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 :

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

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

=\(-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)

=\(-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\right)\)

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

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

A=1/100.99 - 1/99.98 - 1/98.97 -...- 1/3.2 - 1/2.1

A=   - (1/100.99 + 1/99.98 + 1/98.97 +...+ 1/3.2 + 1/2.1)

A=   - (1/2.1+1/3.2 +...+1/98.97+ 1/99.98 +1/100.99 )

A=   - (1/1.2+1/2.3+1/3.4+...+1/97.98+ 1/98.99 +1/99.100)

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

A=   - (1/1-1/100)

A=   - 99/100

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à mình nhầm có phải thế này không

1/100.99 - 1/99.98 - 1/98.97 -...- 1/3.2 - 1/2.1
=-(1/100.99 + 1/99.98 + 1/98.97 +...+ 1/3.2 + 1/2.1)
=-(1/2.1+1/3.2 +...+1/98.97+ 1/99.98 +1/100.99 )
=-(1/1.2+1/2.3+1/3.4+...+1/97.98+ 1/98.99 +1/99.100)
=-(1/1-1/2+1/2-1/3+1/3......-1/98+1/98-1/99+1/99-1/100)
=-(1/1-1/100)=-99/100

-99/100

-49/50                                  

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

= 1/100-(1/100.99+1/99.98+.....+1/3.2+1/2.1)

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

=1/100-(1/1-1/100)

=1/100-99/100

=-98/100

=-49/50

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

\(=\)\(\frac{1}{99}-\left(\frac{1}{1.2}+...+\frac{1}{98.99}\right)\)

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

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

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

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

\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{97.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\) 

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

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

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

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

\(C=C=\frac{1}{50}-1=-\frac{49}{50}\)