}\)

\(=\frac{2}{99}-\frac{101}{100}\)

\(C=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

\(C=\frac{1}{100}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

\(C=\frac{1}{100}-\frac{99}{100}\)

\(C=-\frac{98}{100}=-\frac{49}{50}\)

\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

   \(=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+\frac{1}{98.97}+....+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

    \(=\frac{1}{100}-\left(\frac{1}{100}-\frac{1}{99}+\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)

      \(=\frac{1}{100}-\left(\frac{1}{100}-1\right)\)

      \(=1\)

\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-\frac{1}{3.2}-\frac{1}{2.1}\)

\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=\frac{-98}{100}=\frac{-49}{50}\)

Ta có: \(P=\frac{1}{99}-\frac{1}{99\cdot98}-\frac{1}{98\cdot97}-\frac{1}{97\cdot96}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)

\(=\frac{1}{99}-\left(\frac{1}{98}-\frac{1}{99}\right)-\left(\frac{1}{97}-\frac{1}{98}\right)-\left(\frac{1}{96}-\frac{1}{97}\right)-...-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(1-\frac{1}{2}\right)\)

\(=\frac{1}{99}-\frac{1}{98}+\frac{1}{99}-\frac{1}{97}+\frac{1}{98}-\frac{1}{96}+\frac{1}{97}-...-\frac{1}{2}+\frac{1}{3}-1+\frac{1}{2}\)

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

\(=\frac{-97}{99}\)

a,=(1/3+3/5+1/15)+(3/4+-1/36)+(1/72-2/9)=1+26/36-15/72=1+(52-15)/72=1+37/72=109/72

b,=1/100-(1/1x2+1/2x3+...+1/97x98+1/98x99+1/99x100)

   =1/100-(1/1-1/2+1/2-1/3+...+1/97-1/98+1/98-1/99+1/99-1/100)

   =1/100-(1/1-1/100)=1/100-99/100=-98/100=-49/50

chỉ có mk mk giải thôi đó l-i-k-e đi

mình cũng đang bí bài này

Đặt \(A=\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(-A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)

\(-A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)

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

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

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

Chúc bạn học tốt ~ 

Đặt A = \(\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

=> - A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)

- A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)

- A = \(1-\frac{1}{99}\)

- A = \(\frac{98}{99}\)

=> A = \(-\frac{98}{99}\)

Vậy A = \(-\frac{98}{99}\)

Hok tốt

\(A=\frac{1}{99.98}-...-\frac{1}{2.1}\)

\(-A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)

\(-A=\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)

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

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

     \(\frac{1}{100.99}-\frac{1}{99.98}-......-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=-\left(-\frac{1}{100.99}+\frac{1}{99.98}+...........+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

\(=-\left(-\frac{1}{100}-\frac{1}{99}+\frac{1}{99}-\frac{1}{98}+......+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)

\(=-\left(-\frac{1}{100}-1\right)\)

\(=\frac{1}{100}+1\)

\(=\frac{101}{100}\)

-49

mình giải rồi vòng 11 đó 

\(A=\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{2.1}\)

\(=\frac{1}{100.99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\right)\)

\(=\frac{1}{9900}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\right)\)

\(=\frac{1}{9900}-\left(1-\frac{1}{99}\right)\)

\(=\frac{1}{9900}-\left(\frac{99}{99}-\frac{1}{99}\right)\)

\(=\frac{1}{9900}-\frac{98}{99}=\frac{1}{9900}-\frac{9800}{9900}=\frac{-9799}{9900}\)

Vậy \(A=\frac{-9799}{9900}\).