a) \(\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{99}-\left(\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

đặt \(A=\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\)

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

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

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

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

thay A vào, ta được :

\(\frac{1}{99}-\frac{98}{99}=\frac{-97}{99}\)

b) \(\frac{2}{100.99}-\frac{2}{99.98}-...-\frac{2}{3.2}-\frac{2}{2.1}\)

\(=\frac{2}{100.99}-\left(\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\right)\)

đặt \(A=\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\)

\(A=\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{98.99}\)

\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)\)

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

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

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

\(A=\frac{196}{99}\)

Thay A vào, ta được :

\(\frac{2}{100.99}-\frac{196}{99}=\frac{-19598}{9900}\)

\(A=\frac{1}{2003.2002}-\frac{1}{2002.2001}-\frac{1}{2001.2000}-....-\frac{1}{3.2}-\frac{1}{2.1}\)

    \(=-\left(\frac{1}{2003.2002}+\frac{1}{2002.2001}+\frac{1}{2001.2000}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

     \(=-\left(\frac{1}{2003}-\frac{1}{2002}+\frac{1}{2002}-\frac{1}{2001}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)

      \(=-\left(\frac{1}{2003}-1\right)=-\left(-\frac{2002}{2003}\right)=\frac{2002}{2003}\)

Vậy ....

Đề mk chắc là sai đó bn !!

Bn kiểm tra lại xem

Nhớ đúng mk nha

Ta có:

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

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

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

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

\(=\frac{1}{50}-\frac{50}{50}\)

\(=-\frac{49}{50}\)

Câu này khó quá ta mình suy nghĩ này giờ mà vẫn chưa ra

\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

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

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

\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-\frac{1}{97.96}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=-\left(\frac{1}{99}+\frac{1}{99.98}+\frac{1}{98.97}+\frac{1}{97.96}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

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

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

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

Ta có : \(1-\frac{1}{2014.2013}-\frac{1}{2013.2012}-......-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2013.2014}\right)\)

\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{2013}-\frac{1}{2014}\right)\)

\(=1-\left(1-\frac{1}{2014}\right)\)

\(=1-1+\frac{1}{2014}\)

\(=\frac{1}{2014}\)

\(a,1-\frac{1}{2014.2013}-\frac{1}{2013.2012}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\right)\)

\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}\right)\)

\(=1-\left(1-\frac{1}{2014}\right)\)

\(=1-1+\frac{1}{2014}\)

\(=\frac{1}{2014}\)

Ta có : 

\(A=\frac{1}{2003\cdot2002}-\frac{1}{2002\cdot2001}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)

\(A=-\left(\frac{1}{2003\cdot2002}+\frac{1}{2002\cdot2001}+...+\frac{1}{3\cdot2}+\frac{1}{2\cdot1}\right)\)

\(A=-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2001\cdot2002}+\frac{1}{2002\cdot2003}\right)\)

\(A=-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2002}-\frac{1}{2003}\right)\)

\(A=-\left(1-\frac{1}{2003}\right)\)

\(A=-\frac{2002}{2003}\)

\(A=\frac{1}{2003.2002}-\frac{1}{2002.2001}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2001.2002}\right)+\frac{1}{2002}.\frac{1}{2003}\)

\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2001}-\frac{1}{2002}\right)+\frac{1}{2002}.\frac{1}{2003}\)

\(=-\left(1-\frac{1}{2002}\right)+\frac{1}{2002}.\frac{1}{2003}\)

\(=-1+\frac{1}{2002}.+\frac{1}{2002}.\frac{1}{2003}\)

\(=-1+\frac{1}{2002}\left(1+\frac{1}{2003}\right)\)

\(=-1+\frac{1}{2002}.\frac{2004}{2003}\)

\(=-1+\frac{2}{2003}\)

\(=\frac{-2003+2}{2003}\)

\(=\frac{-2001}{2003}\)

Bạn ơi cho mình hỏi từ sau chỗ \(\frac{1}{2004.2003}\)là dấu trừ hết ạ? Nếu là dấu cộng thì mình làm được :33

đúng rồi bạn ơi thế mới khó

bài này dễ mak bn !tự lm đê!

 1/100‐1/100.99‐1/99.98‐...‐1/3.2‐1/2.1

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

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

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